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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!

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    $\begingroup$ Hint: take the derivative of $\ln{x}\ln(1-x)$, and remember that $x\ln{x}\rightarrow{0}$ as $x\rightarrow{0}$. $\endgroup$ – bgins Feb 11 '12 at 20:31
  • $\begingroup$ 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.) $\endgroup$ – Michael Hardy Feb 11 '12 at 20:38
  • $\begingroup$ @bgins: Can you please extend your comment? I'm not sure I understand what to do. $\endgroup$ – Jozef Feb 11 '12 at 20:46
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    $\begingroup$ @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)! $\endgroup$ – bgins Feb 11 '12 at 22:17
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    $\begingroup$ @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. $\endgroup$ – David Mitra Feb 11 '12 at 23:52
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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 $$ \ln \left( 1-u\right) =-u-\frac{1}{2}u^{2}-\frac{1}{3}u^{3}-\ldots -\frac{ u^{n+1}}{n+1}-\ldots\qquad(\left\vert u\right\vert <1) $$

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    $\begingroup$ 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. $\endgroup$ – Olivier Bégassat Jan 19 '15 at 19:15
  • $\begingroup$ @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. $\endgroup$ – 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 $x$)

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

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

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

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    $\begingroup$ This is exactly Américo's answer (in fact all alnswers are the same :-) ) $\endgroup$ – Mariano Suárez-Álvarez Jan 15 '13 at 4:26
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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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  • $\begingroup$ Where does the result for this last integral come from? $\endgroup$ – Alex Feb 21 '13 at 21:52
  • $\begingroup$ @Alex: See the added. $\endgroup$ – Mhenni Benghorbal Feb 24 '13 at 16:06
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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 $$\frac{\log x}{x-1}=-\frac{\log x}{1-x}=-\log x\sum_{k=0}x^k$$


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