Prove: $\int_0^1 \frac{\ln x }{x-1} d x=\sum_1^\infty \frac{1}{n^2}$

Solution 1:

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 $$ \ln \left( 1-u\right) =-u-\frac{1}{2}u^{2}-\frac{1}{3}u^{3}-\ldots -\frac{ u^{n+1}}{n+1}-\ldots\qquad(\left\vert u\right\vert <1) $$

Solution 2:

$$ \int_0^1 \frac{\log x}{x-1}dx =\lambda$$

Making $x = 1-u$ produces (keep the $x$)

$$-\int_0^1 \frac{\log (1-x)}{x}dx=\lambda$$

$$\frac{\log (1-x)}{x}=-\sum_{n=1}^{\infty} \frac{x^{n-1}}{n}$$

$$-\int_0^1 \frac{\log (1-x)}{x}dx =\left.\sum_{n=1}^{\infty} \frac{x^{n}}{n^2} \right|_0^1 =\sum_{n=1}^{\infty} \frac{1}{n^2}$$

Solution 3:

Write $\ln x = \ln(1 + (x-1))$ and use the log series

Solution 4:

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,

$$ \int_{0}^{\infty}\frac{u^{s-1}}{e^u - 1}=\int_{0}^{\infty}\frac{u^{s-1}}{e^u}(1-e^{-u})^{-1}= \sum_{n=0}^{\infty} \int_{0}^{\infty}{u^{s-1}e^{-(n+1)u}}$$

$$= \sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \int_{0}^{\infty}{y^{s-1}e^{-y}}= \sum_{n=1}^{\infty}\frac{1}{n^s} \Gamma(s)= \zeta(s) \Gamma(s).$$