I would like to evaluate the integral
$$\int_C \frac{e^{-(x^2-a^2)}}{x^2-a^2}\,dx$$
by integration over the complex plane. There are two poles at $x=\pm a$ on the real axis, which I can shift to $x=\pm a\mp i\varepsilon$ (and later consider the limit $\varepsilon\rightarrow 0$), to integrate over the real axis. So my contour $C$ goes along the real axis from $-\infty$ to $+\infty$ and then along a semicircle centered at $0$ from $-\infty$ to $+\infty$.
This is the standard procedure for Feynman propagators, which look similar to my integral, just without the exponential function. However, the exponential function has an essential singularity on the imaginary axis for $x^2\rightarrow-\infty$ (edited in response to comment).
How do I perform a contour integration, which involves poles both on the real axis and an essential singularity at infinity on the imaginary axis? Can I simply apply the residue theorem for the poles at $x=\pm a$ and ignore the singularity of the exponential function? In this case, the integration would simply yield the result $1/(\pm 2a)$ from the residues at $x=\pm a$. But shouldn't also the exponential function and its singularity affect the result of the integration?