# 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!

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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

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

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