TITLE: Integrate $ \int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2$
QUESTION [18 upvotes]: I am trying to evaluate this integral.
$$
I=\int_0^\infty \frac{ \ln^2(1+x)}{x^{3/2}} dx=8\pi \ln 2
$$  Note $$
\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}, \ |x| < 1.
$$
I was trying to do use this series expansion but wasn't sure how to go about it because of the square of the logarithm.  And also it seems than we then will have $I\propto \int_0^\infty x^{n-3/2}dx$ which will diverge. Thanks

REPLY [5 votes]: There is another way to solve. Let
$$ I(\alpha,\beta)=\int_0^\infty\frac{\ln(1+\alpha x)\ln(1+\beta x)}{x^{\frac{3}{2}}}dx. $$
Note $I=I(1,1)$. It is easy to see that
$$ \frac{\partial^2I(\alpha,\beta)}{\partial\alpha\partial\beta}=\int_0^\infty\frac{\sqrt{x}}{(1+\alpha x)(1+\beta x)}dx=\frac{\pi}{\alpha\sqrt{\beta}+\sqrt{\alpha}\beta}  $$
and hence
$$ \frac{\partial I(\alpha,1)}{\partial\alpha}=\int_0^1 \frac{\pi}{\alpha\sqrt{\beta}+\sqrt{\alpha}\beta}d\beta=2\pi\frac{\log(1+\frac{1}{\sqrt{\alpha}})}{\sqrt{\alpha}} $$
and 
$$ I(1,1)=\frac{\partial I(\alpha,1)}{\partial\alpha}=2\pi\int_0^1\frac{\log(1+\frac{1}{\sqrt{\alpha}})}{\sqrt{\alpha}}d\alpha=8\pi\ln 2. $$