For which values of $p \in \mathbb{R}$ is the integral $I = \int_{0}^{\infty} \frac{x^p}{x^2+a^2}dx$ convergent? Evaluate the integral by residue calculations for those values for which it is convergent.
I am making this exercise in preparation for my Complex Analysis exam. I begin by noticing that $lim_{x \rightarrow \infty} \frac{x^p}{x^2+a^2} < \infty$. So we have that $p<2$.
Now we want to use a toy contour to solve the integral. We have the poles $i\sqrt{a}$, $-i\sqrt{a}$ and $0$ if $p<0$. So we make the toy contour that is clockwise the semi circle on the upper half complex plane with radius R, $\gamma_R$. Then the real axis from $-R$ to $R$ but around zero we add a little bow so that when zero would be a pole this would not be a problem,$\gamma_{\epsilon}$.
The integral becomes: $ I = \int_{\gamma_R} \frac{x^p}{x^2+a^2}dx + \int_{\gamma_{\epsilon}} \frac{x^p}{x^2+a^2}dx + \int_{-R}^{R} \frac{x^p}{x^2+a^2}dx$.
Now we want to say $z(t) = e^{it}$. But now I get a little bit lost. Do I then need to solve this three integrals?
I do not get how I need to solve this kind of proves further. Does anyone has a general strategy and can show for values of p that $I$ converges, by applying it on this example?
Thanks in advance.
Near zero, you need $p>-1$ to avoid a non-integrable singularity, while hear infinity, you must have $p<1$ so that the decay is fast enough. Thus your range is $-1<p<1.$