# Evaluate $\ \int_{- \infty}^{\infty} \frac{x}{( x^2 + 4x + 13 )^2}\,dx . \$ using contour integration and the calculus of residues

How do I evaluate $\ \int_{- \infty}^{\infty} \frac{x}{( x^2 + 4x + 13 )^2}\,dx . \$ using contour integration and the calculus of residues?

I have many more problems of this kind need to be done. Please help so I'll have an example to help me to figure out the rest!

To begin with, the roots of $\,x^2+4x+13\,$ are $\,-2\pm 3i\,$ , so

$$x^2+4x+13=\left(x+2-3i\right)\left(x+2+3i\right)$$

and thus the integrand function has two double poles, one in each half plane (upper and lower ones).

Now, if you choose the contour

$$C_R:=[-R,R]\cup \gamma_R:=\{z\in\Bbb C\;;\;|z|=R\,\,,\,\arg z\geq 0\}\,$$

then, as the only pole of the function $\,f(z)=z^2+4z+13\,$ within the domain bounded by $\,C_R\,$ is $\,-2+3i\,$ , we evaluate

$$Res_{z=-2+3i}(f)=\lim_{z\to -2+3i}\left((z+2-3i)^2\frac{1}{(z+2-3i)^2(z+2+3i)^2}\right)'=$$

$$=\lim_{z\to -2+3i}\left(\frac{1}{(z+2+3i)^2}\right)'=\lim_{z\to -2+3i}-\frac{2}{(z+2+3i)^3}=-\frac{2}{(6i)^3}=\frac{1}{108i}$$

Thus, by Cauchy's Residue Theorem:

$$\frac{\pi}{54}=\oint_{C_R}f(z)\,dz=\int_{-R}^R\frac{1}{(x^2+4x+13)^2}dx+\int_{\gamma_R}f(z)\,dz$$

But

$$\left|\int_{\gamma_R}f(z)\,dz\right|\leq \frac{1}{R^2-4R-13}{R\pi}\xrightarrow[R\to\infty]{}0$$

So finally

$$\int_{-\infty}^\infty\frac{dx}{(x^2+4x+13)^2}=\lim_{R\to\infty}\int_{-R}^R\frac{dx}{(x^2+4x+13)^2}=\lim_{R\to\infty}\oint_{C_R}f(z)\,dz=\frac{\pi}{54}$$