Prove that $\int_0^\infty \frac{x\,\log x}{(1+x^2)^2} = 0$ 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.)
 A: 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
A: $I$ converges because 
$$\dfrac{x\ln x}{(1+x^2)^2}\le \dfrac{x^2}{(1+x^2)^2}\le\dfrac{1}{1+x^2} ~~~x>1$$ 
and $$ \lim_{x\to 0}\dfrac{x\ln x}{(1+x^2)^2} = 0$$
that is $$x\mapsto \dfrac{x\ln x}{(1+x^2)^2}~~~\text{is continuous on [0,1]}$$
Now Set $x=\frac{1}{u}$ that is $dx = -\frac{1}{u^2}$
$$I=\int_0^\infty\dfrac{x\ln x}{(1+x^2)^2}dx=-\int^0_\infty\dfrac{\frac{1}{u}\ln \frac{1}{u}}{(1+\frac{1}{u^2})^2}\frac{1}{u^2}du =\int_0^\infty\dfrac{\ln \frac{1}{u}}{(\frac{u^2+1}{u^2})^2}\frac{1}{u^3}du$$
But $$\color{red}{ \ln\frac{1}{u} = -\ln u}$$
hence We have 
$$I=\int_0^\infty\dfrac{x\ln x}{(1+x^2)^2}dx\\=\int_0^\infty\dfrac{\ln \frac{1}{u}}{(\frac{u^2+1}{u^2})^2}\frac{1}{u^3}du\\= -\int_0^\infty\dfrac{u^4\ln u}{(u^2+1)^2}\frac{1}{u^3}du \\=-\int_0^\infty\dfrac{u\ln u}{(u^2+1)^2}du  =-I$$
That $$\color{blue}{ I = -I\implies 2I= 0 \implies I= 0}$$
A: The integral on $[1,\infty]$ converges because 
$$\frac{x \log (x)}{\left(x^2+1\right)^2}<\frac{x^2}{x^4}=\frac{1}{x^2}$$ and the last one converges to $1$. 
Furthermore on $[0,1]$ we have $$x \log (x)\leq \frac{x \log (x)}{\left(x^2+1\right)^2}\leq 0$$ 
that is on $[0,1]$
$$-\frac{1}{4}\leq  \frac{x \log (x)}{\left(x^2+1\right)^2}\leq 0$$ 
So the integral converges on $(0,+\infty)$ and its value is 
$$-\frac{\log (2)}{4}+\frac{\log (2)}{4}=0$$
Hope this helps
A: Hint:
Method $\#1:$  Put $x=\dfrac1y$
Method $\#2:$  Let $x=\tan u\implies x=\sec^2u\ du$
$$I=\int_0^\infty\dfrac{x\ln x}{(1+x^2)^2}dx=\dfrac12\int_0^{\pi/2}\sin2y\ln(\tan y)\ dy$$
Now use the trick described here.
A: A supplement to Raffaele's answer: the substitution $y=-\ln x$ gives $$\int_0^1\frac{x\ln x dx}{(1+x^2)^2}=-\int_0^1\frac{ue^{-2u} du}{(1+e^{-2u})^2}=-\int_0^1u(e^{-2u}-2e^{-4u}+3e^{-6u}-\cdots ) du.$$Since $\int_0^1 nue^{-2nu}du=\frac{1}{4n}$, this sum is $-\frac{1}{4}(1-\frac{1}{2}+\cdots)=-\frac{\ln 2}{4}.$
A: The change of variable $x=e^t$ makes an interesting symmetry appear:
$$\int_0^\infty\frac{x\log x}{(1+x^2)^2}dx=\int_{-\infty}^\infty\frac{te^t}{(1+e^{2t})^2}e^tdt=\frac14\int_{-\infty}^\infty\frac t{\cosh^2t}dt.$$
Even if the integral did not converge (which it does anyway), its Cauchy principal value is zero.
