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Two integrals for Apery's constant $\zeta(3)$ are

$$\zeta(3)=\frac{16}{3} \int_0^1 \frac{x\log^2\left(x\right)}{1+x^2}dx$$

and

$$\zeta(3)=\frac{32}{7} \int_0^1 \frac{x\log^2\left(x\right)}{1-x^4}dx$$

How can series expressions be obtained from them?

Related questions:

A series to prove $\frac{22}{7}-\pi>0$

Series for $\zeta(3)-\frac{6}{5}$

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    $\begingroup$ The "Related questions" mentioned in the question are in fact unrelated. $\endgroup$
    – Did
    Nov 11, 2017 at 8:43
  • 1
    $\begingroup$ BTW, $\log^2(1/x)=\log^2 x$ $\endgroup$
    – FDP
    Nov 11, 2017 at 20:44

3 Answers 3

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From the geometric series we have $$\frac{16}{3}\int_{0}^{1}\frac{x\log^{2}\left(1/x\right)}{1+x^{2}}dx=\frac{16}{3}\sum_{k\geq0}\left(-1\right)^{k}\int_{0}^{1}x^{2k+1}\log^{2}\left(x\right)dx$$ where the exchange of the integral with the series is justified by the dominated convergenge theorem. Then integrating by parts $$\frac{16}{3}\int_{0}^{1}\frac{x\log^{2}\left(1/x\right)}{1+x^{2}}dx=\frac{4}{3}\sum_{k\geq1}\frac{\left(-1\right)^{k-1}}{k^{3}}=\frac{4}{3}\left(1-2^{-2}\right)\sum_{k\geq1}\frac{1}{k^{3}}=\color{red}{\zeta\left(3\right)}$$ where in the penultimate equality we used the relation $$\sum_{k\geq1}\frac{\left(-1\right)^{k-1}}{k^{s}}=\left(1-2^{1-s}\right)\sum_{k\geq1}\frac{1}{k^{s}},\,\mathrm{Re}\left(s\right)>1.$$ The other integral is similar.

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Just to complete Marco's excellent answer,

$$ \int_{0}^{1}\frac{x\log^2 x}{1-x^4}\,dx = \sum_{n\geq 0} \int_{0}^{1}x^{4n+1}\log^2(x)\,dx=\left.\sum_{n\geq 0}\frac{d^2}{ds^2}\int_{0}^{1}x^{4n+1+s}\,dx\right|_{s=0}$$ equals $$ \sum_{n\geq 0}\frac{2}{(4n+2)^3} = \frac{1}{4}\sum_{n\geq 0}\frac{1}{(2n+1)^3}=\frac{1}{4}\cdot\frac{7}{8}\zeta(3) = \frac{7\,\zeta(3)}{32}.$$

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A variant of Marco's answer.

$\displaystyle J=\int_0^1 \frac{x\ln^2 x}{1+x^2}\,dx$

Perform the change of variable $y=x^2$,

$\begin{align} J=\frac{1}{8}\int_0^1 \frac{\ln^2 y}{1+y}\,dy\end{align}$

Let,

$\displaystyle K=\int_0^1 \frac{\ln^2 x}{1+x}\,dx$

$\displaystyle L=\int_0^1 \frac{\ln^2 x}{1-x}\,dx$

$\begin{align} L-K=\int_0^1 \frac{2x\ln^2 x}{1-x^2}\,dx \end{align}$

Perform the change of variable $y=x^2$,

$\begin{align} L-K&=\frac{1}{4}\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &=\frac{1}{4}L \end{align}$

Therefore,

$\displaystyle K=\frac{3}{4}L$

Therefore,

$\displaystyle J=\frac{3}{32}L$

Let,

$\begin{align}M=\int_0^1 \frac{x\ln^2 x}{1-x^4}\,dx\\ \end{align}$

Perform the change of variable $y=x^2$,

$\begin{align}M&=\frac{1}{8}\int_0^1 \frac{\ln^2 x}{1-x^2}\,dx\\ &=\frac{1}{16}\int_0^1 \frac{\ln^2 x}{1-x}\,dx+\frac{1}{16}\int_0^1 \frac{\ln^2 x}{1+x}\,dx\\ &=\frac{1}{16}L+\frac{1}{16}K\\ &=\frac{7}{64}L \end{align}$

$\begin{align}L&=\int_0^1 \frac{\ln^2 x}{1-x}\,dx\\ &=\int_0^1\left(\sum_{n=0}^{\infty} x^n\ln^2 x\right)\,dx\\ &=\sum_{n=0}^{\infty} \left(\int_0^1 x^n\ln^2 x\,dx\right)\\ &=2\sum_{n=0}^{\infty} \frac{1}{(n+1)^3}\\ &=2\zeta(3) \end{align}$

(the exchange of the integral and the series is justified by Fubini-Tonelli theorem)

Therefore,

$\displaystyle J=\frac{3}{16}\zeta(3)$

$\displaystyle M=\frac{7}{32}\zeta(3)$

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