# Visualisation of the residue theorem - contour integration

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?

• Try using the lower plane, as the current solution seems problematic as I discussed in my comment. Jun 29, 2022 at 6:59

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 $$Res(f,E_0+i\pi\zeta_0^2)=\lim_{z\to\infty}\int_{\Gamma R}f(E)dE=\lim_{R\to\infty} (\int_{\gamma R}f(E)dE+\int_{[-R,R]}f(E)dE)=\int_{-\infty}^{\infty}f(E)dE$$

• Thank you so much! I just don't understand how the first integral vanishes, since it has the postive pole? I thought we would have two semicircles, one positive for the positive pole and one negative, for the negative pole.
– bdzh
Jun 28, 2022 at 9:32
• Try it and if you don't get it, i will edit my answer with more hints Jun 28, 2022 at 9:33
• So we would only get the lower half plane?
– bdzh
Jun 28, 2022 at 9:37
• No,no. The contour is rigth. If we want to integrate from the real line we use the semi upper circle or the othe semicircle, each has only one of the poles inside. We dont need the two poles, we need exactly the poles in our contour Jun 28, 2022 at 9:43
• I did this, choosing the positive pole: so I replaced $E=E_{0}+i\pi\xi_{0}^{2}$: $\int_{CR}f(E)dE=2\pi i\cdot Res_{E=E_{0}+i\pi\xi_{0}^{2}}f(E).$. So i got $e^{-iE_{0}t}e^{\pi\xi_{0}^{2}t}$. But i don't think this is right, since the second exponential should have a negative exponent
– bdzh
Jun 28, 2022 at 19:06