Visualisation of the residue theorem - contour integration

Question

I was trying to solve this integral: $$\int_{-\infty}^{+\infty}\frac{\xi_{0}^{2}e^{-iEt}dE}{\left(E-E_{0}\right)^{2}+\left|\xi_{0}\right|^{4}\pi^{2}}$$ For that i used partial fractions to find the poles and then apply the residue theorem, like this: $$\xi_{0}^{2}\int_{-\infty}^{+\infty}e^{-iEt}\frac{1}{2i\pi\xi_{0}^{2}}\left[\frac{1}{E-E_{0}+i\pi\xi_{0}^{2}}-\frac{1}{E-E_{0}-i\pi\xi_{0}^{2}}\right]$$ So we have a positive pole in $E_{0}+i\pi\xi_{0}^{2}$ and a negative one $E_{0}-i\pi\xi_{0}^{2}$.

Assuming this is right, i'm having trouble visualizing this, how can i draw the contour of this specific integral and this specific poles?

Answer 1

The best contour to use is the semi upper circle of some radius enough large with the segment $[-R,R]$. Then the integral over the contour splits in the integral over the semicircle and the integral in $[-R,R]$, whose limit at infinity is the integral you want to compute. You will see that the first integral vanishes when $R\to\infty$, you can use the estimmation lemma to see it, so we have that the integral in $[-R,R]$ only depends on the poles inside the contour. I assume that $E_0>0$, so the only one of the two poles that is inside the contour is $E_0+i\pi\zeta_0^2$. Just calculate the residue as usual knowing that is a simple pole

Edit:

Let $f(E)$ your complex integral. The integral along our contour splits like this:

$$\int_{\Gamma R}f(E)dE=\int_{\gamma R}f(E)dE+\int_{[-R,R]}f(E)dE$$ where $\gamma_R$ is parametrized by $Re^{iz}$, $z\in[0,\pi]$. We want to compute $$\lim_{R\to\infty} \int_{[-R,R]}f(E)dE$$ The integral that vanishes is the following: $$\lim_{R\to\infty} \int_{\gamma R}f(E)dE=0$$ so we have that $$Res(f,E_0+i\pi\zeta_0^2)=\lim_{z\to\infty}\int_{\Gamma R}f(E)dE=\lim_{R\to\infty} (\int_{\gamma R}f(E)dE+\int_{[-R,R]}f(E)dE)=\int_{-\infty}^{\infty}f(E)dE$$