I came across the integral $$ \int_0^1 \frac{-\log x}{1+x}\ \mathrm dx = \frac{\pi^2}{12}, $$ which can be calculated as $\frac 1 2 \zeta(2)$ using analytic number theory.

I'm interested if this integral can be calculated in any other interesting, possibly more elementary ways?

  • $\begingroup$ $x = {\rm e}^{-t}$. Expand $\left(1 + {\rm e}^{-t}\right)^{-1}$ in powers of $\large{\rm e}^{-t}$. $\endgroup$ – Felix Marin Jan 23 '14 at 17:40

Integrating by part, you arrive to

$$-\text{Li}_2(-x)-\log (x) \log (x+1)$$

and, using your bounds, the result is $\frac{\pi ^2}{12}$


You can simply expand the $1/(1+x)$ piece into its equivalent geometric series and get

$$-\sum_{n=0}^{\infty} (-1)^n \int_0^1 dx \, x^n \log{x}$$

One may show using integration by parts that the integral in the sum is simply $-1/(n+1)^2$. The result is simply

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^2} = \frac{\pi^2}{12}$$

  • $\begingroup$ I wouldn't call evaluating this sum much simpler than evaluating the original integral. $\endgroup$ – Christoph Jan 23 '14 at 17:36
  • $\begingroup$ @ChristophPegel: I disagree, although it does depend on what you consider "simple." I consider it simple because everything in the derivation depends on elementary techniques. $\endgroup$ – Ron Gordon Jan 23 '14 at 18:00

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