Question: Evaluate the integral$$\int_{-\infty}^{\infty}\frac{\sin(x)}{x(x^2+a^2)}=Im\left ( \frac{e^{ix}}{x(x^2+a^2)} \right)$$
Attempt: To evaluate this integral, I use the residue theorem (noting that we have poles $z=0$ and $z=ia$ within our contour):
$$2\pi i Res(f(z),z_0)=2\pi i \left (\lim_{z\rightarrow 0}\frac{e^{iz}}{z(z^2+a^2)}+\lim_{z\rightarrow ia}\frac{e^{iz}}{z(z^2+a^2)} \right )$$ $$=2\pi i \left(\frac{i}{a^2}-\frac{ie^{-a}}{2a^2} \right)$$
Therefore, our integral above should be
$$\frac{\pi(2-e^{-a})}{a^2}$$
However, in Mathematica, it says the answer should be
$$\frac{\pi(1-e^{-a})}{a^2}$$
What did I do incorrectly; i.e., where did a factor of 2 come from? Thank you in advance.
Not quite. The pole in $z = 0$ lies on the contour, it is not enclosed by the contour.
The informal short-cut says that the pole lies half in the upper and half in the lower half plane, so it contributes only half its residue to the sum, and
$$\begin{align} \int_{-\infty}^\infty \frac{\sin x}{x(x^2+a^2)}\,dx &= \operatorname{Im} 2\pi i\left(\frac12\operatorname{Res}\left(\frac{e^{iz}}{z(z^2+a^2)};0\right) + \operatorname{Res}\left(\frac{e^{iz}}{z(z^2+a^2)};ia\right)\right)\\ &= \operatorname{Im} 2\pi i \left(\frac{1}{2a^2} + \frac{e^{-a}}{(ia)(2ia)}\right)\\ &= \frac{\pi(1-e^{-a})}{a^2}. \end{align}$$
The formal way modifies the contour of integration by using a small semicircle around the pole in $z=0$ to get a closed contour that doesn't pass through any pole of the integrand, and then subtracts the integral over the small semicircle, and lets the radius of that semicircle tend to $0$. In the limit, you add or subtract $\pi i$ times the residue in $0$, depending on whether you chose the semicircle in the upper or lower half plane to avoid the pole in $0$.