I came across the following integral:
$$\int_{- \infty}^{\infty} \frac{x^2}{((x-t)^2 + \delta^2 )^2((x + t)^2 + \delta^2)^2} \textrm{d}x.$$
I understand that if we turn this into a complex integral along a semi-circular contour in the upper half of the complex plane, denoting the function in the integrand as $f(x)$, we get
$$\oint f(z) \textrm{d}z = 2 \pi i \sum \text{Res} (f(z)) = \int_{\text{arc}} f(z) \textrm{d}z + \int_{- \infty}^{\infty} f(x) \textrm{d}x $$
where $\int_{- \infty}^{\infty} f(x) \textrm{d}x$ is what I'm trying to calculate. The residues can be easily calculated, however, I'm struggling to show that the integral over the arc of the semicircle vanishes as the radius of the semicircle tends to infinity.
Using only algebra
The antiderivative $$I=\int \frac{x^2}{((x-t)^2 + \delta^2 )^2\,\,((x + t)^2 + \delta^2)^2} \,{d}x$$ is easy to compute after partial fraction decomposition (leading as usual to arctangents and logarithms).
Recombining them $$I=\frac{1}{8 t}\log \left(\frac{\delta ^2+t^2-2 t x+x^2}{\delta ^2+t^2+2 t x+x^2}\right)+\frac{1}{4 \delta }\tan ^{-1}\left(\frac{2 \delta x}{\delta ^2+t^2-x^2}\right)$$ Using the above for $$J(r)=\int_{-r}^{+r} \frac{x^2}{((x-t)^2 + \delta^2 )^2\,\,((x + t)^2 + \delta^2)^2} \,{d}x$$ and recombinig again the two functions $$J(r)=\frac{1}{2 \delta }\tan ^{-1}\left(\frac{2 \delta r}{\delta ^2-r^2+t^2}\right)+\frac{1}{4 t}\log \left(\frac{\delta ^2+r^2-2 r t+t^2}{\delta ^2+r^2+2 r t+t^2}\right)$$
Expanding as series for large values of $r$ $$J(r)=-\frac 2 r \left(1+\frac{2 \left(t^2-\delta ^2\right)}{3 r^2}+O\left(\frac{1}{r^4}\right)\right)$$