I have to compute this integral:
$\large{\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}} dx dy \, \LARGE{ \frac{ |\frac{n}{l} (\frac{z-z_{0}}{l})^{n-1}|^{2}}{(1 + \frac{1}{4} |(\frac{z-z_{0}}{l})^{n}|^{2})^2}} $,
$n \in \Bbb{Z}, l \in \Bbb{R}, z_{0} \in \Bbb{C}, z = x+ i y $.
I'm given that the answer is $8\pi n$. I wanted to use the residue theorem, but I'm stuck on what to do with the absolute values, if the theorem is even applicable.
I'm also confused about why they have written an integral over dx and dy when the function is a complex function of z. I tried to parameterize $z$ with $r \, e^{i\theta}$, but again got stuck with the absolute values.
Edit: Thanks, Daniel for your suggestion. I realized there was a typo in the denominator that I fixed, but the suggestion is still helpful
Steps 1 & 2: Redefine $(z-z_0)/l = \tilde{z}$ and switch to polar coordinates.
$\large{\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}} dx dy \,l^{2} \, \Large{ \frac{ |\frac{n}{l} \tilde{z}^{n-1}|^{2}}{(1 + \frac{1}{4} |\tilde{z}^{n}|^{2})^2}} $
$\large{\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}} dx dy \, \Large{ \frac{ n^{2} \tilde{z}^{n-1}{\tilde{z}^{*}}^{n-1}}{(1 + \frac{1}{4} \tilde{z}^{n}{\tilde{z}^{*}}^{n})^2}} $
Since $ \tilde{z}=r e^{i\phi}$, ${\tilde{z}^{*}}=re^{-i\phi}$ and ${\tilde{z}^{*}} = r^{2}/\tilde{z}$.
$\large{\int_{0}^{\infty} \, rdr \int_{0}^{2\pi}} d\phi \, \Large{ \frac{ n^2 \tilde{z}^{n-1}(\frac{r^{2}}{\tilde{z}})^{n-1}}{(1 + \frac{1}{4} \tilde{z}^{n}(\frac{r^{2}}{\tilde{z}})^{n})^{2}}} $
$\large{\int_{0}^{\infty} \, rdr \int_{0}^{2\pi}} d\phi \, \Large{ \frac{ n^2 r^{2n-2}}{(1 + \frac{1}{4} r^{2n})^{2}}} $
$\large{\int_{0}^{\infty} \, dr} 2\pi \, \Large{ \frac{ n^2 r^{2n-1}}{(1 + \frac{1}{4} r^{2n})^{2}}} $
Edit 2:
Define t = $r^{2n}$
$ {\int_{0}^{\infty} \, dt} \pi \, { \frac{ n }{(1 + \frac{1}{4} t)^{2}}} = \frac{-4}{1+\frac{1}{4}t}|_{0}^{\infty} = 4\pi n $
Edit 3:
That the answer--the problem gave me the incorrect solution.