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When I was playing with Wolfram Alpha about the integral $$\int_0^1\frac{x^{s-1}}{1+x^2}dx$$ and its derivatives, since I know the relationship between the Apéry's constant and particular values of the polygamma function, and since I presume that this way will be known, I found playing with the code a closed-form for this $$\int_0^1\frac{\log^2(x)}{1+x^3}dx$$ see this code

integrate 1/(1+x^3)(log^2(x))dx, from x=0 to x=1

in the online calculator. I am not able to get easily the calculations for $$\int\frac{\log^2(x)}{1+x^3}dx$$ and after evaluate it as a definite integral. And you?

Question. This can be a good integral for this friday. Can you prove the closed-form for $$\int_0^1\frac{\log^2(x)}{1+x^3}dx?$$ Thanks in advance.

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  • $\begingroup$ Have you tried reducing this to a series? $\endgroup$
    – Aditya Narayan Sharma
    Mar 10, 2017 at 10:50
  • $\begingroup$ Many thanks for your attention @AdityaNarayanSharma I did the calculation with the online calculator for the indefinite integral, and I've seen that seems very difficutl to me. $\endgroup$
    – user243301
    Mar 10, 2017 at 11:04

2 Answers 2

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Claim: $$ \int_{0}^{1}\frac{\log^2(x)}{1+x^3}\,dx = \color{red}{\frac{5\pi^3}{81\sqrt{3}}+\frac{13\zeta(3)}{18}}.\tag{0}$$

Proof: since $$ \int_{0}^{1}\frac{x^s}{1+x^3}\,dx = \frac{1}{6}\left[\psi\left(\frac{s+4}{6}\right)-\psi\left(\frac{s+1}{6}\right)\right]\tag{1} $$ the answer just depends on $\psi''\left(\frac{2}{3}\right)-\psi''\left(\frac{1}{6}\right)$. In particular, $(0)$ follows by applying the differential operator $\frac{d^3}{dz^3}\log(\cdot)$ to the reflection formula and the duplication formula for the $\Gamma$ function, then evaluating at $z=\frac{1}{3}$. This appears to be a further generalization of the result shown here about the values of the $\eta$ function at odd positive integers.

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  • $\begingroup$ how do you get $\zeta(3)$ from $\sum_{k=0}^{\infty} \frac{(-1)^k}{(3 k+1)^3}$ (Ron Gordon's answer) ? $\endgroup$
    – reuns
    Mar 10, 2017 at 14:24
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    $\begingroup$ @user1952009: it is not trivial to get $\zeta(3)$ from there, I got it through $-\frac{1}{2}\psi''(1)$, but if you want to continue Ron's approach, it is probably best to ask him. In order to derive $\zeta(3)$ from such series, I would write it as a linear combination of Dirichlet $L$-functions at $s=3$, but I did not take such path. $\endgroup$
    – Jack D'Aurizio
    Mar 10, 2017 at 14:45
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    $\begingroup$ The reflection and duplication formula follows from $\psi(z)= C+\sum_{n \ge 0} (\frac{1}{z+n}-\frac{1}{n})$ so I don't see how you'll get something non-trivial as (0) from it. Hence there should be a "more trivial" derivation of (0) $\endgroup$
    – reuns
    Mar 10, 2017 at 15:09
  • $\begingroup$ I hope understand better your answer, and Gordon's answer in next hours many thanks for your attention. And many thanks also/always to @user1952009 $\endgroup$
    – user243301
    Mar 10, 2017 at 19:35
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    $\begingroup$ What I can get is a closed-form for $\sum_{n=0}^\infty (-1)^n(\frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3})$ $\quad$ ( $f(z) = \frac{\pi}{\sin(\pi z)}= \sum_{n=-\infty}^\infty \frac{(-1)^n}{z+n}$,$ f''(z) = 2\sum_{n=-\infty}^\infty \frac{(-1)^n}{(z+n)^3}$,$ f''(1/3)=2 \sum_{n=-\infty}^\infty \frac{(-1)^n}{(n+1/3)^3} = 54\sum_{n=0}^\infty (-1)^n(\frac{1}{(3n+1)^3}-\frac{1}{(3n+2)^3})$). $\endgroup$
    – reuns
    Mar 11, 2017 at 8:43
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By expanding the denominator and performing the integral

$$\int_0^1 dx \, x^{3 k} \log^2{x} $$

on each term in the sum, we find that the integral is equal to

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{(3 k+1)^3} $$

This is as far as I will take this.

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  • $\begingroup$ Many thanks for your calculations. The code that I've provided integrate 1/(1+x^3)(log^2(x))dx, from x=0 to x=1 tell me using the online calculator, that is expressible in terms of $\zeta(3)$. Then I understand your calculations and claim. $\endgroup$
    – user243301
    Mar 10, 2017 at 13:25
  • $\begingroup$ @JackD'Aurizio: I made no false statement. (I wasn't thinking that.) You don't need to be sorry. You are right. $\endgroup$
    – Ron Gordon
    Mar 10, 2017 at 14:08
  • $\begingroup$ @JackD'Aurizio $\int_0^1 x^{3 k} \log^2{x}dx = -\int_0^1 \frac{x^{3 k+1}}{3k+1} \frac{2\log{x}}{x}dx=\int_0^1 dx \frac{x^{3 k+1}}{(3k+1)^2} \frac{2}{x}dx=\frac{2}{(3k+1)^3}$ so that $\int_0^1\frac{\log^2(x)}{1+x^3}dx = 2\sum_{k=0}^\infty \frac{(-1)^k}{(3k+1)^3}$ $\endgroup$
    – reuns
    Mar 10, 2017 at 14:28
  • $\begingroup$ @user1952009: I understand that, I was just remarking that such series can be expressed in terms of $\pi^3$ and $\zeta(3)$. $\endgroup$
    – Jack D'Aurizio
    Mar 10, 2017 at 14:46
  • $\begingroup$ Many thanks everybody, you are the bests, Jack, Gordon and user1952009 this was a good friday. $\endgroup$
    – user243301
    Mar 10, 2017 at 19:26

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