How can I prove that $$\int_{0}^{\infty }\frac{\log(1+x)}{x(1+x)}dx=\sum_{n=1}^{\infty }\frac{1}{n^2}$$


closed as off-topic by Jeel Shah, Jack D'Aurizio, Ali Caglayan, user2345215, user26857 Jan 6 '15 at 0:14

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The substitution $u = \log(1+x)$ gives

$$I = \int_0^\infty \frac{\log(1+x)}{x(x+1)} dx = \int_0^\infty \frac{udu}{e^u-1}$$

Now by using $\frac{1}{1-x} = 1+x+x^2+\ldots$ with $x=e^{-u}$ we get

$$I = \sum_{k=0}^\infty\int_0^\infty ue^{-(k+1)u}du$$

and by integration by parts it finally follows that

$$I = \sum_{k=0}^\infty\int_0^\infty \frac{1}{k+1}e^{-(k+1)u}du = \sum_{k=0}^\infty\frac{1}{(k+1)^2}$$

Your integral can be generalized to

$$I(s) = \int_0^\infty \frac{\log^{s-1}(1+x)}{x(x+1)} dx = \int_0^\infty \frac{u^{s-1}du}{e^u-1} = \zeta(s)\Gamma(s)$$

where $\zeta(s)$ is the Riemann zeta function and $\Gamma(s)$ is the gamma function.

  • $\begingroup$ I should note that switching the the order of integration and summation (as I did above) is not always justified, but it is allowed here. A justification for this can be found in this answer. $\endgroup$ – Winther Jan 5 '15 at 22:35
  • $\begingroup$ one thing: to use the Binomial expansion we need $|x|<1$, the inequality is strict, while $e^u =1$ when $x=0$. Why is this not an issue? $\endgroup$ – Alex Jan 6 '15 at 0:11
  • $\begingroup$ @Alex Because the integrand is multiplied by $u$ which cancels the divergence at $u=0$. Another approach here that avoids this issue is to split the integral into two parts $[0,\epsilon]$ plus $[\epsilon,\infty)$ and then take the limit $\epsilon\to 0$ in the end. $\endgroup$ – Winther Jan 6 '15 at 0:24
  • $\begingroup$ Thanks. I understand that $\lim_{u} \frac{u}{e^u -1} = 1$, still a bit unclear why it justifies $\frac{1}{e^u -1} = \sum_k e^{-uk}$ expansion. $\endgroup$ – Alex Jan 6 '15 at 0:29