# Show that $\int^{\infty}_{-\infty}\frac{dx}{(x^2-4x+5)^2} = \frac{\pi}{2}$ using residue theory

I'm trying to evaluate some complex integrals using residue theory. I've read a number of articles with different examples here on Stack Exchange, but I'm still really lost and could use some help.

Show that $$\int^{\infty}_{-\infty}\frac{dx}{(x^2-4x+5)^2} = \frac{\pi}{2}$$.

First, I found the poles of the function as $$2-i$$ and $$2+i$$. Both of these poles are of order $$2$$. I am also considering my region as the semicircle of radius $$R$$ in the upper half-plane with the line segment between $$x=−R$$ and $$x=R$$ on the real axis. So I know that I need to evaulate

$$\int_{\Gamma}f(z)dz = 2\pi iRes(f,2+i) + 2\pi iRes(f,2-i)$$

At this point, I'm really stuck on what to do from here. I tried evaluating the residues and was getting some really weird answers. My work was really messy and likely completely wrong, so hopefully it's OK if I don't reproduce it here. I'm not sure how to finish solving this.

Similarly, I'm having trouble with this integral:

$$\int_{|z| = 1}z^3e^{1/z}\sin(\frac{1}{z}) dz$$

I know that there is a pole at $$z = 0$$. So that means I need to evaluate

$$\int_{|z| = 1}z^3e^{1/z}\sin(\frac{1}{z}) dz = 2\pi i Res(f,0)$$

Do I consider this pole of order $$1$$ or order $$2$$ and how do I find this residue?

For the first problem, to compute the resides of $$\frac1{(z^2-4x+5)^2}$$, we evaluate the limits

$$\lim_{z\to (2\pm i)}\frac{d}{dz}\left(\frac{z-(2\pm i)}{(z^2-4x+5)^2}\right)$$

But only the pole at $$z=2+i$$ is in the upper half plane. Can you finish now?

For the second problem, there is no pole at $$z=0$$. Rather, the point $$z=0$$ is an essential singularity. Expand both $$e^{1/z}$$ and $$\sin(1/z)$$ in Taylor series of powers of $$1/z$$ (the Laurent series) and use their Cauchy Product to determine the coefficient on the term $$\frac1{z^4}$$. Can you wrap this up now?

Alright, so your curve is the semi-circle of radius $$R$$ in the upper half-plane, so here it is:

Since the poles are $$2 \pm i$$ (pink points in the pic), it only contains one of them, so the integral over your curve $$\Gamma$$ will simply be:

$$\displaystyle\int_\Gamma f(z) \mathrm{d}z = 2\pi i \text{Res}(f, 2 + i)$$.

Now, your curve $$\Gamma$$ can be split into $$\gamma_1$$ and $$\gamma_2$$, where $$\gamma_1 = [-R, R]$$, and $$\gamma_2$$ is the half-circle.

So $$\displaystyle\int_\Gamma f(z)\mathrm{d}z = \displaystyle\int_{\gamma_1} f(z)\mathrm{d}z + \displaystyle\int_{\gamma_2} f(z)\mathrm{d}z = \displaystyle\int_{-R}^R f(x)\mathrm{d}x + \displaystyle\int_{\gamma_2} f(z)\mathrm{d}z$$

So you want to know the integral over the real-axis, you can just isolate it:

$$\displaystyle\int_{-R}^R f(x)\mathrm{d}x = \displaystyle\int_\Gamma f(z) \mathrm{d}z - \displaystyle\int_{\gamma_2} f(z) \mathrm{d}z = 2\pi i \text{Res}(f, 2 + i) - \displaystyle\int_{\gamma_2} f(z) \mathrm{d}z$$.

So... all you have to do now is compute the integral over $$\gamma_2$$ (hint: it's going to be zero, can you see why?)