Problem: Analyze the convergence of an improper integral $I=\int_{0}^\infty x^a \ln{(1+\frac{1}{x^2})} dx$.

It seems to me that 'I' converges for $0<a<1$.

My work:
I wrote integral 'I' as a sum of two integrals ($\int_0^1+\int_1^\infty$), since $0$ and $\infty$ are 'problematic' points. Also, since $x^a\ln(1+\frac{1}{x^2})$ behaves as $x^{a-2}$ for large x, second integral in the sum converges for $a<1$. In first integral, $x^a\ln{(1+\frac{1}{x^2})}$ behaves as $x^a\ln{x}$ as $x\to 0$ and I'm not sure which condition should I get from it? (Is it $a>0$)?

Any help is welcome. Thanks in advance!


1 Answer 1


For $a\ne -1$, we have using integration by parts

$$\begin{align} I(a)&=\lim_{\epsilon \to 0^+}\int_\epsilon^1 x^a \log(x)\,dx\\\\ &=\lim_{\epsilon \to 0^+}\left(\left.\frac{x^{a+1}\log(x)}{a+1}\right|_{\epsilon}^1-\frac{1}{a+1}\int_\epsilon^1 x^{a}\,dx\right)\\\\ &=\lim_{\epsilon \to 0^+}\left(-\frac{\epsilon^{a+1}\log(\epsilon)}{a+1}-\frac{1}{(a+1)^2}(1-\epsilon^{a+1})\right)\\\\ \end{align}$$

It is easy to see that if $a>-1$, then $I(a)$ converges, while for $a\le -1$, $I(a)$ fails to converge.

The OP notes that the integral $J(a)$ as given by

$$J(a)=\lim_{L \to \infty}\int_1^L x^a \log\left(1+\frac1{x^2}\right)\,dx$$

converges for all $a<1$.

Putting it together, we find that the integral $\int_0^\infty x^a \log\left(1+\frac1{x^2}\right)\,dx$ converges for $|a|<1$.

Aside, it can be shown (using contour integration, for example) that

$$\int_0^\infty x^a \log\left(1+\frac1{x^2}\right)\,dx=\frac{\pi}{(a+1)\cos\left(\frac{\pi a}{2}\right)}$$

for $|a|<1$. To show this, we integrate by parts to reveal

$$\int_0^\infty x^a \log\left(1+\frac1{x^2}\right)\,dx=\frac{2}{a+1}\int_0^\infty \frac{x^a}{x^2+1}$$

Now, we move to the complex plane and analyze the integral

$$F(a)=\oint_C \frac{z^a}{z^2+1}\,dz$$

where $C$ is the classical keyhole contour. Then, we find that

$$(1-e^{i2\pi a})\int_0^\infty \frac{x^a}{x^2+1}\,dx=2\pi i \left(\frac{e^{i\pi a/2}}{2i}+\frac{e^{i3\pi a/2}}{-2i}\right)$$

Solving for the integral we find that

$$\frac{2}{a+1}\int_0^\infty \frac{x^a}{x^2+1}\,dx=\frac{\pi}{(a+1)\cos\left(\frac{\pi a}{2}\right)}$$

as was to be shown!

  • $\begingroup$ Shouldn't the upper limit in the definition of $J(a)$ be $L$? Good one (+1). $\endgroup$ Jun 28, 2016 at 22:58
  • $\begingroup$ @Jack'swastedlife Jack, nice catch! Thank you. Se, you're not wasting your life. -Mark $\endgroup$
    – Mark Viola
    Jun 28, 2016 at 23:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .