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I have to test convergence of improper integral

$$ \int_{0}^{\infty} \frac{x\log(x)}{(1+x^2)^2}\,\mathrm dx$$

I write as $\log(x) \leq x$ . So $x\log(x) \leq x^2$. So $ \frac{x\log(x)}{(1+x^2)^2} \leq \frac{x^2}{(1+x^2)^2}$ . Now using comparison test with integral $\frac{1}{x^2}$ to get original integral convergent. Is that right ? Not sure Thanks

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  • $\begingroup$ That works at infinity. Also note that $\lim_{x \to 0} x \log{x} = 0$ and you are done because there are no poles on the $x$-axis. BTW the integral may be evaluated using any number of techniques. $\endgroup$ – Ron Gordon Feb 9 '16 at 6:04
  • $\begingroup$ Try with $x=1/y$ or $x=\tan u$ $\endgroup$ – lab bhattacharjee Feb 9 '16 at 6:07
  • $\begingroup$ @RonGordon so in that case i have to split integral from 0 to 1 and 1 to infinity. first integral is proper and for second one i have posted my solution. is that okay $\endgroup$ – Taylor Ted Feb 9 '16 at 6:09
  • $\begingroup$ @TaylorTed: you don't need to do that just to prove convergence. That may be a useful technique to evaluate the integral though. I personally prefer using the residue theorem. $\endgroup$ – Ron Gordon Feb 9 '16 at 6:10
  • $\begingroup$ @RonGordon Gordon Textbook asks to just state whether it is convergent or not. $\endgroup$ – Taylor Ted Feb 9 '16 at 6:11
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This is probably not an answer but it is too long for a comment.

$$I=\int_{0}^{\infty} \frac{x\log(x)}{(1+x^2)^2}\, dx=\int_{0}^{1} \frac{x\log(x)}{(1+x^2)^2}\, dx+\int_{1}^{\infty} \frac{x\log(x)}{(1+x^2)^2}\, dx$$ As lab bhattacharjee commented, change variable $x=\frac 1y$ for the second integral; so $$\int_{1}^{\infty} \frac{x\log(x)}{(1+x^2)^2}\, dx=\int_1^0 \frac{y\log(y)}{(1+y^2)^2}\, dy=-\int_0^1 \frac{y\log(y)}{(1+y^2)^2}\, dy$$ which makes the beautiful result $I=0$.

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  • $\begingroup$ Can you show me using comparison test $\endgroup$ – Taylor Ted Feb 9 '16 at 6:59
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I will show that the integral converges by actually evaluating it. Firstly, consider the following complex contour integral: $$J:=\oint_C \frac{z \ln^2 z}{\left ( 1+z^2 \right )^2}dz$$ where $C$ is a dumbell contour pictured here, with the branch cut along the positive part of the real axis: enter image description here We see that $$J=\int _{c1}+\int _{c2}+\int_{C_{R}} +\int _{C \epsilon}$$ I want to show that the integral along the outer circle of radius R tends to zero for large R. Recall the estimation lemma: $$\left | \int _{C_R}f\left ( z \right ) dz\right |\leq 2 \pi \frac{R^2\left ( \ln^2 R+\theta \right )}{\left ( R-1 \right )^4}, \theta \in \left [0, 2 \pi \right )$$ Taking the limit as $R\rightarrow \infty$ and knowing that $ \ln x \leq x$ for large $x$ we get that $\left | \int _{C_R}f\left ( z \right ) dz\right |\rightarrow 0$ Similarly, we show that $\left | \int _{C\epsilon}f\left ( z \right ) dz\right |\rightarrow 0$, acknowledging the fact that $x \ln x \rightarrow 0$ as $x \rightarrow 0$. Now we are left with only two integrals, along C1 and along C2. $$\int _{C1}f\left ( z \right )dz=\int_{0}^{\infty}\frac{x \ln ^2 x}{\left ( 1+x^2 \right )^2}dx$$ The argument of the log has a phase of $2 \pi i$ along C2, so : $$\int _{C2}f\left ( z \right )dz=\int_{0}^{\infty}\frac{x \left ( \ln x +i2 \pi \right )^2}{\left ( 1+x^2 \right )^2}dx$$ Adding them together leaves us with : $$J:=-4 i \pi\int_{0}^{\infty}\frac{x \ln x}{\left ( 1+x^2 \right )^2}dx+4 \pi^2\int_{0}^{\infty}\frac{x dx}{\left ( 1+x^2 \right )^2} (*)$$ On the other hand $$J:=2 \pi i \sum Res f\left ( z \right )$$ The function has poles of order 2, at $z= \pm i$, so $$J:=2 \pi i \left ( i\frac{\pi}{4} -i\frac{\pi}{4}\right )=0(**)$$ By equating $(*)$ and $(**)$ we get that $$\int_{0}^{\infty}\frac{x \ln x}{\left ( 1+x^2 \right )^2}dx=0$$

and also $$\int_{0}^{\infty}\frac{x}{\left ( 1+x^2 \right )^2}dx=0$$

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Letting $u=x^2+1$ then $\frac{1}{2}\,\mathrm du=x\,\mathrm dx$ and $\log(x)=\log((u-1)^\frac{1}{2})=\frac{1}{2}\log(u-1)$. With that substitution I think the convergence is easier to see. Leibovici's answer is the most beautiful answer, imo.

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