I'd like your help with proving that $$\int_0^1 \frac{\ln x }{x-1}d x=\sum_{n=1}^\infty \frac{1}{n^2}.$$ I tried to use Fourier series, or to use a power series and integrate it twice but it didn't work out for me.
Any suggestions?
Thanks!
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$\begingroup$Hint: use the substitution $u=1-x$ to obtain $$ I:=\int_{0}^{1}\frac{\ln x}{x-1}dx=-\int_{0}^{1}\frac{\ln \left( 1-u\right) }{u}\,du $$
and the following Maclaurin series$$ \ln \left( 1-u\right) =-u-\frac{1}{2}u^{2}-\frac{1}{3}u^{3}-\ldots -\frac{ u^{n+1}}{n+1}-\ldots\qquad(\left\vert u\right\vert <1) $$
$\endgroup$ 2 $\begingroup$$$ \int_0^1 \frac{\log x}{x-1}dx =\lambda$$
Making $x = 1-u$ produces (keep the $x$)
$$-\int_0^1 \frac{\log (1-x)}{x}dx=\lambda$$
$$\frac{\log (1-x)}{x}=-\sum_{n=1}^{\infty} \frac{x^{n-1}}{n}$$
$$-\int_0^1 \frac{\log (1-x)}{x}dx =\left.\sum_{n=1}^{\infty} \frac{x^{n}}{n^2} \right|_0^1 =\sum_{n=1}^{\infty} \frac{1}{n^2}$$
$\endgroup$ 1 $\begingroup$Write $\ln x = \ln(1 + (x-1))$ and use the log series
$\endgroup$ $\begingroup$Related problem: I, II. Using the change of variables $u=-\ln(x)$ and the identity
$$ \int_{0}^{\infty}\frac{u^{s-1}}{e^u -1}=\zeta{(s)}\Gamma{(s)} $$
we reach to the deisred result
$$ \int_0^1 \frac{\ln x }{x-1}= \int_{0}^{\infty}\frac{u}{e^u -1}=\zeta{(2)}\Gamma{(2)} =\sum_{n=1}^\infty \frac{1}{n^2}. $$
Added: Note that,
$\endgroup$ 2 $\begingroup$$$ \int_{0}^{\infty}\frac{u^{s-1}}{e^u - 1}=\int_{0}^{\infty}\frac{u^{s-1}}{e^u}(1-e^{-u})^{-1}= \sum_{n=0}^{\infty} \int_{0}^{\infty}{u^{s-1}e^{-(n+1)u}}$$
$$= \sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \int_{0}^{\infty}{y^{s-1}e^{-y}}= \sum_{n=1}^{\infty}\frac{1}{n^s} \Gamma(s)= \zeta(s) \Gamma(s).$$
Hint: Use a geometric sum and a partial integration $$\int_0^1x^n\log x \,dx=\frac{x^{n+1}}{n+1}\log x \bigg|_0^1-\int_0^1\frac{x^{n}}{n+1}$$
Edit: The first step is $$\frac{\log x}{x-1}=-\frac{\log x}{1-x}=-\log x\sum_{k=0}x^k$$
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