TITLE: Improper Integral Question: $ \int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx $
QUESTION [9 upvotes]: I have to test the convergence of the integral : 
$$ \int_0 ^ \infty\frac{x\log x}{(1+x^2)^2} dx $$
Please suggest. Also, have to show that the value of the integral is zero ?

REPLY [3 votes]: There is a simple solution for this question. Note
\begin{eqnarray}
\int_0^\infty\frac{x\ln x}{(1+x^2)^2}dx&=&\int_0^1\frac{x\ln}{(1+x^2)^2}dx+\int_1^\infty\frac{x\ln x}{(1+x^2)^2}dx.
\end{eqnarray}
But for the second integral, we use the substitute $u=\frac{1}{x}$:
\begin{eqnarray} 
\int_1^\infty\frac{x\ln x}{(1+x^2)^2}dx=-\int_0^1\frac{u\ln u}{(1+u^2)^2}du.
\end{eqnarray}
Thus 
$$ \int_0^\infty\frac{x\ln x}{(1+x^2)^2}dx=0.$$