TITLE: Integrate: $\int_{a - i\infty}^{a + i\infty} \frac{e^{tz}}{z^2 + p^2}dz$
QUESTION [5 upvotes]: Q. Show that : $$\int_{a - i\infty}^{a + i\infty} \frac{e^{tz}}{z^2 + p^2}dz = \frac{\sin pt}{p}$$

I considered the following contour

$$\int_\Gamma \frac{e^{tz}}{z^2 + p^2}dz + \int_{a - i\infty}^{a + i\infty} \frac{e^{tz}}{z^2 + p^2}dz  = 2 \pi i (\text{Residue}[f(z), z=ip] + \text{Residue}[f(z), z=ip]) \; \; \; \;\; \;(1)$$
The first integral goes to zero. 
$$\text{Residue}[f(z), z=ip]  + \text{Residue}[f(z), z=-ip]= \frac{\sin (pt)}{p} \hspace{1cm} (2)$$
From $(1)$ and $(2)$ we get 
$$\int_{a - i\infty}^{a + i\infty} \frac{e^{tz}}{z^2 + p^2}dz = 2 \pi i \frac{\sin pt}{p}$$
Did I make any mistake or (answer according to) question is wrong?

REPLY [2 votes]: I do not use the contour shown in the figure.  I prefer going, from the top of the vertical line at $a+i R$, to the left horizontally to $z=i R$, and then going in a semicircular arc between $\theta \in [\pi/2,3 \pi/2]$.  In this case, the integral over the horizontal portion to the left goes as 
$$\frac{e^{i R t}}{R^2-p^2} \frac{1-e^{a t}}{t} $$
as $R \to \infty$.  We get a similar expression for the horizontal segment at $z=-i R$.
It then becomes crystal clear that the integral along the circular arc vanishes, because the integral there looks like
$$i R \int_{\pi/2}^{3 \pi/2} d\theta e^{i \theta} \frac{e^{R t \cos{\theta}} e^{i R \sin{\theta}}}{R^2 e^{i 2 \theta}+p^2}$$
The magnitude of this integral vanishes as $R \to \infty$ because the cosine term is negative over the integration region. So, at least along this contour, the integral vanishes.