Polylogs in Integrals

Calculus Level 5

∫ 0 1 Li 3 ( x ) ln ⁡ ( 1 − x ) x d x \large \int_0^1 \dfrac{ \text{Li}_3(x) \ln(1-x)} x \, dx

The value of the integral above can be expressed in the form of

A B π C ζ ( D ) − E F π G ζ ( H ) , \dfrac AB \pi^C \zeta (D) - \dfrac EF \pi^G \zeta (H) \; ,

where A , B , C , D , E , F , G , H A,B,C,D,E,F,G,H are non-negative integers, where gcd ⁡ ( A , B ) = gcd ⁡ ( E , F ) = 1 \gcd(A,B) = \gcd(E,F) = 1 .

Find A + B + C + D + E + F + G + H A+B+C+D+E+F+G+H .

Notations :

  • Li n ( a ) { \text{Li} }_{ n }(a) denotes the polylogarithm function, Li n ( a ) = ∑ k = 1 ∞ a k k n . { \text{Li} }_{ n }(a)=\displaystyle\sum _{ k=1 }^{ \infty }{ \frac { { a }^{ k } }{ { k }^{ n } } }.

  • ζ ( ⋅ ) \zeta(\cdot) denotes the Riemann zeta function .


The answer is 21.

Pi Han Goh by Pi Han Goh , 30, Malaysia

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2 solutions

Aditya Kumar
Mar 16, 2016

I = ∫ 0 1 L i 3 ( x ) ln ⁡ ( 1 − x ) x d x I=\int _{ 0 }^{ 1 }{ \frac { { Li }_{ 3 }\left( x \right) \ln { \left( 1-x \right) } }{ x } dx }

Now, on using the definition of polylogarithms, we can write it as: I = ∑ k = 1 ∞ [ 1 k 3 ∫ 0 1 x k − 1 ln ⁡ ( 1 − x ) d x ] I=\sum _{ k=1 }^{ \infty }{ \left[ \frac { 1 }{ { k }^{ 3 } } \int _{ 0 }^{ 1 }{ { x }^{ k-1 }\ln { \left( 1-x \right) } dx } \right] }

Now, I'll use the following identity (readers can prove it by repeated use of IBP): − ∫ x k − 1 ln ⁡ ( 1 − x ) d x = 1 − x k k ln ⁡ ( 1 − x ) + 1 k ∑ j = 1 k x j j -\int { { x }^{ k-1 }\ln { \left( 1-x \right) } dx } =\frac { 1-{ x }^{ k } }{ k } \ln { \left( 1-x \right) } +\frac { 1 }{ k } \sum _{ j=1 }^{ k }{ \frac { { x }^{ j } }{ j } }

Now on plugging it into the main equation and on applying the limits, we get: I = − ∑ k = 1 ∞ ∑ j = 1 k 1 k 4 j I=-\sum _{ k=1 }^{ \infty }{ \sum _{ j=1 }^{ k }{ \frac { 1 }{ { k }^{ 4 }j } } }

Now, on using the definition of harmonic numbers, I'll re-write it as: I = − ∑ k = 1 ∞ H k k 4 I=-\sum _{ k=1 }^{ \infty }{ \frac { { H }_{ k } }{ { k }^{ 4 } } }

We can evaluate this using Generalised Harmonic Sum which I learnt from here .

So the final answer comes out to be: I = π 2 6 ζ ( 3 ) − 3 ζ ( 5 ) I=\frac { { \pi }^{ 2 } }{ 6 } \zeta \left( 3 \right) -3\zeta \left( 5 \right) \\

In this question, A = 1 , B = 6 , C = 2 , D = 3 , E = 3 , F = 1 , G = 0 , H = 5 A=1, B=6, C=2, D=3, E=3,F=1,G=0,H=5 .

Therefore, A + B + C + D + E + F + G + H = 21 \boxed{A+B+C+D+E+F+G+H=21}

2 Helpful 0 Interesting 2 Brilliant 0 Confused

Simply do it like this...

I = ∫ 0 1 ∑ n = 1 ∞ f r a c 1 n 3 ∫ 0 1 log ⁡ ( 1 − x ) x n − 1 d x \ = lim ⁡ p 0 \del \del ( p ) ∫ 0 1 ( ( 1 − x ) p ) x n − 1 ∑ n = 1 ( inf ⁡ ) 1 n 3 d x \ = s u m n = 1 ( inf ⁡ ) 1 n 3 lim ⁡ p 0 \del \del ( p ) γ ( p + 1 ) × γ ( n ) γ ( p + n + 1 ) I = \int_{0}^1 \sum_{n=1}^{\infty} frac{1}{n^3} \int_{0}^1 \log(1-x) x^{n-1} dx \ = \lim_{p^0} \frac{\del}{\del (p)} \int_{0}^1 ((1-x)^p) x^{n-1} \sum_{n=1}^(\inf) \frac{1}{n^3} dx \ = \ sum_{n=1}^(\inf) \frac{1}{n^3} \lim_{p^0} \frac{\del}{\del (p)} \frac{\gamma(p+1) \times \gamma(n)}{\gamma(p+n+1)}

We get , after some simple simplification , = ∑ n = 1 ( inf ⁡ ) − d i g a m m a ( n + 1 ) + \eulergamma n = − ∑ n = 1 ( inf ⁡ ) 1 n 3 H n n = \sum_{n=1}^(\inf) \frac{ - digamma(n+1) + \eulergamma}{n} = - \sum_{n=1}^(\inf) \frac{1}{n^3} \frac{H_n}{n}

As we know ,

\H_n = \digamma(n+1) + \eulergamma

Using this , we get ,

= − ∑ n = 1 ( inf ⁡ ) H n n 4 = - \sum_{n=1}^(\inf) \frac{H_n}{n^4}

As we know , 

  \\\[  \\sum\_\{n=1\}^\(\\inf\) \\frac\{H\_n\}\{n^4\} =  3 \\times \\zeta\(5\)  \-  \\zeta\(2\) \\times \\zeta\(3\)\\\]

Using this , we get ,

I = ζ ( 2 ) × ζ ( 3 ) − 3 × ζ ( 5 ) I = \zeta(2) \times \zeta(3) - 3 \times \zeta(5)

So , we can directly compare given question and answer , we finally get ,

A + B + C + D + E + F + G + H = 21 

                                      (Q.E.D)

Shivam Sharma - 4 years, 2 months ago
Naren Bhandari
Oct 23, 2020

∫ 0 1 Li ⁡ 3 ( x ) ln ⁡ ( 1 − x ) x d x = ∑ k ≥ 1 1 k 3 ∫ 0 1 x k − 1 ln ⁡ ( 1 − x ) d x = − ∑ k ≥ 1 H k k 4 \int_0^1\frac{\operatorname{Li}_3(x)\ln(1-x)}{x}dx=\sum_{k\geq 1}\frac{1}{k^3}\color{#D61F06}{\int_0^1 x^{k-1}\ln(1-x)dx}=-\sum_{k\geq 1}\frac{H_k}{k^4} the integral highlighted with red which I solved here in MSE without derivatives of beta function. To evaluate the last sum we use generalized Euler sum (equation 21) giving us − ∑ k ≥ 1 H k k 4 = π 2 6 ζ ( 3 ) − 3 ζ ( 5 ) -\sum_{k\geq 1 }\frac{H_k}{k^4}= \frac{\pi^2}{6}\zeta(3)-3\zeta(5)

1 Helpful 1 Interesting 1 Brilliant 0 Confused

Ah I see that you're active on MSE too. Hi there!

Pi Han Goh - 7 months, 3 weeks ago

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I also have seen some of your postings on MSE however, you seems less active there. A friend of mine , Ali Shather ( most Harmonic problem poster on MSE) suggested me to participate on MSE. :D Are you on facebook sir? I guess you dont mind. :D

Naren Bhandari - 7 months, 3 weeks ago

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Yeah, I don't particularly fancy MSE as there's plenty of moderators silencing users for not following a strict arbitrary rule.

But I do make comments here and there, which as you know, does not "score me any points".

No, I don't use Facebook.

Pi Han Goh - 7 months, 3 weeks ago

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@Pi Han Goh – I see. Thank you for the reply. :)

Naren Bhandari - 7 months, 3 weeks ago

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