Integration in finite terms with dilogarithmic integrals, logarithmic integrals and error functions

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.1 (1-24)

Journal of Symbolic Computation ••• (••••) •••–•••

Contents lists available at ScienceDirect

Journal of Symbolic Computation www.elsevier.com/locate/jsc

Integration in finite terms with dilogarithmic integrals, logarithmic integrals and error functions Yashpreet Kaur 1 , Varadharaj R. Srinivasan 2 IISER Mohali, Department of Mathematical Sciences, SAS Nagar, Punjab 140306, India

a r t i c l e

i n f o

Article history: Received 21 January 2018 Accepted 28 August 2018 Available online xxxx Keywords: Differential fields Differential algebra Integration in finite terms Elementary functions Liouville’s Theorem Dilogarithmic integrals Error functions Logarithmic Integrals

a b s t r a c t We extend the theorem of Liouville on integration in finite terms to include dilogarithmic integrals, logarithmic integrals and error functions along with transcendental elementary functions. We also generalise a result of Baddoura on integration in finite terms with dilogarithmic integrals. © 2018 Elsevier Ltd. All rights reserved.

1. Introduction In this article we shall extend Liouvillie’s Theorem on integration in finite terms to include dilogarithmic integrals, error functions and logarithmic integrals. Let F be a field of characteristic zero equipped with a single derivation map, which we denote by  . The kernal of the map  will be denoted by C F and a easy computation shows that C F is a subfield of F . We call E = F (θ1 , · · · , θn ) a DEL-extension of F if C E = C F and there is a tower of differential fields F i such that

E-mail address: [email protected] (V.R. Srinivasan). The author is supported by PhD fellowship from IISER Mohali and the results of this paper are part of author’s doctoral dissertation. 2 The author is supported by the DST-SERB grant: EMR/2016/001516. 1

https://doi.org/10.1016/j.jsc.2018.08.004 0747-7171/© 2018 Elsevier Ltd. All rights reserved.

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.2 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

2

F = F 0 ⊆ F 1 ⊆ F 2 ⊆ · · · ⊆ Fn = E and for each i, F i = F i −1 (θi ) and one of the following holds: (i) (ii) (iii) (iv)

θi is algebraic over F i −1 . θi = u  θi for some u ∈ F i −1 (i.e. θi = e u ). θi = u  /u for some u ∈ F i −1 (i.e. θi = log(u )).  θi = u  / v, where v  = u  /u for some u , v ∈ F i −1 (i.e. θi = u  / log(u ), also denoted by i (u )).  2 (v) θi = u  v, where v  = (−u 2 ) v for some u , v ∈ F i −1 (i.e. θi = u  e −u , also denoted by erf (u )).  u     (vi) θi = vu /u, where v = (1 − u ) /(1 − u ) (i.e. θi = u log(1 − u ), also denoted by −2 (u )) for some u , v ∈ F i −1 .

We say that v ∈ F admits a DEL-expression over F if there are finite indexing sets I , J , K and elements r i , g i ∈ F for all i ∈ I , elements u j , log(u j ) ∈ F and constants a j for all j ∈ J , elements v k , e − v k ∈ F and constants bk for all k ∈ K , and an element w ∈ F such that 2

v=

 g i ri

i∈I

gi

+

 j∈ J

aj

u j log(u j )

+



bk v k e − v k + w  ,

k∈ K

2

(1.1)

n

where for each i ∈ I , there is an integer ni such that r i = l=i 1 c il hil /h il for some constants c il and elements h il ∈ F . A DEL-expression will be called a special DEL-expression if for each i ∈ I , r i = c i (1 − g i ) /(1 − g i ) for some constant c i . The main result of this paper is the following theorem. Theorem. Let F be a field of characteristic zero, E = F (θ1 , · · · , θn ) be a DEL-extension of F and for each i, θi be transcendental over F i −1 . Suppose that there is an element u ∈ E with u  ∈ F . Then u  admits a special DEL-expression over F (log(h1 ), · · · , log(hm )), where each hi belongs to F and u  admits a DEL-expression over F . In a nutshell, the proof uses only standard techniques from differential algebra and many calculations involved boils down to comparing terms of certain partial fraction expansions. A differential field extension E = F (θ1 , · · · , θn ) of F is called an elementary field extension if each θi satisfies either case i, ii or iii. In Rosenlicht (1968), Rosenlicht provided a purely algebraic proof of Liouville’s Theorem on integration in finite terms. He showed that (see Rosenlicht, 1968, Theorem, p. 157) if E is an elementary field extension of F with C E = C F and there is an element u ∈ E such that u  ∈ F then there are constants r1 , · · · , rn and elements w , g 1 , · · · , gn ∈ F such n that u  = i =1 r i ( g i / g i ) + w  . That is, upto an element of F , u must be a constant linear combination of logarithms of elements of F . The problem of extending Liouville’s Theorem to allow special functions was first studied by Moses (1969). Later in Cherry (1986, 1985), Cherry proved an extension of the Theorem to include logarithmic integrals and error functions. In Singer et al. (1985), Singer, Saunders and Caviness extended Liouville’s theorem to include a large class of functions which they called E L-elementary functions. In particular, if θi satisfies any one of the cases i–v then it is an E L-elementary function. However, case vi was not covered under the E L-class of functions. A DEL-extension E = F (θ1 , · · · , θn ) is called a transcendental dilogarithmic-elementary extension if C E = C F and for each i, θi is transcendental over F i −1 and satisfies either case ii or iii or vi. In Baddoura (2006), Baddoura extended Liouville’s Theorem to include dilogarithmic integrals. He proved that if E is a transcendental dilogarithmic-elementary extension of F having an algebraically closed field of constants C F and if F is a liouvillian extension of C F then any u ∈ E with u  ∈ F has the following form over F :

u=w+

m  i =1

r i log( g i ) +

n  j =1

c j D (h j ),

(1.2)

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.3 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

3

where each r i , g i , h j , w ∈ F , log( g i ) and D (h j ) belong to some differential field extension of F and 

h D (h j ) = −21 h j log(1 − h j ) + j

 1 (1−h j ) 2 1−h j

log(h j ). Baddoura’s proof involves producing equation (1.2) over

F when one such equation for u is given over F (θ), where θ satisfies ii, iii or vi. The problematic terms here are r i log( g i ), where both r i and g i are arbitrary elements of F (θ). Lengthy and involved calculations, along with a new dilogarithmic identity (see Baddoura, 2006, Proposition 2, p. 923) were needed to obtain the desired expression over F . In the spirit of Liouville’s theorem as extended by Singer, Saunders and Caviness, and from an algorithmic view point, it is desirable to obtain an expression for u  in terms of elements of F . However, no such expression for u  was produced in Baddoura (2006). Our main theorem restricted to transcendental dilogarithmic-elementary extensions will yield the following expression for u  over F : 

u =

 g i ri

i∈I

gi



+w,



ri =

ni  l =1

c il

hil h il

(1.3)

,

where c il are constants and elements r i , g i , w , h il belong to F . For a transcendental exponential θ over F , a DEL-expression for u  ∈ F over F (θ) does not in general reduce to a similar expression over F , however, when it is a special DEL-expression, it does reduce (see Theorem 5.1). We utilize this fact and set up a special induction procedure to prove our main theorem. The problematic

n

h

terms that appear in our proof are those r i ( g i / g i ), where 0 = r i = l=i 1 c il hil . However, we only need il basic dilogarithmic identities (see Remark 2 and Proposition 3.2), in particular, we do not require Baddoura’s dilogarithmic identity, to handle these terms. Consequently, we obtain a simpler proof of Baddoura’s Theorem which neither requires that F is a liouvillian extension of C F nor that C F is an algebraically closed field. Note that our main theorem handles only purely transcendental extension and at present, the authors do not know how to generalise the Theorem to include algebraic elements (see Remark 3). The first author hopes to address this issue and extend results of this paper to polylogarithmic integrals (see Baddoura, 2011, Conjecture 12, p. 234) in her doctoral dissertation. Organisation of this paper: We divide this paper into 6 sections. In Section 2 several standard results from differential algebra are recorded for later use. In Sections 3 and 4 we set up our special induction procedure along with few propositions and lemmas to tackle our main theorem, which appears as Theorems 5.2 and 5.3 in Section 5. We prove a generalized version of Baddoura’s Theorem in Section 6 and provide an example to show why our results are optimal. Acknowledgements: The authors would like to thank the anonymous referee of this article for valuable comments and corrections. The second author would like to thank DST-SERB, Govt. of India for supporting him with the grant: EMR/2016/001516. 2. Preliminaries and basic results In this section we shall record several standard results and terminologies from differential algebra that we need in our proofs. All the differential fields considered in this paper are of characteristic zero. We denote the field of constants of a differential field F by C F . If F is any algebraic closure of F then there is a unique derivation on F making F a differential field extension of F (see Rosenlicht, 1968) and it should be noted that C F = C F if and only if C F is an algebraically closed field (see Proposition 2.1). A differential extension field E = F (θ1 , · · · , θn ) is called a liouvillian extension of F if for each i, either θi is algebraic over F i −1 = F (θ1 , · · · , θi −1 ) or θi ∈ F i −1 or θi /θi ∈ F i −1 . A liouvillian extension F (θ1 , · · · , θn ) of F will be called transcendental if each θi is transcendental over F i −1 and observe that DEL-extensions are special cases of liouvillian extensions. Proposition 2.1. Let F  F (θ) be differential fields and θ be algebraic over F . Then the following statements hold: (a) If θ  ∈ F then there is an element x ∈ F such that x = θ  and C F (θ)  C F .

JID:YJSCO

4

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.4 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

(b) If θ  /θ ∈ F then there is an element x ∈ F − {0} and an integer n such that x /x = nθ  /θ . Furthermore, if C F (θ) = C F then the minimal monic polynomial of θ over F is of the form P ( X ) = X n + cx for some c ∈ CF . (c) Every c ∈ C F (θ) is algebraic over C F .

n

Proof. Let P ( X ) = i =0 ai X i , an = 1 and n ≥ 2 be the minimal monic polynomial of θ over F . Differn i entiating i =0 ai θ = 0, we obtain that θ is also a root of the polynomial





P  ( X ) = (nθ  + an −1 ) X n−1 + · · · + iai θ  + ai −1 X i −1 + · · · + a1 θ  + a0 ∈ F [ X ]. If θ  ∈ F then by minimality of P ( X ), P  ( X ) must be the zero polynomial. In particular, (−an−1 /n) = θ  and θ + (an−1 /n) is constant that is not in C F . A similar calculation with the minimal monic polynomial over F of c ∈ C F (θ) would give us that ai = 0 for all i and thus c is algebraic over C F . If θ  /θ = α ∈ F then we shall rewrite

P  ( X ) = nα X n + (an −1 + (n − 1)an−1 α ) X n−1 + · · · + a0 and observe that P  ( X ) = nα P ( X ). Then for each i ∈ {1, · · · , n}, we have

ai −1 = (n − (i − 1))αai −1 . In particular, a0 = nαa0 and since a0 = 0, we have nθ  /θ = a0 /a0 . Finally, if C F (θ) = C F then ai = 0 for all i ∈ {1, · · · , n − 1}. Otherwise, (θ n−i /ai ) = 0 and therefore θ n−i + cai = 0 for some non zero constant c ∈ C F . This contradicts the assumption that P ( X ) is of degree n. 2 Proposition 2.2. Let F ⊂ F (θ) be differential fields, θ be transcendental over F , θ  ∈ F and v = F [θ] be a polynomial in θ over F . Suppose that there is a w ∈ F (θ) − F such that w  = v.

s

i =0 βi θ

i

∈

t

(a) If C F (θ) = C F then w = i =0 αi θ i ∈ F [θ], αt = 0 and t ≥ 1. (b) If v = 0, that is C F (θ)  C F , then there is a non zero constant c ∈ C F and α0 ∈ F such that (c θ + α0 ) = 0.  0, C F (θ) = C F and s = deg( v ) then either deg( w ) = s or s + 1. In the former case αt = βs and in (c) If v = the latter case αt ∈ C F and (t αt θ + αt −1 ) = βs . (d) If α ∈ F , x = α for all x ∈ F and θ  = α then C F (θ) = C F . In general, if α1 , · · · , αn ∈ F are non zero elements then there is a differential field extension E of F such that C E = C F and E = F (θ1 , · · · , θn ), where θi = αi . Proof. Let there be an element w ∈ F (θ) such that w  = v. Then there are relatively prime polynomials P , Q ∈ F [θ], where Q is monic, such that w = P / Q . Taking derivatives, we obtain

Q 2v = P Q − Q  P.

(2.1)

From the above equation, it is immediate that Q divides Q  . Since Q is monic and θ  ∈ F we have deg Q  < deg Q . This forces Q  = 0. If C F (θ) = C F , then Q = 1 and thus P = w ∈ F [θ] − F . Now suppose that v = 0. If Q = 1 then deg P ≥ 1 and P  = 0 and if Q = 1 then deg Q ≥ 1 and as observed earlier t  i  Q  = 0. Thus we have = 0, αt = 0 and t ≥ 1. Now we compare coefficients and obtain i =0 αi θ   that αt = 0 and (t αt θ + αt −1 ) = t αt θ  + αt−1 = 0. This proves (b). From (a), we have

αt θ t + (t αt θ  + αt−1 )θ t −1 + · · · + α1 θ  + α0 =

s 

βi θ i = v .

(2.2)

i =0

If deg( w ) = s ≥ 0 and C F (θ) = C F then it is easy to see that t = s or t = s + 1. If t = s then αt = βs , where αt ∈ F and if t = s + 1 then αt ∈ C F and (t αt θ + αt −1 ) = t αt θ  + αt−1 = βs . Suppose that θ  = α and x = α for all x ∈ F . If w ∈ F (θ) − F and w  = 0 then from (b) there is a nonzero constant c ∈ C F and an element α0 ∈ F such that 0 = (c θ + α0 ) = c α + α0 . Thus (−α0 /c ) = α

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.5 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

5

and this is a contradiction. Finally, let F 0 = F and F n−1 be a differential field extension of F such that C F n−1 = C F and F n−1 = F (θ1 , · · · , θn−1 ), where θi = αi for all 1 ≤ i ≤ n − 1. If there is no element x ∈ F n−1 such that x = αn then let θn be a transcendental and define a derivation on E := F n−1 (θ) by defining θn = αn . Clearly, C E = C F n−1 = C F . On the other hand if there is an element x ∈ F n−1 such that x = αn then take θn to be x. 2 Proposition 2.3. Let F (θ) ⊃ F be a transcendental liouvillian extension of F with C F (θ) = C F . Let v ∈ F (θ), s F be an algebraic closure of F and v = η j =1 (θ − α j )m j , where η ∈ F , 0 = α1 , · · · , αs are distinct elements in F and m j are integers. (a) Suppose that θ  ∈ F . Then each θ  − α j is a non zero element of F and

v v

=

  s η  θ − α j + mj η j =1 θ − α j

(2.3)

is the partial fraction expansion of v  / v. (b) Suppose that θ  /θ ∈ F . Then (b.1)

v v

=μ+

s 

mj

j =2

where μ = (η /η) +

μj θ − αj

s

j =1 m j (θ

expansion of v  / v.

,  /θ) ∈ F and μ j

(2.4)

= α j (θ  /θ) − α j ∈ F − {0}, is the partial fraction

(b.2) If v 0 is the constant term of the partial fraction expansion of v in F (θ) then the constant term of v  is v 0 . (c) If v ∈ F (θ) has a pole of order m ≥ 1 and θ  ∈ F then v  has a pole of order m + 1. Similarly, if v ∈ F (θ) has a non-zero pole of order m ≥ 1 and θ  /θ ∈ F then v  has a pole of order m + 1. Proof. If θ  ∈ F and θ  = α j for some

α j then by Proposition 2.1, there is an element x ∈ F such

that θ  = x . Now θ − x ∈ / F is a constant of F (θ) and this contradicts our assumption that C F (θ) = C F . Similarly, if θ  /θ = x ∈ F and α j = xα j then again from Proposition 2.1, there are an integer n and an

element y ∈ F such that nx = n(θ  /θ) = y  / y. Thus θ n / y ∈ F (θ) − F is a constant which again contradicts our assumption. A straightforward calculation shows that Equations (2.3) and (2.4) represents the partial fraction expansion of v  / v. Let

v=

mi l   i =1 j =1

vij

(θ − αi ) j

+ v 0 + v 1θ + · · · + vnθ n,

αi , v i and v i j belong to F be the partial fraction expansion of v over F . Note that ⎧ v i j − jv i j (x − αi ) ⎪   ⎪ + if θ  = x ∈ F ⎨ vij j j +1 (θ − α ) (θ − α ) i i (2.5) = v i j ⎪ − jv i j (xαi − αi ) − jv i j x (θ − αi ) j ⎪  /θ = x ∈ F ⎩ + + if θ (θ − αi ) j (θ − αi ) j +1 (θ − αi ) j

where elements

and

 i 

(v i θ ) =

i v i xθ i −1 + v i θ i

(v  i

+ i v i x)θ

i

if θ  = x ∈ F

if θ  /θ = x ∈ F .

From this observation it follows that when θ  /θ ∈ F , the constant term of v  is v 0 . Suppose that v has a pole at αi of order mi . Then, −mi v imi (x − αi ) = 0 when θ  ∈ F and −mi v imi (xαi − αi ) = 0 when

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.6 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

6

αi =  0 and θ  /θ ∈ F . Therefore, from Equation (2.5), we obtain that v  has a pole of order mi + 1 αi . 2

at

Theorem 2.4. Let E = F (θ = θ1 , θ2 , · · · , θn ) be a liouvillian extension of F with C E = C F .

n

(a) (Kolchin–Ostrowski) If y ∈ E, y  ∈ F and for each i, θi ∈ F then y = i =1 c i θi + η , where c i are constants and η ∈ F . (b) If θi ∈ F for all i ≥ 2 and θ is transcendental over F then there are elements y 1 , · · · , yt ∈ {θ2 , · · · , θn } such that θ, y 1 , · · · , yt are algebraically independent over F and E = F (θ, y 1 , · · · , yt ). (c) Suppose that n = 2 and E = F (θ, θ2 ) is a transcendental liouvillian extension of F such that θ  ∈ F and θ2 = v  / v for some v ∈ F (θ) − F . If y ∈ E and y  ∈ F then y ∈ F (θ) and y = c θ + η for some constant c and η ∈ F . Proof. (a) follows from Kolchin–Ostrowski Theorem which was first proved by A. Ostrowski and later generalised by Kolchin (see Kolchin, 1968, p. 1155). An elementary proof of this result can be found in Rubel and Singer (1988), Appendix, p. 366. Let { y 1 , · · · , yt } ⊂ {θ2 , · · · , θn } be such that θ, y 1 , · · · , yt is a transcendence base of E over F . Suppose that there is a smallest integer i such that θi ∈ / F ∗ := F (θ, y 1 , · · · , yt ). Then θi must be algebraic over F ∗ and since θi ∈ F for all i ≥ 2, the field F ∗ is a differential field. Now from Proposition 2.1, we obtain C F ∗ (θi )  C F and this contradicts our assumption that C E = C F . Finally, if n = 2, θ2 = v  / v for some v ∈ F (θ) − F and y ∈ E with y  ∈ F then we shall use (a) to find some constant c and an element f ∈ F (θ) such that y = c θ2 + f . Taking derivatives, we obtain

y = c(v  / v ) + f  . As in Proposition 2.3, we write v = η

(2.6)

s

− α j )m j , where η ∈ F , 0 = α1 , · · · , αs are distinct elements in F and m j are integers. Since v ∈ F (θ) − F , we must have a j such that m j = 0. Now since y  ∈ F ⊂ F (θ) and f  ∈ F (θ), from Equation (2.3), we conclude that c must be zero for Equation (2.6) to hold. Thus y − f ∈ C F ⊂ F and that y ∈ F (θ). Now we apply (a) to obtain that y = c θ + η for some c ∈ C F and η ∈ F . 2 j =1 (θ

Proposition 2.5. Let F (θ) ⊃ F be a liouvillian extension with C F (θ) = C F . Suppose that v , u 1 , · · · , un ∈ F (θ) and w ∈ F are elements such that

v +

n 

c i (u i /u i ) = w ,

i =1

where c 1 , · · · , cn are Q-linearly independent constants then (a) If θ  ∈ F then u i ∈ F for all i and v = c θ + η for some constant c and η ∈ F . (b) If θ  /θ ∈ F then v ∈ F and for each i, u i = ηi θ mi where ηi ∈ F and mi is an integer. (c) If θ is algebraic over F and F is a liouvillian extension of C F then v ∈ F and there is a non zero integer m such that um i ∈ F for all i. Proof. (a) and (b) follows from Rosenlicht (1968), Lemma, p. 155 and (c ) follows from Rosenlicht and Singer (1977), Lemma, p. 338. 2 3. Special expressions and identities A DEL-expression will be called (a) a special DEL-expression if for each i ∈ I , r i = c i (1 − g i ) /(1 − g i ) for some constant c i , (b) a L-expression if for all i , j , k, r i = 0, a j = 0 and bk = 0,

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.7 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

7

(c) a D -expression if it is special and for all j , k, a j = bk = 0, (d) a DL-expression if bk = 0 for all k. An L-expression constants of F .

p



c (v i /v i ) i =1 i

over F is said to be reduced if c 1 , · · · , c p are Q-linearly independent

Remark 1 (Reduced L-Expressions). Note that if u 1 , u 2 ∈ F and c ∈ C F then for any p /q ∈ Q, we have

c

u 1 u1

+c

p u 2 q u2

c (u 1 u 2 ) q

=

p

q u q1 u 2p

.

In general, if u 1 , · · · , um ∈ F and a1 , · · · , am ∈ C F then

m



i =1 ai (u i /u i )

=

p



c ( v i / v i ), where i =1 i m qj v i = j =1 u j , q j ∈ Z.

c 1 , · · · , c p is a Q-basis for the vector space spanned by a1 , · · · , am over Q and Thus if v ∈ F admits a L-expression over F then it also admits a reduced L-expression over F . To this m  end, whenever we write i =1 ai (u i /u i ), we shall assume that a1 , · · · , am are Q-linearly independent. In particular,

n

i c h /h il that appear in the definition of (a) we assume that for each i, the L-expression l=1 il il DEL-expression is reducedand  (b) if I 1 = {i ∈ I | r i = 0} then i ∈ I 1 r i ( g i / g i ) is also reduced.

Remark 2 (D -Expressions). Let F ⊂ E be differential fields and y ∈ E be such that y  = rg  / g, r  = c (1 − g ) /(1 − g ), g , r ∈ F and c ∈ C F . Then we have



−g 1− g (−r ) −g 1− g

 = −r

g g

+

1 2c

and thus we have a dilogarithm z =



−g 1 −g z = −r −g 1− g

(r 2 )

r2 2c

− y ∈ E of F where



 and

(−r ) = c

1



1− g 1 1− g



−g 1− 1− g =c −g 1− 1− g

 .

Thus if we denote rˆ := −r and gˆ := − g /(1 − g ) then

r

g g

= −ˆr

gˆ  gˆ

+

1 2c

(r 2 ) .

(3.1)

We observe that if F = F ∗ (θ), where θ is transcendental over F ∗ and

1− g =η

P Q

,

where P and Q are monic relatively prime polynomials over F ∗ and of gˆ , we note that

1 − gˆ =

1 1− g

=

1 Q

η P

η ∈ F ∗ then from the definition

.

Proposition 3.1. Let F (θ) ⊃ F be differential fields where θ is transcendental, C F (θ) = C F and either θ  ∈ F or θ  /θ ∈ F and v ∈ F (θ). Suppose that

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.8 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

8

v=



r i ( g i / g i ) + w 

(3.2)

i∈I

is a D -expression over F (θ). (a) If θ  ∈ F then for each i such that r i = 0, we have g i ∈ F . (b) For each i, r i = ai θ + ηi for some constant ai ∈ C F and θ  /θ ∈ F .

ηi ∈ F . Furthermore, each ri belong to F when

Proof. We have r i − c i (1 − g i ) /(1 − g i ) = 0 for some c i ∈ C F . Suppose that c i = 0 for some i. If θ  ∈ F then since c i = 0, we apply Proposition 2.5 and obtain 1 − g i ∈ F and consequently, g i ∈ F for all i. On the other hand if θ  /θ ∈ F then we have 1 − g i = ξi θ mi for some integer mi and elements ξi and r i in F . Thus for each i, r i ∈ F . From Proposition 2.5 it follows that r i = ai θ + ηi for some ai ∈ C F and ηi ∈ F and that each ri ∈ F (that is, ai = 0) when θ  /θ ∈ F . 2 A differential field extension E of F will be called a logarithmic extension of F if C E = C F and there are elements h1 , · · · , hm ∈ F such that E = F (log(h1 ), · · · , log(hm )). Proposition 3.2. Let F (θ) ⊃ F be a transcendental DEL-extension and suppose that

v :=



r i ( g i / g i ) + w 

(3.3)

i∈I

is a D -expression over a logarithmic extension E of F (θ). Then v admits a D -expression:

v=



˜ r˜i ( g˜ i / g˜ i ) + w

i∈I

over some logarithmic extension E˜ of F (θ), containing E, such that for each i, r˜i is a constant or g˜ i ∈ F (θ) and 1 − g˜ i = ηi P i / Q i , where P i and Q i are monic relatively prime polynomials over F [θ] and ηi ∈ F having the following properties: (a) θ is neither a factor of P i nor a factor of Q i and deg( Q i ) ≥ deg ( P i ). (b) If ηi = 1 then deg( Q i − ηi P i ) = deg( Q i ) and if ηi = 1 then log (ηi ) ∈ C F . (c) If ξi is the leading coefficient of Q i − ηi P i then either deg( P i ) = deg( Q i ) or ξi = 1. Furthermore, either ηi = 1 or ξi = 1 or ξi = 1 − ηi and in any event, log(ηi )(ξi /ξi ) is a D -expression over F (log(ηi )). Proof. Let E = F (θ)(log( y 1 ), · · · , log( yn )) for y 1 , · · · , yn ∈ F (θ) and  p = F (θ)(log( y 1 ), · · · , log( y p −1 ), log( y p +1 ), · · · , log( yn )). Observe that r i − c i (1 − g i )/(1 − g i ) = 0 ∈  p [log( y p )] and that  p (log( y p )) = E. Apply Proposition 2.5 and obtain that 1 − g i , and therefore g i , belongs to  p . Thus g i ∈ F (θ) = ∩ p  p for each i ∈ I . Let 1 − g i = ξi P i / Q i , where P i and Q i are relatively prime monic polynomials over F and ξi ∈ F . Then θ can either divide P i or Q i but not both. Suppose that θ divides P i . Then over the differential field E (log( g i )), we have r i ( g i / g i ) = r˜i g˜ i / g˜ i + (log( g i )r i ) , where r˜i = −c i log( g i ) and g˜ i = 1 − g i . Then

v=



˜ i + w ) , r j ( g j / g j ) + r˜i ( g˜ i / g˜ i ) + ( w

j ∈ I , j =i

˜ i = log( g i )r i , is a D -expression over E (log( g i )). Note that Q i − ξi P i and Q i are relatively where w prime polynomials such that θ neither divides Q i − ξi P i nor Q i . Since 1 − g˜ i = g i = ( Q i − ξi P i )/ Q i , we shall factor the leading coefficient η˜ i of Q i − ξi P i and obtain for all such i, relatively prime monic polynomials P˜ i and Q i such that η˜ i P˜ i = Q i − ξi P i , 1 − g˜ i = g i = η˜ i P˜ i / Q i and that θ neither dividing P˜ i nor Q i . Now we suppose that θ divides Q i . We make use of Remark 2 and write r i g i / g i = −ˆr i gˆ i / gˆ i + ((1/2c i )r i2 ) , where rˆi = c i (1 − gˆ i ) /(1 − gˆ i ). Since 1 − gˆ i = 1/(1 − g i ) = (1/ξi )( Q i / P i ), we have θ

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.9 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

9

dividing the numerator polynomial Q i . Therefore, we shall proceed as in the previous case and ob˜ i . This proves the first part of (a). To prove the second part of (a), we simply apply tain r˜i , g˜ i and w the “hat operation”, that is Remark 2, to those terms that have deg( P j )  deg( Q j ). Since deg( Q i ) ≥ deg( P i ), if ηi = 1 then the leading coefficient of deg( Q i − ηi P i ), which we shall call ξi is non zero and therefore deg( Q i − ηi P i ) = deg( Q i ). Since log(ηi ) = ηi /ηi , if ηi = 1 then log(ηi ) must be a constant. Note that if deg( Q i )  deg( P i ) then the polynomial Q i − ηi P i must be monic, that is, ξi = 1. If neither ηi = 1 nor ξi = 1 then it is clear that deg( Q i ) = deg( P i ) and therefore ξi = 1 − ηi . Thus we have the following observations on log(ηi )(ξi /ξi ):

⎧  ⎪ ⎨(log(ηi ) log(ξi )) if ηi = 1; i if ξi = 1; log(ηi ) = 0 ξi ⎪ ⎩log(η ) (1−ηi ) if ξ = 1 − η . ξ

i

i

1 −η i

i

Thus, in any event, log(ηi )(ξi /ξi ) is a D -expression over F (log(ηi )).

2

Lemma 3.3. Let F (θ) ⊃ F be a transcendental DEL-extension of F and v ∈ F (θ) be an element such that

v=

 j∈ J

aj

u j log(u j )

where u j , log(u j ), v k , e

+

− v k2



bk v k e − v k + 2

s 

w i

δi

wi

i =1

k∈ K

+ w,

, w i , w ∈ F (θ), δi ∈ F and a j , bk are constants.

(i) Suppose that θ  ∈ F . Then each u j , v k and e − v k belong to F . Furthermore, if v ∈ F [θ] then w ∈ F [θ], v − w  ∈ F and there is a subset J 1 ⊂ J and elements ξi ∈ F such that 2

v=



aj

j∈ J 1

u j log(u j )

+



bk v k e − v k + 2

s  ξ δi i + w  . ξi i =1

k∈ K

Finally, if v ∈ F then w = c θ + w 0 for some constant c and w 0 ∈ F . (ii) Suppose that θ  /θ ∈ F . Then each log(u j ) and v k belong to F and there are elements

η j and ζk in F

and integers m j and nk such that u j = η j θ m j and e − v k = ζk θ nk . Furthermore, if v ∈ F , θ  /θ = x for some x ∈ F and for each i, δi is a constant or the constant term of the partial fraction expansion of the corresponding w i / w i is zero, then there are sets J 1 = { j ∈ J | m j = 0} and K 1 = {k ∈ K | nk = 0} and an ˜ ∈ F such that element w 2

v=

 j∈ J 1

aj

u j log(u j )

+



bk v k e − v k + 2

s  ξ ˜ . δi i + w ξi i =1

k∈ K 1

Proof. Fix an algebraic closure F of F and write w i = ξi integers mil as in Proposition 2.3. Consider the equations



e−vk

2

e−vk



2

  = − v k2 and

log(u j ) =

u j uj

n

l=1 (θ

− αl )mil with αl ∈ F , ξi ∈ F and

.

(i). From Proposition 2.5, we have for each k ∈ K , e − v k ∈ F and − v k2 = ck θ + ηk for some constant ck and ηk ∈ F . Since θ is transcendental, the latter equation holds only if ck = 0 and that v k ∈ F . Similarly, we have u j ∈ F and log(u j ) = c j θ + ζ j , where c j is a constant and ζ j ∈ F . Now we further suppose that v ∈ F [θ]. Write w i as in Proposition 2.3 and observe that the partial s  fraction expansion of i =1 δi ( w i / w i ) contains only poles of order at most 1 and a constant term ζ , where 2

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.10 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

10

s



i =1 δi (ξi /ξi )

ζ = s



i =1 δi (ξi /ξi ) +

s

i =1

n

l=1 mil δi



when θ  ∈ F

(3.4)

(θ  /θ) when θ  /θ ∈ F

Let J 1 = { j ∈ J | log(u j ) ∈ F } and write

v=



aj

j∈ J 1

u j log(u j )



+

aj

j∈ J − J 1

u j c jθ + ζ j

+



bk v k e − v k + 2

s  i =1

k∈ K

δi

w i wi

+ w.

(3.5)

Since v ∈ F [θ], it is clear from the above equation that v − w  has poles of order at most 1. Therefore, from Proposition 2.3, we obtain that w has no poles and thus w ∈ F [θ]. Consequently, all the poles of Equation (3.5) must cancel out and we obtain

v=



aj

j∈ J 1

u j log(u j )

+



bk v k e − v k + 2

s  ξ δi i + w  . ξi

(3.6)

i =1

k∈ K

If v ∈ F then w  ∈ F and we have w = c θ + w 0 for some constant c and w 0 ∈ F . 2 (ii). In this case we have each − v k2 , and therefore v k , belong to F and e − v k = ηk θ nk for elements ηk ∈ F and integers nk . We have log(u j ) ∈ F and each u j = ζ j θ m j for some ζ j ∈ F and integers m j . Let J 1 = { j ∈ J | m j = 0} and K 1 = {k ∈ K | nk = 0} and rewrite

v=



u j

aj

log(u j )

j∈ J 1

+

s 

δi

i =1

where

w i

wi



+

aj

j∈ J − J 1

μ jζ jθmj log(u j )

+



bk v k e − v k + 2

k∈ K 1



bk v k ηk θ nk

k∈ K − K 1

+ w,

(3.7)

μ j = (ζ j /ζ j ) + m j (θ  /θ) ∈ F .

s

 that δi ∈ C F for 1 ≤ i ≤ p and Let v ∈ F . We rearrange the terms of i =1 δi ( w i / w i ) and assume  s   the constant term of w i / w i is zero for p + 1 ≤ i ≤ s. By assumption, i = p +1 δi ( w i / w i ) is a sum of poles. Now use Equation (3.4), Proposition 2.3 and compare the constant terms of Equation (3.7) and obtain for some w 0 ∈ F that

v=



aj

j∈ J 1

=



aj

j∈ J 1

where

u j log(u j ) u j log(u j )

 p n i =1

l=1 mil δi





+

 − v k2

bk v k e

i =1

k∈ K 1



+

 p n  p   ξi θ + δi + mil δi + w 0 , ξi θ

bk v k e − v k + 2

k∈ K 1

˜ 0 ∈ F. x + w0 = w

(3.8)

i =1 l =1

p  ξ ˜ 0 when θ  /θ = x , δi i + w ξi

(3.9)

i =1

2

4. Preparation for main results We recall that if E is an algebraic extension of F , u ∈ E and P ( X ) = X m + αn−1 X m−1 + · · · + α0 is the minimal monic polynomial of u over F then tr(u ) := −αn−1 and nr(u ) = (−1)m α0 . Let N be a finite Galois extension of F containing u with Galois group G and n := [ N : K ]. Define Tr(u ) :=   σ ∈G σ (u ) and Nr(u ) := σ ∈G σ (u ). It is easy to see that Tr(u ) and Nr(u ) belong to F and

Tr(u ) =

n m

tr(u )

and

n

Nr(u ) = nr(u ) m .

Theorem 4.1. Let E = F (θ1 , · · · , θm ) ⊃ F be a DEL-extension of F and u ∈ E be an element with u  ∈ F .

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.11 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

11

(i) If each θi is neither algebraic over F i −1 nor an exponential of an element of F i −1 then u  admits a DEL-expression over F . (ii) If F be a liouvillian extension of C F and each θi is neither an exponential of F i −1 nor an error function of F i −1 then u  admits a DL-expression over F . Proof. We prove the result using an induction on m. When m = 1, we apply Proposition 2.2 (c) and get u = c θ + η for some constant c ∈ C F and η ∈ F . Differentiating this equation we obtain the desired expression for u  . Let I , J , K be finite indexing sets such that

u =

 g i ri

i∈I

gi

+



aj

j∈ J

u j log(u j )

+



bk v k e − v k + w  ,

r i =

2

ni 

c it hit /h it ,

(4.1)

t =1

k∈ K

where elements r i , g i , w , h it , u j , log(u j ), v k and e − v k all belong to F 1 := F (θ1 ), a j and bk are constants and c i1 , · · · , c ini are Q-linearly independent constants for each i, be a DEL-expression for u  over F 1 := F (θ1 ). Let F be an algebraic closure of F and β j ∈ F and ξi ∈ F be elements such that 2

g i / g i = ξi /ξi +

p 

mil

l =1

θ1 − βl , θ1 − βl

(4.2)

where each mil is an integer. (i). We have θ1 ∈ F . Then from Proposition 2.5 each h it belongs to F and from Lemma 3.3 each u j ,

e − v k and v k belong to F . If all r i and log(u j ) belong to F then we shall apply Lemma 3.3 and obtain 2

u =

 ξ   u j 2 ri i + aj bk v k e − v k + c 1 θ1 + w 0 . + ξi log(u j ) i∈I

j∈ J

(4.3)

k∈ K

From the definition of θ1 , it is clear that the above expression is a DEL-expression of u  over F . Now suppose that there is an r ∈ {r i , log(u j ) | i ∈ I , j ∈ J } and r ∈ / F . Since F (θ1 ) = F (r ), we shall find constants c i and element ηi ∈ F such that r i = c i r + ηi ∈ F [θ1 ]. We shall take θ1 = r in Equation (4.2) and rewrite Equation (4.1) over F (r ) as

u =

p p   ξ  ξ  r  − βl  r  − βl ci i r + mil c i r+ ηi i + mil ηi ξi r − βl ξi r − βl i∈I

+



aj

j∈ J

i ∈ I l =1

u j log(u j )

+



i∈I

(4.4)

i ∈ I l =1

bk v k e − v k + w  . 2

k∈ K

Thus,

u −

p  ξ  ξ   u j r  − βl  2 ci i r − ηi i = δl + aj bk v k e − v k + w  , + ξi ξi r − βl log(u j ) i∈I

where δl =



i∈ I

i∈I

mil (c i βl + ηi ) and w is replaced with w +

Note that u  −



ξ

i∈ I

ci ξ r − i

i



the following expression for u  :

u =

j∈ J

l =1

ξ

i∈ I



(4.5)

k∈ K

i∈ I

p

l=1

mil c i (r − βl ).

ηi ξi ∈ F [r ] and apply Lemma 3.3 to the fields F (r ) ⊃ F and obtain i

 ξ  ξ   u j 2 ci i r + ηi i + aj bk v k e − v k + w  , + ξi ξi log(u j ) i∈I

i∈I

j∈ J 1

(4.6)

k∈ K

where for j ∈ J 1 , log(u j ) ∈ F . But we know that log(u j ) ∈ F (r ) for all j and thus log(u j ) ∈ F for all j ∈ J 1 . We already know that each v k , e − v k belongs to F . Since u  − 2



ξ

i∈ I

c i ξi r − i



ξ

i∈ I

ηi ξii ∈ F [r ],

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.12 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

12

which is a polynomial of degree one and that w ∈ F (r ), we obtain w  ∈ F [r ] ⊂ F (r ). Now from Proposition 2.2, we shall replace w with an element of F (r ) that satisfies Equation (4.6) and we also conclude w = dr 2 + w 1 r + w 0 , where w 1 , w 0 ∈ F and d ∈ C F . Comparing the coefficients of r of  Equation (4.6), we obtain w 1 = −2dr  − i ∈ I c i ξi /ξi and comparing constant terms, we obtain

u =



ηi

i∈I

  u j ξi 2 + w 1r + aj bk v k e − v k + w 0 . + ξi log(u j ) j∈ J 1

k∈ K

(4.7)

q

Since, for some j, r = r j or r = log(u j ), we have r  = l=1 el (hl /hl ) for some constants el and elements hl ∈ F . Therefore, rewriting Equation (4.7), we obtain the following DEL-expression for u  :

u =





 h   uj ξi 2 + w1 el l + aj bk v k e − v k + w  , + ξi hl log(u j ) q

ηi

i∈I

j∈ J 1

l =1

(4.8)

k∈ K

where

ηi =

ni 

c it

t =1

hit

− ci

h it

q  hl

el

l =1

hl

, w 1 = −2d

q  hl

el

l =1

hl

−

 ξ ci i . ξi

(4.9)

i∈I

(ii). Let θ1 be algebraic over F and bk = 0 for all k ∈ K . From Proposition 2.5, each r i , log(u j ) belong to F and each u j /u j and hit /h it belong to F . For every element v ∈ {u j , h it | j ∈ J , i ∈ I , 1 ≤ t ≤ ni }, we can choose the smallest integer n v ≥ 0 such that v n v ∈ F (see Proposition 2.1 (b)). Let N be a finite algebraic Galois extension of F containing F (θ1 ) with Galois group G. Then either 0 = tr( v ) = Tr( v ) or v ∈ F and Tr( v ) = [ N : F ] v. Moreover for σ ∈ G we have σ ( y  ) = σ ( y ) , and thus T r ( y  ) = T r ( y )  Nr(u ) σ (u ) for all y ∈ N and Nr (u ) = σ ∈G σ (u ) . Let J 1 be the subset of J such Tr(u j ) = 0 for all j ∈ J 1 . Then u j ∈ F and Tr(u j ) = [ N : F ]u j for all j ∈ J 1 . For each σ ∈ G we have

u =

 σ ( g i )  σ (u j ) ri dj + + (σ w ) . σ ( gi ) log(u j ) i∈I

j∈ J

σ ∈ G and obtain   σ ( g i )   σ (u j )  [ N : F ]u  = ri dj (σ w ) and thus + + σ ( gi ) log(u j ) σ σ σ i∈I j∈ J

Therefore, we sum over all

u =

 Nr( g i )  u j 1 ri dj Tr( w ) . + + [N : F ] Nr( g i ) log(u j ) [N : F ] 1

i∈I

(4.10)

j∈ J 1

Using Proposition 2.5 (c), we choose an integer n ≥ 0 such that hnit ∈ F for all i , t and observe that

r i =

ni  c it (hnit ) t =1

n

hnit

.

Thus Equation (4.10) provides a DL-expression for u  over F .

2

Remark 3 (Problem of Algebraic Elements). When dealing with expressions involving algebraic elements, the only method the authors know, to obtain a similar expression over the base field is the standard method of taking “Trace” as done in Theorem 4.1(ii). This method fails when we deal with error 2 functions: If u , e −u belongs to an algebraic extension of F with the same field of constants as F then from equation

(e −u ) 2

(−u 2 ) =

e −u

2

,

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.13 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

13

2

we only obtain u 2 ∈ F and that (e u )2 ∈ F (see Singer et al., 1985, p. 977, Theorem 4.1). Moreover, 2 such error functions do exists (see Singer et al., 1985, p. 968, Example 1.1). Thus Tr(u  e −u ) can’t be simplified further to obtain a similar expression over F . Similarly, if

g

v =r

for some v ∈ F and r  = c

g

(1 − g ) 1− g

is a D -expression over an algebraic extension of F then r must be in F. However, we only know that some power of 1 − g belongs to F . Now taking Trace, we obtain

r Nr ( g )

v=

and

m Nr ( g )

 r  m

=

c Nr (1 − g ) m2 Nr (1 − g )

(4.11)

.

Since Nr (1 − g ) need not equal a constant multiple of 1 − Nr ( g ), the above equation does not provide a D -expression over F . Nonetheless, Equation (4.11) does provide a DL-expression over F and indeed, this argument was used in the proof of Theorem 4.1 (ii). 4.1. Induction step Let E = F (θ1 , · · · , θn ) be a DEL-extension of F . The proof of our main theorem uses an induction on n and the crucial argument (the induction step) is the following: First we show that if

u =

 g i ri

i∈I

gi

+



 j∈ J

aj

u j log(u j )

+



bk v k e − v k + w  , 2

k∈ K



r i = c i (1 − g i ) /(1 − g i ) is a special DEL-expression over a logarithmic extension L = F l−1 (log( f 1 ), · · · , log( f p )) of F l−1 = F l−2 (θl−1 ), where f 1 , · · · , f p ∈ F l−1 then one can find elements h1 , · · · , hq ∈ F l−2 such that u  admits a special DEL-expression over F l−1 (log(h1 ), · · · , log(hq )). Next, we show that there is an element h ∈ F l−2 such that u  admits a special DEL-expression over F l−2 (log(h), log(h1 ), · · · , log(hq )) which then completes the induction argument. The following lemmas will help us to accomplish this task. Lemma 4.2. Let E ⊃ F (θ) ⊃ F be differential fields, θ be transcendental over F and C E = C F . Let

α1 , · · · , αt ∈ F be distinct elements such that E = F (θ)(log(θ − α1 ), · · · , log(θ − αt )) and that log(θ − α1 ), · · · , log(θ − αt ) are algebraically independent over F (θ). Let there be elements v ∈ F , u 1 , · · · ut ∈ F (θ), and w ∈ E such that t 

u i log(θ − αi ) + w  − v =: T ∈ F (θ)

i =1

˜ ∈E and that each u i admits a L-expression over F (θ). Then one can find elements f 1 , · · · , f t in F (θ) and w such that t  i =1

fi

θ  − αi ˜  − v = T ∈ F (θ) +w θ − αi

and that each f i equals the L-expression −u i −

t

θ  −αl

l=1 c il θ−αl

, where each c il is a constant.

Proof. Let E 0 = F (θ) and E i = E i −1 (log(θ − αi )) for 1 ≤ i ≤ t. Observe that ut log(θ − αt ) + w  ∈ E t −1 . Therefore by Proposition 2.2, w = ct log2 (θ − αt ) + f t log(θ − αt ) + w t , where ct ∈ C F and f t , w t ∈ E t −1 and we have

v=

t −1  i =1

  u i log(θ − αi ) + f t (θ  − αt )/(θ − αt ) + ut + 2ct (θ  − αt )/(θ − αt ) + f t log(θ − αt )

− T + w t

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.14 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

14

Compare constant coefficients to get

v = f t (θ  − αt )/(θ − αt ) +

t −1 

u i log(θ − αi ) − T + w t ,

i =1

where f t = −ut − 2ct (θ  − αt )/(θ − αt ) ∈ F (θ). Noting that f t admits a L-expression over F (θ) of desired form, for n  t, we suppose that there are elements w n+1 , gn+1 , · · · , gt ∈ E n such that each g i admits a L-expression over F (θ) of the form t 

g i = −u i +

c il

l=n+1

v=

t 

θ  − αl θ − αl

and

g i (θ  − αi )/(θ − αi ) +

i =n+1

n  i =1

u i log(θ − αi ) − T + w n +1 .

(4.12)

Then since g i ∈ F (θ), there are constants ai ∈ C F and f i ∈ E n−1 such that f i = g i − ai log(θ − αn ). Note that

⎛

⎞

t 

⎝u n +

i =n+1

ai (θ  − αi )/(θ − αi )⎠ log(θ − αn ) + w n +1 ∈ E n−1

and therefore w n+1 = cn log(θ − αn )2 + f n log(θ − αn ) + w n , where cn ∈ C F and f n , w n ∈ E n−1 . Now we shall compare the constant coefficients of the equation (4.12) and obtain

v=

n −1

t 

fi

i =n

θ  − αi  + u i log(θ − αi ) − T + w n , θ − αi i =1

where for each i , n + 1 ≤ i ≤ t, t 

f i = −u i +

c il

l=n+1

f n = −un − 2cn

θ  − αl θ  − αn − ai θ − αl θ − αn

and

t  θ  − αi θ  − αn − ai θ − αn θ − αi i =n+1

are L-expressions over F (θ) of desired form. Thus by induction, we have shown that

v=

t 

˜ , f i (θ  − αi )/(θ − αi ) − T + w

i =1

where each f i ∈ F (θ), f i = −u i −

t

θ  −αl

l=1 c il θ−αl

(4.13)

˜ ∈ F (θ). , c il ∈ C F and w

2

Lemma 4.3. Let F (θ) ⊃ F be a transcendental DEL-extension of F . If v ∈ F admits a special DEL-expression over the differential field F (θ)(log( y 1 ), · · · , log( yn )), each y i ∈ F (θ), having the same field of constants as F then there is a differential field M = F (log(h1 ), · · · , log(hm ), θ), where h i ∈ F and having the same field of constants as F such that v admits a special DEL-expression over M.



g i i∈ I 1 ri gi





u

a j log(uj ) + k∈ K bk v k e − v k + w  be a special DEL-expression over j some logarithmic extension E = F (θ)(log( y 1 ), · · · , log( yn )) of F (θ). For convenience, we shall rewrite Proof. Let v =

v=

 g i ri

i∈I

gi

+

 j∈ J

+

aj

j∈ J

u j log(u j )

+

 l∈ I 1 − I

rl

gl gl

2

+

 k∈ K

bk v k e − v k + w  , 2

(4.14)

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.15 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

15

where I = {i ∈ I 1 | r i is not a constant}. We can apply Proposition 3.2 to the D -expression



g i i∈ I 1 ri gi

+ w  , enlarge E to a logarithmic extension of F (θ) to include log( g i ) and assume that 1 − g i = ηi P i / Q i , where P i , Q i ∈ F [θ] are relatively prime monic polynomials, θ neither divides P i nor Q i , deg( Q i ) ≥ deg( P i ) and ηi ∈ F . Now since r i ∈ F (θ), there are constants c ip n such that r i − )] for each i, where  p = p =1 c ip log( y p ) ∈ F (θ) and in particular, r i ∈  p [log( y p   F (θ)(log( y 1 ), · · · , log( y p −1 ), log( y p +1 ), · · · , log( yn )). Observe that v − i ∈ I r i ( g i / g i ) ∈  p [log( y p )] and that

v−





r i ( g i / g i ) =

i∈I

aj

j∈ J

u j log(u j )

+

 l∈ I 1 − I

rl

gl gl

+



bk v k e − v k + w  . 2

k∈ K



Let γ := v − i ∈ I r i ( g i / g i ) − w  and apply Lemma 3.3 to get that γ ∈  p for each p. Thus γ ∈ F (θ). Now a repeated application of Lemma 3.3 to the field extension E of F (θ), with v = γ and w = 0,

would tell us that we could assume u j , log(u j ), v k , e − v k and gl belong to F (θ). We enlarge E to  include log( gl ) and replace w with w + l∈ I 1 − I rl log( gl ) and write 2

v=

 g i ri

i∈I

gi

+ S + w,

(4.15)

where r i , w is in some logarithmic extension E of F (θ), g i ∈ F (θ) and S =



k∈ K



u

j∈ J

a j log(uj ) + j

bk v k e − v k . 2

Fix F an algebraic closure of F . It is easy to see that there is a subset A = {0 = α1 , · · · , αt } of F such that the following holds:

t

t

t

(i) P i = j =1 (θ − α j )li j , Q i = j =1 (θ − α j )mi j and Q i − ηi P i = ξi j =1 (θ − α j )ni j for some ξi ∈ F , where li j , mi j and ni j are non negative integers. (ii) w ∈ M 1 ({log(θ − α ) | α ∈ A }), where M 1 = F ({log(ηi ), log(ξi ), θ |i ∈ I }). Let M = F ({log(ηi ), log(ξi ), θ | i ∈ I }) and choose a differential subfield M ∗ of M 1 such that F ⊂ M ∗ , θ is transcendental over M ∗ and M ∗ (θ) = M 1 (see Theorem 2.4 (b)). Let ai j = li j − mi j and b i j = ni j − mi j t t and observe that j =1 ai j = deg( P i ) − deg( Q i ) and j =1 b i j = deg( Q i − ηi P i ) − deg( Q i ). We have

r i = c i log(ηi ) +

t 

c i ai j log(θ − α j ) + e i

for some e i ∈ C F

(4.16)

j =2

g i / g i = ξi /ξi +

t  j =1

v=

 g i ri

i∈I

=

t  j =2



gi

bi j

θ  − α j θ − αj

and therefore

(4.17)

+ S + w

u j log(θ − α j ) +

 i∈I

c i log(ηi )

g i gi

+ S + w,

(4.18)

where u j := i ∈ I c i ai j ( g i / g i ) is a L-expression over F (θ) and w ∈ M 1 ({log(θ − α ) | α ∈ A }). Consider the differential fields M 1 = M ∗ (θ) ⊃ M ∗ . It is easy to see that log(θ − α2 ), · · · , log(θ − αt ) are algebraically independent over M ∗ (see Propositions 3 and 4, pp. 931–933 of Baddoura, 2006).

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.16 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

16

Thus, we apply Lemma 4.2 to differential fields M 1 = M ∗ (θ) ⊃ M ∗ with T := − and get

v=

t 

θ  − α j

fj

θ − αj

j =2

+



c i log(ηi )

i∈I

g i gi

+ S + w,

f j = −u j −

t 

c jl

l =2



g

i∈ I

c i log(ηi ) gi − S i

θ  − αl , c jl ∈ C F , θ − αl (4.19)

where w and each f j belong to M 1 = M ∗ (θ). Substituting Equation (4.17) in Equation (4.19) we obtain the following expression for v over M ∗ (θ):

v=

t 

 fj +

j =1



 b i j c i log(ηi )

θ  − α j

i∈I

θ − αj

+



c i log(ηi )(ξi /ξi ) + S + w  , where f 1 = 0 and

i∈I

(4.20) f j = −



c i ai j

i∈I

ξi − ξi

t   l =1 i ∈ I

c i ai j b il

θ  − αl − θ − αl



t 

c jl

l =2

θ  − αl . θ − αl

(4.21)

 From Proposition 3.2, we have i ∈ I c i log(ηi )(ξi /ξi ) is a D -expression over M and that S +   /ξ ) is a special DEL-expression over M. We only need to the handle the first and c log ( η )(ξ i i i i∈ I i last term appearing on the above expression of v.     First we suppose that θ  ∈ F . Since i ∈ I b i j c i log(ηi ) = i ∈ I b i j c i (ηi /ηi ), from Equation (4.21)  and Proposition 2.5, we have f j + i ∈ I b i j c i log(ηi ) = d j θ + h j for some d j ∈ C F and h j ∈ M ∗ . Similarly, log(ηi ) − λi θ ∈ M ∗ for some λi ∈ C F . Thus

v=

t  θ  − α j  ξ (d j θ + h j ) + c i log(ηi ) i + S + w  θ − αj ξi j =1

=

i∈I

t 

(d j α j + h j )

j =1

θ

− α

j

θ − αj

+



c i log(ηi )

i∈I

ξi ˜ , +S+w ξi

t

(4.22)

˜ = w + j =1 d j (θ − α j ) ∈ M ∗ (θ) and d j α j + h j ∈ F . Since v − where w we shall apply Lemma 3.3 to the fields M ∗ (θ) ⊃ M ∗ and obtain v=



c i log(ηi )

i∈I

p

c i =1 i

log(ηi )(ξi /ξi ) ∈ M ∗ [θ],

 u j ξi  2 ˜ , + aj v k e − v k + w + ξi log(u j ) j∈ J 1

(4.23)

k∈ K

˜  is a special DEL-expression over M = F ({log(ηi ), log(ξi ), θ | i ∈ I }). This implies that where v − w ˜  ∈ M. Since w ˜ ∈ M ∗ [θ] ⊂ M 1 , which is an algebraic extension of M, we shall apply Proposition 2.1 w ˜  . This settles the case when θ  ∈ F . ˜ with an element w ∈ M such that w  = w and replace w   Now we suppose that θ /θ = x ∈ F for some x ∈ F and consider Equations (4.20) and (4.21). We have for each i and j both f j and log(ηi ) belong to M ∗ . Using the fact that

θ  − α j θ − αj

= x +

x α j − α j

θ − αj

,

we rewrite Equation (4.20) as

v=

t  j =1

 fj +

 i∈I

 b i j c i log(ηi )



x +

x α j − α j

θ − αj

 +

 i∈I

c i log(ηi )(ξi /ξi ) + S + w  .

(4.24)

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.17 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

17

As in Equation (3.7), we find sets J 1 = { j ∈ J | m j = 0} and K 1 = {k ∈ K | nk = 0} and we rewrite

v=

t  

fj +



j =1

+

t   b i j c i log(ηi ) x +

i∈I



 fj +



j =1

 b i j c i log(ηi )

x α j − α j



θj − αj

i∈I

c i log(ηi )(ξi /ξi )

i∈I

+



u j

aj

log(u j )

j∈ J 1

where

+

 j∈ J − J 1

aj

μ jζ jθmj log(u j )

+



bk v k e − v k +

k∈ K 1

2



bk v k ηk θ nk + w  ,

k∈ K − K 1

μ j = (ζ j /ζ j ) + m j (θ  /θ) ∈ F . Now comparing the constant coefficients, we obtain v=

t  j =1

⎛ ⎞ t   u j ξ  f j x + ⎝ b i j c i log(ηi )⎠ x + c i log(ηi ) i + aj ξi log(u j ) i ∈ I j =1



+

bk v k e

− v k2

i∈I

j∈ J 1

+ w 0

(4.25)

k∈ K 1

t



ηi = 1 then deg( Q i − ηi P i ) = deg( Q i ) and therefore tj=1 bi j = 0 t and thus j =1 b i j c i log(ηi ) = 0. If ηi = 1 then log(ηi ) ∈ C F and we have j =1 b i j c i log(ηi ) ∈ C F . Thus,  t in any event, d := i ∈ I b c log ( η ) ∈ C . Therefore F i j i i j =1 Consider

j =1 b i j c i

t

v = f x +



log(ηi ). If

c i log(ηi )

i∈I

t

 u j ξi  2 + aj bk v k e − v k + ( w 0 + dx) , + ξi log(u j ) j∈ J 1

t

(4.26)

k∈ K 1



 where f = i ∈ I c i log(ηi )(ξi /ξi ) is a dilogarithmic expression j =1 f j = j =2 f j . As observed earlier,   over M. Now we will show that f is either a constant or f x = h for some h ∈ F (θ), and w 0 ∈ M. Comparing the constant term of the equation (4.21), we obtain

f =−

t  i ∈ I j =2

ai j

ξi − cx , for some constant c ∈ C F . ξi

t

= 0 then since ai1 = 0 for all i ∈ I , we have deg( Q i )  deg( P i ) and as observed in Propo t ξi   sition 3.2, ξi = 1 for all such i. Thus i∈ I j =2 ai j ξi = 0 and therefore f = −cx . Now it follows that ˜ := w 0 + dx − f 2 /(2c ) and observe from Equation (4.26) that either f  = 0 or (− f 2 /(2c )) = f x . Let w ˜  ∈ M. Since w ˜ ∈ M 1 which is an algebraic extension of M, we shall apply Proposition 2.1 and find w ˜ . 2 an element w ∈ M such that w  = w If

j =2 ai j

5. Main results We recall the following definitions: A differential field extension E of F will be called a logarithmic extension of F if C E = C F and there are elements h1 , · · · , hm ∈ F such that E = F (log(h1 ), · · · , log(hm )). A differential field extension E = F (θ1 , · · · , θn ) is a transcendental dilogarithmic-elementary extension of F if for each i, θi is transcendental over F i −1 and satisfies either case ii or iii or vi. Theorem 5.1. Let F (θ) ⊃ F be transcendental DEL-extension of F such that F (θ) = F (log(h)) for any h ∈ F . If v ∈ F admits a special DEL-expression over F (θ) then v admits a special DEL-expression over F .

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.18 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

18

Proof. As in Equation (4.14), let

v=

 g i ri

i∈I

gi

+

 j∈ J

aj

u j log(u j )

+

 l∈ I 1 − I

rl

gl gl

+

r i = c i (1 − g i ) /(1 − g i ) where each c i = 0,



bk v k e − v k + w  , 2

(5.1)

k∈ K

(5.2)

be a special DEL-expression over F (θ). Suppose that θ  /θ = x for some x ∈ F . We use Proposition 2.5 to Equation (5.2) and if necessary, Remark 2 and obtain r i ∈ F and 1 − g i = ηi /θ mi for some integer mi ≥ 0 and ηi ∈ F . Now

g i gi

  m  θ m i − ηi θ i mi x ηi − ηi − m = m m θ i − ηi θ i θ i − ηi

 =

and therefore g i / g i has no constant term when mi  0. We now apply Lemma 3.3 and obtain a   special DEL-expression for v over F . On the other hand, if θ  ∈ F then  (r i /c i ) = (1 − g i ) /(1 − g i ) and therefore g i ∈ F . From our hypothesis r i ∈ F . Now we have v − i ∈ I r i ( g i / g i ) ∈ F and we can apply Lemma 3.3 to conclude that v admits a special DEL-expression over F . 2 Theorem 5.2. Let E = F (θ1 , · · · , θn ) be a transcendental DEL-extension of F with C E = C F . Suppose that there is an element u ∈ E with u  ∈ F then u  admits a special DEL-expression over some logarithmic extension of F . Furthermore, if E is a transcendental dilogarithmic-extension of F then u  admits a D -expression over some logarithmic extension of F . Proof. We shall use an induction on n to prove the theorem. The case when n = 1, we have u ∈ F (θ) and that u = c θ + w for some c ∈ C F and w ∈ F and therefore u  = c θ  + w  . Now the definition of θ  proves that, in fact, u  admits a special DEL-expression over F itself. Note that u  ∈ F ⊂ F (θ) and suppose that u  admits a special DEL-expression (respectively a D -expression) over some logarithmic extension of F (θ) having the same field of constants as F . Then we shall apply Lemma 4.3 and obtain that u  admits a special DEL-expression (respectively a D -expression) over M = F (θ, log(h1 ), · · · , log(hm )), where h1 , · · · , hm ∈ F and C M = C F . Let F ∗ = F (log(h1 ), · · · , log(hm )) and suppose that F ∗ (θ) = F ∗ (log(h)) for some h ∈ F ∗ . If there is an integer p such that h ∈  p (log(h p )) −  p , where  p is the field generated by F and all log(hi ) except log(h p ), then we shall apply Theorem 2.4 to the fields M =  p (log(h p ))(log(h)) ⊃ F ∗ =  p (log(h p )) ⊃  p and obtain that θ ∈ F ∗ . This implies M is a logarithmic extension of F . If no such p exists then h ∈ F and we have M = F (log(h), log(h1 ), · · · , log(hm )), which is again a logarithmic extension of F . On the other hand if for any h ∈ F ∗ , log(h) ∈ / F ∗ (θ) − F ∗ then we shall apply Theorem 5.1 and show that u  admits a special DEL-expression (respectively a D -expression) over the logarithmic extension F ∗ . This completes the induction argument. 2 Theorem 5.3. Let E = F (θ1 , · · · , θn ) be a transcendental DEL-extension of F with C E = C F . Suppose that there is an element u ∈ E with u  ∈ F then u  admits a DEL-expression over F . Proof. From Theorem 5.2 we know that u  admits a (special) DEL-expression over a logarithmic extension of F . We shall now apply Theorem 4.1(i) and obtain a DEL-expression for u  over F . 2 6. Baddoura’s theorem Using techniques from Proposition 1 of Baddoura (2006) and our Theorem 5.2, we shall generalise and provide a proof of Baddoura’s Theorem. We recall that a DEL-extension E = F (θ1 , · · · , θn ) is called a transcendental dilogarithmic-elementary extension if C E = C F and for each i, θi is transcendental over F i −1 and satisfies either case (ii) or (iii) or (vi). Before we proceed to the proof of

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.19 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

19

Baddoura’s Theorem, we recall the definitions of 2 ( g ) and D ( g ) from Baddoura (2006). Let E be a differential field extension of F and g ∈ F be an element of F . If y ∈ E is an element such that

y  = − log(1 − g )

g g

then we shall pick one such element y and denote it by 2 ( g ). Note that any other element in E whose derivative equal 2 ( g ) differs from 2 ( g ) by some constant of E. The element 2 ( g ) + (1/2) log( g ) log(1 − g ) will be denoted by D ( g ), which is sometimes called the Bloch–Wigner–Spence function of g and its derivative is:

D ( g ) = −

1 g 2 g

log(1 − g ) +

1 (1 − g ) 2 1− g

log( g ).

Theorem 6.1. Let E = F (θ1 , · · · , θn ) be a transcendental dilogarithmic-elementary extension of F . Suppose that there is an element u ∈ E with u  ∈ F . Then

u=

m 

c j D(g j ) +

n 

j =1

f i log(h i ) + w ,

(6.1)

i =1

where each f i , h i , g j , w ∈ F , c j are constants and log(h i ) and D ( g j ) belong to some dilogarithmic-elementary extension of F .



g

Proof. From Theorem 5.2 we have the D -expression u  = i ∈ I 1 r i gi + w  over a logarithmic extension i E = F (log(h1 ), · · · , log(hm )) of F . As done in Equation (4.14), we shall rewrite this expression as:

u =

 g i ri

i∈I

gi

+

q 

dl

l =1

w l wl

+ w,

(6.2)

r i = −c i (1 − g i ) /(1 − g i ),

(6.3)

where each c i is a non zero constant.3 From Remark 1, we shall assume that {d1 , · · · , dq } are Q-linearly independent constants of F . Claim. Each g i and w l belong to F and E can be chosen to be the differential field F ({log(1 − g i ) | i ∈ I }). As denoted earlier, let  p be the field generated by F and all log(h i ) except log(h p ). We know from Proposition 3.1 that each g i ∈  p and each r i is a polynomial in log(h p ) of degree one. Thus u  −



g i i∈ I ri gi

∈  p [log(h p )]. Now, from Proposition 2.3 and from the fact that d1 , · · · , dq are Q-linearly independent, it follows that w l ∈  p . Since p ∈ I is arbitrary, we have g i ∈ F and that w l ∈ F . Note that each r i ∈ E and consider the differential subfield F ∗ := F ({log(1 − g i ) | i ∈ I }) of E. Then r i ∈ F ∗ and observe that v := u  − w=



g i i∈ I ri gi

q

w d l l=1 l w l

∈ F ∗ . Since w  = v, we apply Theorem 2.4 and write ˜ for some constants a j ∈ C F and w ˜ ∈ F ∗ . Thus j =1 a j log(h j ) + w,

m

u =

 g i ri

i∈I

gi

+

q  l =1

dl

w l wl

+

−

m  j =1

aj

hj hj

˜ +w

and this proves the claim.

3

Presence of negative sign is to smoothen out calculations.

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.20 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

20

Thus g i and w l belong to F and we shall assume E = F ({log(1 − g i ) | i ∈ I }). “Taking integrals” we see that there is some dilogarithmic elementary extension E ∗ of F containing E such that

u=



c i 2 ( g i ) +

i∈I

q 

dl log( w l ) + w + c ,

(6.4)

l =1

where c ∈ C F and each 2 ( g i ) and log( w l ) belong to E ∗ . We shall first show that w ∈ F [log(1 − g 1 ), · · · , log(1 − gm )] is a polynomial of total degree at most 2 and then show how to combine terms of w with 2 ( g i ) of Equation (6.4) to obtain Equation (6.1).4 Without loss of generality, assume that I = {1, 2, · · · , m} and {log(1 − g 1 ), · · · , log(1 − gn )} is a transcendence base of E over F for some n, 1 ≤ n ≤ m. Then E = F (log(1 − g 1 ), · · · , log(1 − gn )) and since r i ∈ F , we have constants c i and e i such that

r i = −c i log(1 − g i ) + e i , for 1 ≤ i ≤ m

(6.5)

and we shall also rewrite r i for n + 1 ≤ i ≤ m as

ri =

n 

c is log(1 − g s ) − zi ,

(6.6)

s =1

where c is are constants and elements zi ∈ F . In particular, r i is a polynomial in log(1 − g p ) over  p of degree at most 1 and w ∈  p [log(1 − g p )] for any p, 1 ≤ p ≤ n. Now from Proposition 2.2, we have w is a polynomial in log(1 − g p ) over  p of degree at most 2 whose leading coefficient is a constant. Since p is arbitrary, we have w ∈ F [log(1 − g 1 ), · · · , log(1 − gn )] is a polynomial of total degree at most 2. Write

w=

n 

n 

as log(1 − g s )2 +

s =1

xst log(1 − g s ) log(1 − gt ) +

s=1,t =1 st

n 

xs log(1 − g s ) + w 0 , (6.7)

s =1

where as ∈ C F and xst , xs , w 0 belong to F . Then

w =

n  s =1

⎛

g s )

⎝2as (1 − + xs + 1 − gs

n 

+

 n 

xst log(1 − gt ) + xst

t =1,st

gt )

(1 − 1 − gt

⎞ ⎠ log(1 − g s )

(6.8)

 (1 − g s ) (1 − g s ) log(1 − gt ) + xs + w 0 . 1 − gs 1 − gs n

xst

s=1,t =1 st

s =1

Substituting Equations (6.6) and (6.8) in Equation (6.2) and comparing the coefficients of log(1 − g s ), we obtain

− cs

+

g s gs

+

n  t =1 t s

4

m  i =n+1

xts

c is

g i gi

+ 2as

n   (1 − g s ) (1 − gt ) + xs + xst log(1 − gt ) + xst 1 − gs 1 − gt t =1 st

(1 − gt ) = 0. (1 − gt )

The technique used to combine terms is taken from Proposition 1, in particular pp. 920–921, of Baddoura (2006).

(6.9)

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.21 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

Therefore

n

t =1,st

21

xst log(1 − gt ) ∈ F and since {log(1 − g i ) | i = 1, 2, · · · , n} is algebraically indepen-

dent over F , we must have xst = 0 for all s  t. Now it follows that there is a constant a ∈ C F such that

as log(1 − g s ) +

n  xst

2

t =1 st

=

cs 2

log(1 − gt ) +

n  xts t =1 t s

m  c is

log( g s ) −

i =n+1

2

2

log(1 − gt )

log( g i ) −

xs 2

+ a.

Now we multiply the above equation by log(1 − g s ) and sum over all of s to obtain n 

n 

as log(1 − g s )2 +

s =1

=

n 

⎛ ⎝

s =1

cs 2

2

i =n+1

2

s =1

m  c is

log( g s ) −

n   xs

−

xst log(1 − gt )

t =1 , s =1 st

⎞ log( g i )⎠ (log(1 − g s ))

 − a log(1 − g s ).

(6.10)

We have

⎛ u=

n 

c i 2 ( g i ) +

i =1

+

m 

c i 2 ( g i ) +

s =1

i =n+1

n 

xs log(1 − g s ) +

s =1

n 

q 

⎞

⎜ ⎜as log(1 − g s ) + ⎝

n  t =1 st

dl log( w l ) + w 0 + c

l =1

and we shall use Equation (6.10) to rewrite the above equation and get

u=

n  

c i 2 ( g i ) +

i =1



m 

+

i =n+1

+

n  xs s =1

2

ci 2

n  −c is s =1

2



log( g i ) log(1 − g i ) +

m 

c i 2 ( g i )

i =n+1

 log(1 − g s ) log( g i )

log(1 − g s ) + a

n 

log(1 − g s ) +

s =1

q 

dl log( w l ) + w 0 + c

l =1

Now from Equations (6.6) and (6.5), we have n  −c is s =1

2

log(1 − g s ) = −

ri 2

for each i, n + 1 ≤ i ≤ m. Therefore

−

zi 2

=

ci 2

log(1 − g i ) −

zi + e i 2

⎟

xst log(1 − gt )⎟ ⎠ log(1 − g s )

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.22 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

22

u=

n  

c i 2 ( g i ) +

i =1

+

n  xs s =1

+

q 



ci 2

c i 2 ( g i ) +

i =n+1

log(1 − g s ) −

2

m  

log( g i ) log(1 − g i ) + m  zi + e i

2

i =n+1

log( g i ) + a

n 

ci 2



log( g i ) log(1 − g i )

log(1 − g s )

s =1

dl log( w l ) + w 0 + c

l =1

and thus we shall rewrite

u=

m 

c j D(g j) +

j =1

n 

f i log(h i ) + w

i =1

for suitable f i , h i and w in F .

2

6.1. Example Let log( z), log( z − 1), log( z + 1) and log( z2 + z − 1) be designated solutions of the differential equations y  = 1/ z, y  = 1/( z − 1), y  = 1/( z + 1) and y  = (2z + 1)/( z2 + z − 1) respectively. Denote 2 log( z) + log( z − 1) + log(z2 + z − 1) and log( z) + log( z + 1) by log  (z(z − 1)(z + z − 1)) and log(z(z + 1)) 2 respectively. Let F = C z, log( z + 1), log( z( z − 1)( z + z − 1)) be the ordinary differential field with the derivation  := d/dz and E = F 3 ⊃ F 2 ⊃ F 1 ⊃ F 0 = F be the dilogarithmic-elementary extension of F , where (a) F 1 = F (log z), z) (b) F 2 = F 1 (2 (1 − z)), 2 (1 − z) = − (11− −z log z and (c) F 3 = F 2 (2 (1 − z( z + 1))), 2 (1 − z( z + 1)) = −

(1−z(z+1)) log( z( z + 1)). 1− z( z+1)

Note that

log( z) + log( z − 1) + log( z2 + z − 1) = log( z) + log(1 − z) + log(1 − z − z2 ) + c ,

(6.11)

for some constant c ∈ C F and consider the element

− log( z + 1)

 z  (1 − z( z + 1)) + v 0 =: v ∈ F , + log z( z − 1)( z2 + z − 1) 1 − z ( z + 1) z

(6.12)

where v 0 ∈ F is arbitrary. Over the field F 1 , we rewrite v as

 (1 − z) (1 − z( z + 1)) (1 − z) (1 − z( z + 1)) log( z( z + 1)) − log z + log z + 1 − z ( z + 1) 1−z 1 − z ( z + 1) 1−z  z  + log z( z − 1)( z2 + z − 1) + v 0 . (6.13)

v =−

2



z



Let w = −(1/2) log ( z) + log z( z − 1)( z + z − 1) log( z) + v 0 and observe that

v =−

2

(1 − z( z + 1)) (1 − z) log( z( z + 1)) − log z + w  . 1 − z ( z + 1) 1−z

(6.14)

Thus we have u := 2 (1 − z( z + 1)) + 2 (1 − z) + w ∈ E, and from Equations (6.12) and (6.14), we have u  = v ∈ F . From Equations (6.12), (6.14) we see that v admits a DEL-expression over F and a special DEL-expression over the extension field F 1 = F (log(z)) respectively.

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.23 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

23

Representation of u  in terms of Bloch Wigner Spence function: Observe that

u = D (1 − z( z + 1)) −

1

1

log z log(1 − z) 2  1 − log2 ( z) + log z( z − 1)( z2 + z − 1) log( z) + v 0 . 2   1 = D (1 − z( z + 1)) + D (1 − z) − log( z) log(1 − z( z + 1)) + log(1 − z) + log( z) 2  1  2 + log( z) log z( z − 1)( z + z − 1) − log( z + 1) log(1 − z( z + 1)) + v 0 . 2

log( z( z + 1)) log(1 − z( z + 1)) + D (1 − z) −



2

Now we shall substitute Equation (6.11) in the above equation to obtain

u = D (1 − z( z + 1)) + D (1 − z) +

−

1 2

1 2





1

log( z) log z( z − 1)( z2 + z − 1) +

2

c log( z)

(6.15)

log( z + 1) log(1 − z( z + 1)) + v 0 .





Since the elements log z( z − 1)( z2 + z − 1) and log( z + 1) belong to F , the above equation provides the Bloch–Wigner–Spence function representation of u over F . Now we will show that u  does not admit a special DEL-expression over F of the form:

v = u =

 g i ri

i∈I

+ w 0 ,

gi

r i = c i

(1 − g i ) , ri , g i , w 0 ∈ F and c i ∈ C. 1 − gi

(6.16) 

(1 − g i ) Suppose that u  does admit such an expression over F . Then since r i − c i = 0, which belong to 1 − gi   2 2 the rings C ( z, log( z + 1)) [log( z( z − 1)( z + z − 1))] as well as C z, log( z( z − 1)( z + z − 1)) [log( z + 1)], we apply Proposition 2.5 and obtain that 1 − g i ∈ C( z). On the other hand, by Theorem 2.4, there are constants c i1 , c i2 ∈ C and an element di ∈ C( z) such that r i = c i1 log( z + 1) + c i2 log( z( z − 1)( z2 + z − 1)) + di .

(6.17)

Taking derivatives, we obtain

 (1 − g i ) 1 1 1 1 1 + di , = mi1 + mi2 + + + 1 − gi z+1 z z−1 z − ω1 z − ω2

where

√

(6.18)

√

ω1 = (−1 + 5)/2, ω2 = (−1 − 5)/2 and mi j = c i j /c i for j = 1, 2. For any x ∈ C(z) we write n c z − αi x = mi =1 , z j =1 − β j

where c , αi , β j ∈ C and observe that x has no poles of order 1 and that x /x is either zero (that is x ∈ C) or a sum of poles of order 1. Thus from Equation (6.18), we obtain di ∈ C, mi1 and mi2 are m integers and that 1 − g i = ai ( z + 1)mi1 z( z − 1)( z2 + z − 1) i2 for some constant ai ∈ C. We shall use Remark 2 and assume that mi2 ≥ 0. From Equations (6.12), (6.16), (6.17) and (6.18), we obtain



(1 − z( z + 1))  g i w = log( z + 1) c i1 + 1 − z ( z + 1) gi i∈I    g g z  × − + c i2 i + di i , 

z

i∈I

gi

i∈I

gi



  + log z( z − 1)( z2 + z − 1) (6.19)

where w = v 0 − w 0 . It follows that w must be a polynomial in log( z + 1) and log( z( z − 1)( z2 + z + 1)) over C( z) of total degree 2. Write

JID:YJSCO

AID:1891 /FLA

[m1G; v1.242; Prn:3/09/2018; 17:22] P.24 (1-24)

Y. Kaur, V.R. Srinivasan / Journal of Symbolic Computation ••• (••••) •••–•••

24





w = e 1 log2 ( z( z − 1)( z2 + z − 1)) + e 2 log2 ( z + 1) + e 3 log z( z − 1)( z2 + z − 1) log( z + 1)

  + α1 log z( z − 1)( z2 + z − 1) + α2 log( z + 1) + β

and substitute in Equation (6.19) and obtain e 1 , e 2 and e 3 are constants. Moreover, from algebraic independence of logarithms, we also obtain that the coefficients of log( z + 1) and log( z( z − 1)( z2 + z − 1)) must be zero. Thus

1

− + z

 i∈I

c i2

g i gi



= 2e 1

1 z

+

1 z−1

+

1 z − ω1

+

1 z − ω2



+ e3

1 z+1

+ α1 ,

where e 1 and e 3 are constants, α1 ∈ C( z). Note that α1 has no poles of order 1. Any pole of    c 0. If for some i, c i2  0 (that is, when i ∈ I i2 ( g i / g i ) is a pole of g i / g i for some i such that c i2 =  mi2 = 0) then g i / g i has no poles at 0, 1, ω1 and ω2 . Thus, i ∈ I c i2 ( g i / g i ) has no poles at 0, 1, ω1 and ω2 . Now by comparing poles of the above equation, one arrives at a contradiction. References Baddoura, J., 2006. Integration in finite terms with elementary functions and dilogarithms. J. Symb. Comput. 41 (8), 909–942. Baddoura, J., 2011. A note on symbolic integration with polylogarithms. Mediterr. J. Math. 8 (2), 229–241. Cherry, G., 1986. Integration in finite terms with special functions: the logarithmic integral. SIAM J. Comput. 15 (1), 1–21. Cherry, G., 1985. Integration in finite terms with special functions: the error function. J. Symb. Comput. 1 (3), 283–302. Kolchin, E.R., 1968. Algebraic groups and algebraic dependence. Am. J. Math. 90 (4), 1151–1164. Moses, J., 1969. The integration of a class of special functions with the Risch algorithm. ACM SIGSAM Bull. 13, 14–27. Rubel, L., Singer, M., 1988. Autonomous functions. J. Differ. Equ. 75, 354–370. Rosenlicht, M., 1968. Liouville’s theorem on functions with elementary integrals. Pac. J. Math. 24, 153–161. Rosenlicht, M., Singer, M., 1977. On elementary, generalized elementary, and liouvillian extension fields. In: Bass, H., et al. (Eds.), Contributions to Algebra. Academic Press, pp. 329–342. Singer, M., Saunders, B., Caviness, B., 1985. An extension of Liouville’s theorem on integration in finite terms. SIAM J. Comput. 14 (4), 966–990.