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

  • $\begingroup$ 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

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 $I=0$.

  • $\begingroup$ Can you show me using comparison test $\endgroup$ – Taylor Ted Feb 9 '16 at 6:59

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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.