Q:

Complex analysis on real integral

Use complex analysis to compute the real integral
$$\int_{-\infty}^\infty \frac{dx}{(1+x^2)^3}$$
Attempt
I think I want to consider this as the real part of
$$\int_{-\infty}^\infty \frac{dz}{(1+z^2)^3}$$
and then apply the residue theorem. However, I am not sure how that is the complex form and the upper integral is the real part and how to apply.

A:

Define a contour that is a semicircle in the upper half plane of radius R.  Plus the real line from $-R$ to $R$
Then let R get to be arbitrarily large.
There is one pole at $z = i$ inside the contour
Cauchy integral formula says:
$$f^{(n)}(a) = \frac {n!}{2\pi i}\oint \frac {f(a)}{(z-a)^{n+1}}\ dz$$
$$\oint \frac {\frac {1}{(z+i)^3}}{(z-i)^3}\ dz = \pi i \frac{d^2}{dz^2} \frac {1}{(z+i)^3} \text{ evaluated at } z=i.$$
Next you will need to show that the integral along contour of the semi-cricle goes to $0$ as $R$ gets to be large.  
$$z = R e^{it}, dz = iR e^{it}\\
\displaystyle\int_0^{\pi} \frac {iRe^{it}}{(R^2 e^{2it} + 1)^3} \ dt\\
\left|\frac {iRe^{it}}{(R^2 e^{2it} + 1)^3}\right| < R^{-5}\\
\left|\int_0^{\pi} \frac {iRe^{it}}{(R^2 e^{2it} + 1)^3} \ dt\right| < \int_0^{\pi} R^{-5}\ dt\\
\lim_\limits{R\to \infty}\int_0^{\pi} R^{-5}\ dt = 0$$

