TITLE: Prove that $\int_0^\infty \frac{x\,\log x}{(1+x^2)^2} = 0$
QUESTION [5 upvotes]: I have been asked to prove that $$\int_0^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx = 0$$
I tried proceeding as follows:
\begin{align}
\int_0^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx &= \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx + \int_1^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx \\
& = \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx + \int_{u=1}^0 \frac{u\,\log u}{(1+u^2)^2}\,du\quad (\text{where }u=1/x) \\
&= \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx - \int_0^1 \frac{u\,\log u}{(1+u^2)^2}\,du \\
&=^? \int_0^1 0 \,dx \\
&= 0
\end{align}
But I think that I should first show that the improper integral $\int_0^1 {x\,\log x}/{(1+x^2)^2}\,dx$ is convergent.
Do I need to show it? If yes, how to do so?
(I have been taught basic methods such as comparison, limit comparison and Dirichlet test.)

REPLY [3 votes]: Your proof is good.  Here is a proof of the convergence

For $x\ge1$
$$
\begin{align}
\log(x)
&=\int_1^x\frac1t\,\mathrm{d}t\\
&\le\int_1^x1\,\mathrm{d}t\\[9pt]
&=x-1\\[9pt]
&\le x-\frac1x\tag1
\end{align}
$$
Since the transformation $x\mapsto\frac1x$ negates both sides of $(1)$, we get that for all $x\gt0$,
$$
\bbox[5px,border:2px solid #C0A000]{|\log(x)\,|\le\left|\,x-\frac1x\,\right|}\tag2
$$

$$
\begin{align}
\left|\,\int_0^\infty\frac{x\log(x)}{\left(1+x^2\right)^2}\,\mathrm{d}x\,\right|
&\le\int_0^\infty\frac{\left|\,x^2-1\,\right|}{\left(1+x^2\right)^2}\,\mathrm{d}x\tag3\\
&\le\int_0^\infty\frac{x^2+1}{\left(1+x^2\right)^2}\,\mathrm{d}x\tag4\\
&\le\int_0^\infty\frac1{1+x^2}\,\mathrm{d}x\tag5\\[6pt]
&=\frac\pi2\tag6
\end{align}
$$
Explanation:
$(3)$: apply $(2)$
$(4)$: triangle inequality
$(5)$: simplify
$(6)$: arctan integral