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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!

  • 0
    [Inverse question](http://math.stackexchange.com/questions/100495): how to compute $\int_0^1 \frac{x-1}{\ln x} d x$?2013-06-28

5 Answers 5

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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) $

  • 0
    @Olivier Bégassat Thanks for your comment. I've just assumed, without justification, that the given integral could be evaluated by expanding the integrand in a series and integrating term by term.2015-01-19
12

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

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

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

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

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

  • 7
    This is exactly Américo's a$n$swer (in fact all alnswers are the same :-) )2013-01-15
11

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

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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}}$

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

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    @Alex: See the added.2013-02-24
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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 $\frac{\log x}{x-1}=-\frac{\log x}{1-x}=-\log x\sum_{k=0}x^k$