SCIENTIA Series A: Mathematical Sciences, Vol. 18 (2009), 77-84 Universidad T´ecnica Federico Santa Mar´ıa Valpara´ıso, Chile ISSN 0716-8446 c Universidad T´ecnica Federico Santa Mar´ıa 2009 The integrals in Gradshteyn and Ryzhik. Part 12: Some logarithmic integrals Victor H. Moll and Ronald A. Posey Abstract. We present the evaluation of some logarithmic integrals. The inte- grand contains a rational function with complex poles. The methods are illus- trated with examples found in the classical table of integrals by I. S. Gradshteyn and I. M. Ryzhik. 1. Introduction The classical table of integrals by I. Gradshteyn and I. M. Ryzhik [3] contains many entries from the family 1 (1.1) R(x) log x dx Z0 where R is a rational function. For instance, the elementary integral 4.231.1 1 log x dx π2 (1.2) = , 1+ x − 12 Z0 is evaluated simply by expanding the integrand in a power series. In [1], the first author and collaborators have presented a systematic study of integrals of the form b log t dt (1.3) h (b)= , n,1 (1 + t)n+1 Z0 as well as the case in which the integrand has a single purely imaginary pole b log t dt (1.4) h (a,b)= . n,2 (t2 + a2)n+1 Z0 arXiv:1004.2437v1 [math.CA] 14 Apr 2010 The work presented here deals with integrals where the rational part of the inte- grand is allowed to have arbitrary complex poles. 2000 Mathematics Subject Classification. Primary 33. Key words and phrases. Logarithmic integrals, Clausen functions, Chebyshev polynomials. The first author wishes to thank the partial support of nsf-dms 0713836. 1 2 VICTORH.MOLLANDRONALDA.POSEY 2. Evaluations in terms of polylogarithms In this section we describe the evaluation of 1 log x dx (2.1) g(a)= , x2 2ax +1 Z0 − under the assumption that the denominator has non-real roots, that is, a2 < 1. The first approach to the evaluation of g(a) is based on the factorization of the quartic as (2.2) x2 2ax +1=(x + r )(x + r ), − 1 2 where r = a + i√1 a2 and r = a i√1 a2. The partial fraction expansion 1 − − 2 − − − 1 1 1 1 (2.3) = , (x + r )(x + r ) r r x + r − x + r 1 2 2 − 1 1 2 yields 1 1 log x dx 1 1 log x dx (2.4) g(a)= . r r x + r − r r x + r 2 − 1 Z0 1 2 − 1 Z0 2 These integrals are computed in terms of the dilogarithm function defined by x log(1 t) (2.5) PolyLog[2, x] := − dt. − t Z0 A direct calculaton shows that log x dx (2.6) = log x log(1 + x/a) + PolyLog[2, x/a], x + a − Z and thus 1 log x dx 1 (2.7) = PolyLog 2, . x + a −a Z0 It follows that 1 1 1 (2.8) g(a)= PolyLog 2, PolyLog 2, . r r −r − −r 2 − 1 1 2 Observe that the real integral g(a) appears here expressed in terms of the poly- logarithm of complex arguments. Example 2.1. The case a =1/2 yields 1 log x dx i √ √ (2.9) 2 = PolyLog 2, (1 + i 3)/2 PolyLog 2, (1 i 3)/2 . 0 x x +1 √3 − − Z − h i h i The polylogarithm function is evaluated using the representation (2.10) (1 + i√3)/2= eiπ/3, LOGARITHMIC INTEGRALS 3 to produce 1 k ∞ (1 + i√3) ∞ eiπk/3 PolyLog 2, (1 + i√3)/2 = 2 = k2 k2 k=1 k=1 h i X X πk πk ∞ cos + i sin = 3 3 . k2 k=1 X Similarly πk πk ∞ cos i sin PolyLog 2, (1 i√3)/2 = 3 − 3 , − k2 k=1 h i X and it follows that 1 log x dx i = PolyLog 2, (1 + i√3)/2 PolyLog 2, (1 i√3)/2 x2 x +1 √3 − − Z0 − hπk i h i 2 ∞ sin = 3 . −√ k2 3 k=1 X The function sin(πk/3) is periodic, with period 6, and repeating values √3 , √3 , 0, √3 , √3 , 0. 2 2 − 2 − 2 Therefore πk ∞ sin √3 ∞ 1 ∞ 1 ∞ 1 ∞ 1 3 = + . k2 2 (6k + 1)2 (6k + 2)2 − (6k + 4)2 − (6k + 5)2 k=1 k=0 k=0 k=0 k=0 ! X X X X X To evaluate these sums, recall the series representatin of the polygamma function ψ(x)=Γ′(x)/Γ(x), given by 1 ∞ x (2.11) ψ(x)= γ + . − − x k(x + k) kX=1 Differentiation yields ∞ 1 (2.12) ψ′(x)= , − (x + k)2 Xk=0 and we obtain ∞ 1 1 ∞ 1 = . (6k + j)2 36 j 2 (k + 6 ) Xk=0 kX=0 This provides the expression πk ∞ sin 3 √3 1 2 4 5 (2.13) = ψ′ + ψ′ ψ′ ψ′ . k2 72 6 6 − 6 − 6 k=1 X 4 VICTORH.MOLLANDRONALDA.POSEY The integral (2.9) is 1 log x dx 1 1 2 4 5 (2.14) = ψ′ + ψ′ ψ′ ψ′ . x2 x +1 −36 6 6 − 6 − 6 Z0 − The identities (2.15) ψ(1 x)= ψ(x)+ π cot πx, − and 1 (2.16) ψ(2x)= ψ(x)+ ψ(x + 1 ) + log 2, 2 2 produce 2 2 2 1 1 4π 2 1 4π 5 1 16π ψ′ =5ψ′ , ψ′ = ψ′ + , ψ′ = 5ψ′ + . 6 3 − 3 3 − 3 3 6 − 3 3 Replacing in (2.14) yields 1 log x dx 2π2 1 1 (2.17) = ψ′ . x2 x +1 9 − 3 3 Z0 − This appears as formula 4.233.2 in [3]. Note. The method described in the previous example evaluates logarithmic integrals in terms of the Clausen function ∞ sin kx (2.18) Cl (x) := . 2 k2 kX=1 Note. An identical procedure can be used to evaluate the integrals 4.233.1, 4.233.3, 4.233.4 in [3] given by 1 log x dx 4π2 2 1 (2.19) = ψ′ , x2 + x +1 27 − 9 3 Z0 1 x log x dx 7π2 1 1 (2.20) = + ψ′ , x2 + x +1 − 54 9 3 Z0 and 1 x log x dx 5π2 1 1 (2.21) = ψ′ , x2 x +1 36 − 6 3 Z0 − respectively. LOGARITHMIC INTEGRALS 5 3. An alternative approach In this section we present an alternative evaluation for the integral 1 log x dx (3.1) g(a)= , x2 2ax +1 Z0 − based on the observation that d 1 xs dx (3.2) g(a) = lim . s 0 ds x2 2ax +1 → Z0 − The proof discussed here is based on the Chebyshev polynomials of the second kind Un(a), defined by sin[(n + 1)t] (3.3) U (a)= , n sin t where a = cos t. The relation with the problem at hand comes from the generating function 1 ∞ (3.4) = U (a)xk. 1 2ax + x2 k − Xk=0 This appears as 8.945.2 in [3]. Observe that 1 s 1 x dx ∞ k+s ∞ Uk(a) 2 = Uk(a) x dx = . 0 x 2ax +1 0 k + s +1 Z − kX=0 Z Xk=0 It follows that 1 log x dx ∞ Uk(a) (3.5) 2 = 2 . 0 x 2ax +1 − (k + 1) Z − Xk=0 Replacing the trigonometric expression (3.3) for the Chebyshev polynomial, it follows that 1 log x dx 1 ∞ sin kt Cl2(t) (3.6) 2 = 2 = . 0 x 2ax +1 −sin t k − sin t Z − kX=0 This reproduces the representation discussed in Section 2. Note. The methods presented here give the value of (3.1) in terms of the dilogarithm function. The classical values ∞ ( 1)k (3.7) Cl π = Cl 3π = − = Catalan, 2 2 − 2 2 (2k + 1)2 k=0 X are easy to establish. More sophisticated evaluations appear in [5]. These are given in terms of the Hurwitz zeta function ∞ 1 (3.8) ζ(s, q)= . (k + q)s Xk=0 6 VICTORH.MOLLANDRONALDA.POSEY For instance, the reader will find s 2π 3− 1 s 1 (3.9) Cl = √3 − ζ(2)+3− ζ(2, ) , 2 3 2 3 and s π 3− 1 s 1 1 (3.10) Cl = √3 − ζ(2)+6− ζ(2, )+ ζ(2, ) . 2 3 2 6 3 Note. Integrals of the form 1 1 (3.11) R(x) log log dx x Z0 present new challeges. The reader will find some examples in [4]. The current version of Mathematica is able to evaluate 1 x log log 1/x π (3.12) dx = 6 log 2 3 log 3 + 8 log π 12 log Γ( 1 ) , x4 + x2 +1 12√3 − − 3 Z0 but is unable to evaluate 1 x log log 1/x π (3.13) dx = 7 log π 4 log sin π 8 log Γ( 1 ) . x4 √2x2 +1 8√2 − 8 − 8 Z0 − 4. Higher powers of logarithms The method of the previous sections can be used to evaluate integrals of the form 1 (4.1) R(x) logp x dx, Z0 where R is a rational function. The ideas are illustrated with the verification of formula 4.261.8 in [3]: 1 1 x 8√3π3 + 351ζ(3) (4.2) − log2 x dx = . 1 x6 486 Z0 − Define 1 log2 x dx 1 log2 x dx J = , J = , 1 1+ x 2 1 x + x2 Z0 Z0 − 1 x log2 x dx 1 log2 x dx J = , J = . 3 1 x + x2 4 1+ x + x2 Z0 − Z0 The partial fraction decomposition 1 x 1 1 1 1 1 x 1 1 − = + + , 1 x6 3 1+ x 6 1 x + x2 − 3 1 x + x2 2 1+ x + x2 − − − gives 1 1 x 1 1 1 1 (4.3) − log2 x dx = J + J J + J .