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$$\int_{0}^{1} \frac{\log x}{x-1}dx$$

I was wondering: is it possible to evaluate this integral with real methods? Playing around with a series expansion I was able to recognize that the integral is equal to $\zeta(2)$, but since I don't know how to evaluate that without the parseval identity (I haven't really studied it yet, so it would be cheating :P) that road isn't feasible. I was thinking maybe of some tricks using differentiation under the integral sign or the method used to evaluate $\displaystyle\int_{0}^{\infty}\frac{\sin x}{x}$ (recognize that $\displaystyle\int_{0}^{a}\frac{\sin x}{x}dx=\int_{0}^{a}dx\int_{0}^{\infty}e^{-xy}\sin x\ dy$ and use the Fubini-Tonelli Theorem) but I couldn't get anywhere with those.

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    $\begingroup$ $\frac{\log x}{x-1} = - \sum_{k = 0}^\infty x^k \log x$ for $0 < x < 1$. These terms are easily integrated. The result is the series for $\zeta(2)$, as you already know. This avoids the explicit use of Parseval's identity. But it does not give a direct proof. $\endgroup$ May 30, 2014 at 13:26
  • $\begingroup$ Yeah. Mostly, I would like a method to evaluate the integral directly, but if the road "evaluate $\zeta(2)$" is easier it's ok, just as long as it doesn't use Fourier analysis. It mostly is just a curiosity. :P $\endgroup$ May 30, 2014 at 13:36
  • $\begingroup$ To be clear: Are you looking for a way to show that $I = \frac{\pi^2}{6}$ without using that $\zeta(2) = \sum 1/n^2 = \frac{\pi^2}{6}$ ? $\endgroup$
    – Winther
    May 30, 2014 at 13:36
  • $\begingroup$ Either that or a method to evaluate $\zeta (2)$ not using Fourier analysis. $\endgroup$ May 30, 2014 at 13:38
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    $\begingroup$ OK. You can find several proofs here. See "A rigorous elementary proof" $\endgroup$
    – Winther
    May 30, 2014 at 13:39

2 Answers 2

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Note that $$ \int_z^1\frac{\ln x}{1-x}\ dx=-\text{Li}_2(1-z)=-\sum_{n=1}^\infty\frac{(1-z)^n}{n^2}. $$ In this case, we have $z=0$ and $$ \int_0^1\frac{\ln x}{1-x}\ dx=-\sum_{n=1}^\infty\frac{1}{n^2}=-\zeta(2)=\large\color{blue}{-\frac{\pi^2}{6}}. $$


Another way to evaluate the integral is using substitution $u=1-x$ and the integral turns out to be $$ \int_0^1\frac{\ln (1-u)}{u}\ du.\tag1 $$ Use Maclaurin series for Natural logarithm: $$ \ln(1-u)=-\sum_{n=1}^\infty \frac{u^{n}}{n}.\tag2\\ $$ Substituting $\,(2)$ to $\,(1)$, yields $$ \begin{align} \int_0^1\frac{\ln(1-u)}{u}\,du&=-\int_0^1\sum_{n=1}^\infty \frac{u^{n}}{un}\,du\\ &=-\sum_{n=1}^\infty\int_0^1 \frac{u^{n-1}}{n}\,du\\ &=-\sum_{n=1}^\infty \left.\frac{u^{n}}{n^2}\right|_{u=0}^1\\ &=-\sum_{n=1}^\infty \frac{1}{n^2}.\tag3 \end{align} $$ The infinite series in $(3)$ is defined as Riemann zeta function $\,\zeta (2)=\large\dfrac{\pi^2}{6}$.

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Another way: it's actually easier to expand the denominator: $-\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k$ and since the bounds on the integral are $0$ and $1$, $x^k$ converges uniformly, and you can interchange summation and integration. You get an integral of the form $$ I_k = - \int_{0}^{1}x^k \log x dx $$ which are easily solved by parts, and then sum over $k$.

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