Integrating $\frac{\ln x^2}{e^{x^2}}$

Question

I am stuck when trying to find $$\int_0^\infty\frac{\ln x^2}{e^{x^2}}\,dx$$ The context of the question is asking if the integral is convergent or divergent?

Answer 1

The function is equivalent to $2\ln{|x|}$ at $x$ goes to $0$ so is integrable there. The function is $o(e^{-|x|})$ as $|x|$ goes to infinity, so is integrable there as well.

Answer 2

Near $0$, your function is equivalent to $\ln(x^2)=2 \ln(x)$ which is integrable.

Near $+\infty$, your function is a $o \left( \frac{x^2}{e^{x^2}} \right)$ which is integrable.

Answer 3

If you want the explicit value, well, $$ \int_{0}^{+\infty}2\log(x)e^{-x^2}\,dx \stackrel{x\mapsto\sqrt{x}}{=}\frac{1}{2}\int_{0}^{+\infty}x^{-1/2}\log(x)e^{-x}\,dx $$ equals $\frac{1}{2}\Gamma'\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}\psi\left(\frac{1}{2}\right)$ by the dominated convergence theorem and the integral definition of the $\Gamma$ function, together with the well known $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$. Since $\psi(1)=\Gamma'(1)=-\gamma$ and $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b},$$ by picking $a=\frac{1}{2}$, $b=1$ and rearranging we get $$ \int_{0}^{+\infty}\log(x^2)e^{-x^2}\,dx = \color{blue}{-\frac{\sqrt{\pi}}{2}(\gamma+2\log 2)}$$ which is approximately $-\frac{7}{4}$.