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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!

Américo Tavares
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Jozef
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    Hint: take the derivative of $\ln{x}\ln(1-x)$, and remember that $x\ln{x}\rightarrow{0}$ as $x\rightarrow{0}$. – bgins Feb 11 '12 at 20:31
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    Fiddling with this led me to think about $\lim\limits_{x\to0+}((\log x)(\log(1-x)))$. Maybe if I ever have a class of bright students taking calculus, I'll assign this. (...and I see "bgins" also thought of that.) – Michael Hardy Feb 11 '12 at 20:38
  • @bgins: Can you please extend your comment? I'm not sure I understand what to do. – Jozef Feb 11 '12 at 20:46
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    @Jozef: $[\ln(x)\ln(1-x)]'=\frac{\ln(1-x)}{x}+\frac{\ln{x}}{x-1}$. Integrating, the LHS is zero because of the limit mentioned. Then, we have the same situation as americo-tavares and peter below, except their routes are better because more they are direct (no pulling rabbits out of hats)! – bgins Feb 11 '12 at 22:17
  • I think it's appropriate to mention that, if one wishes to use the power series representation of $f(x)={\ln(1-x)\over x}$ on $(-1,1)$, the following theorem (which is not usually mentioned in an introductory Calculus sequence) is needed: Suppose a power series $f(x)=\sum\limits_{n=0}^\infty a_n x^n$ converges on $[a,b)$ and suppose $\sum\limits_{n=0}^\infty a_n {b^{n+1}\over n+1}$ converges. Then $f(x)$ is improperly integrable on $[a,b]$ and $\int_a^b f(x),dx= \sum\limits_{n=0}^\infty a_n\int_a^b x^n, dx$. – David Mitra Feb 11 '12 at 22:38
  • @DavidMitra: If we prove that the original integral is convergent, do we still need applying this theorem? – Américo Tavares Feb 11 '12 at 23:45
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    @AméricoTavares I'm not sure (I shouldn't have said "needed")... The fact in my previous comment was the only theorem I could find that justifies the method. I'd be interested to know the answer to your question too. – David Mitra Feb 11 '12 at 23:52
  • @AméricoTavares You could apply Lebesgue Dominated Convergence, no? – David Mitra Feb 12 '12 at 01:09
  • @DavidMitra You may be right, but honestly I really don't know. – Américo Tavares Feb 12 '12 at 13:06
  • You're missing a $\textrm{d}x$ in your integral. – JavaMan Jan 15 '13 at 05:32
  • Inverse question: how to compute $\int_0^1 \frac{x-1}{\ln x} d x$? – sdcvvc Jun 28 '13 at 13:33

6 Answers6

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

Américo Tavares
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    In case anybody reads this answer, I think it is necessary to say something like for all $u\in(0,1)$ and for all $n\in\Bbb N$, $$\left|1+\frac12u+\frac13u^2+\cdots\frac1nu^{n-1}\right|\leq-\frac{\ln(1-u)}u=g(u)$$ so that, since $g$ is integrable on $(0,1)$, one can apply dominated convergence, and swap the sum with the integral. – Olivier Bégassat Jan 19 '15 at 19:15
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    @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. – Américo Tavares Jan 19 '15 at 19:48
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$$ \int_0^1 \frac{\log x}{x-1}dx =\lambda$$

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

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

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

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

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

kiwi
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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,

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

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

Community
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Mhenni Benghorbal
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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 $$\frac{\log x}{x-1}=-\frac{\log x}{1-x}=-\log x\sum_{k=0}x^k$$

AD - Stop Putin -
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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*}

Said Attaoui
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