Integrating $ \int_0^\infty\frac{(\log(1+x))^2}{x^2}dx$

Question

I would like to calculate $$ \int_0^\infty\frac{(\log(1+x))^2}{x^2}dx\,. $$ Integrating by parts, $$\begin{aligned} \int_0^\infty\frac{(\log(1+x))^2}{x^2}dx&= -\frac{(\log(1+x))^2}{x}\Bigg|_0^\infty+2\int_0^\infty \frac{\log(1+x)}{x(1+x)}dx\\ &=0+2\int_1^\infty\frac{\log x}{x(x-1)}dx\, , \end{aligned}$$ then, letting $x=e^{t}$, we get $$ 2\int_0^\infty\frac{t}{e^t-1}dt=2\zeta(2)=\frac{\pi^2}{3}\,; $$ alternatively we can set $x=1/s$, so we arrive at $$ -2\int_0^1 \frac{\log(1-s)}{s}ds = 2\sum_{n=1}^\infty \int_0^1\frac{s^n}{n}dx=2\sum_{n=1}^\infty\frac{1}{n^2}=2\zeta(2)=\frac{\pi^2}{3}\, . $$ Is there a way of tackling the original integral using contour integration?

Answer 1

Notice first that the function

$$f (z) = \frac {\log^2 (1+z)}{z^2}$$ Has a removable singularity near 0 hence we don't have any residues to calculate. You can consider a contour of a semi-circle in the upper half plain indented at the branch point $-1$. Note that we will have

$$\int_{-\infty}^{-1}\frac {(\log |1+x|+i\pi)^2}{x^2}dx+\int_{-1}^{0}\frac {\log^2 (1+x)}{x^2}dx=-\int^{\infty}_{0}\frac {\log^2(1+x)}{x^2}dx$$

Note the imaginary part Note that $$\int_{-\infty}^{-1} \frac{(\log |1+x|)}{x^2}dx=0$$

for the real parts we have

$$\int^{\infty}_{1}\frac {\log^2(x-1)}{x^2}dx+\int_{0}^{1}\frac {\log^2 (1-x)}{x^2}dx-\pi^2=-\int^{\infty}_{0}\frac {\log^2(1+x)}{x^2}dx$$

This is no better than solving the original integral !