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

Question

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!

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

Answer 2

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

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

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

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

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

Answer 3

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

Answer 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,

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

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

Answer 5

Hint: Use a geometric sum and a partial integration $$\int_0^1x^n\log x \,dx=\frac{x^{n+1}}{n+1}\log x \bigg|_0^1-\int_0^1\frac{x^{n}}{n+1}$$

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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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Answer 6

Using the expansion series $\displaystyle\frac 1{1-x}=\displaystyle\sum_{n=0}^\infty x^n,\, |x|<1$, we get \begin{eqnarray*} \int_0^1\frac{x^a}{x-1}\, dx&=& -\int_0^1 x^a\left(\sum_{n=0}^\infty x^{n}\, dx\right)\\ &=& -\sum_{n=0}^\infty\left(\int_0^1 x^{n+a}\, dx\right)\\ &=& -\sum_{n=0}^\infty \frac 1{(n+a+1)}\\ &=& -\sum_{n=1}^\infty \frac 1{(n+a)} \end{eqnarray*} Differentiate wrt $a$, we obtain \begin{eqnarray*} \int_0^1\frac{x^a\ln(x)}{x-1}\, dx &=& \sum_{n=1}^\infty \frac 1{(n+a)^2} \end{eqnarray*} Fix $a=0$ in both sides, we have \begin{eqnarray*} \int_0^1\frac{\ln(x)}{x-1}\, dx &=& \sum_{n=1}^\infty \frac 1{n^2}\\ &=&\frac{\pi^2}6. \end{eqnarray*}