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.

  • $\begingroup$ Only one integral and $dxdy$? Double integral? $\endgroup$ – Martín-Blas Pérez Pinilla Oct 8 at 18:38
  • $\begingroup$ I think they meant integrate over all x and y, so I updated for clarity $\endgroup$ – pqr7345 Oct 8 at 19:46
  • 1
    $\begingroup$ Step 1: Say why you can without loss of generality assume that $z_0 = 0$. (This can be omitted, at the cost of less convenient notation in the following.) Step 2: Write the integral in polar coordinates, outer integral over $r$, inner over $\varphi$. Step 3: Apply the residue theorem to the inner integral. For that, write $d\varphi$ in terms of $dz$ and note that for $\lvert z\rvert = r > 0$ you have $\overline{z} = r^2/z$. $\endgroup$ – Daniel Fischer Oct 8 at 21:10
  • $\begingroup$ Thanks, @DanielFischer for your suggestion. I tried it out and my answer is still incorrect, and I'm unsure where I went wrong $\endgroup$ – pqr7345 Oct 9 at 15:54
  • $\begingroup$ By my outline the residue theorem would have been applied to the integral over the circle $\lvert \tilde{z}\rvert = r$ (somewhat silly, because the integrand doesn't depend on $\varphi$, computing the integral in a purely real way is the natural thing to do). To use the residue theorem for $\int_0^{\infty} \ldots\,dr$ you need to use a roundabout trick like introducing an $r^{\varepsilon}$ ($0 < \lvert \varepsilon\rvert < 1$) and let $\varepsilon \to 0$. That works, but evaluating it the real way is much simpler. $\endgroup$ – Daniel Fischer Oct 9 at 17:50

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