Q:

$PV \int_{ia-\infty}^{ia+\infty}\frac{e^{ikt}}{\sqrt{t^2-b^2}}dt$

How would I calculate 
\begin{align}
PV \int_{ia-\infty}^{ia+\infty}\frac{e^{ikt}}{\sqrt{t^2-b^2}}dt
\end{align}
The square root should be dealt with as is most appropriate, e.g. by taking a branch cut from $t=-b$ to $t=+b$.

A:

You do not need the $PV$.  Assume $k, b \gt 0$ and $a \lt 0$.  Consider
$$\oint_C dz \frac{e^{i k z}}{\sqrt{z^2-b^2}} $$
where $C$ consists of a number of segments:
$$C_1 = \{ z = i a + t | t \in [-R,R] \} $$
$$C_2 = \{ z = R e^{i \theta} | \theta \in [0,\pi] \}$$
$$C_3 = \{ z = -b+\epsilon e^{i \phi} | \phi \in (0, 2 \pi) \} $$
$$C_4 = \{ z = x + i \delta | x \in [-b+\epsilon,b-\epsilon] \} $$
$$C_5 = \{ z = b+\epsilon e^{i \phi} | \phi \in (-\pi, \pi) \} $$
$$C_6 = \{ z = x - i \delta | x \in [-b+\epsilon,b-\epsilon] \} $$
where $\delta \gt 0$ depends on $\epsilon$ and goes to zero with $\epsilon$.  
We are interested in the limits as $R \to \infty$ and $\epsilon \to 0$.  In these limits, the integral vanishes over $C_2$, $C_4$, and $C_6$.  The result is, for the contour integral,
$$\int_{i a -\infty}^{i a + \infty} dt \frac{e^{i k t}}{\sqrt{t^2-b^2}} - i \int_b^{-b} dx \frac{e^{i k x}}{\sqrt{b^2-x^2}} + i \int_{-b}^b dx \frac{e^{i k x}}{\sqrt{b^2-x^2}}$$ 
A little explanation.  On $C_4$, $\arg{(x+b)} = 0$ and $\arg{(x-b)} = \pi$.  However, on $C_6$, $\arg{(x+b)} = 0$ and $\arg{(x-b)} = -\pi$.  Thus, because the contour integral is zero by Cauchy's theorem,
$$\int_{i a -\infty}^{i a + \infty} dt \frac{e^{i k t}}{\sqrt{t^2-b^2}} = -i 2 \int_{-b}^b dx \frac{e^{i k x}}{\sqrt{b^2-x^2}} = -i 2 \pi J_0(b k)$$
The above assumed that $k \gt 0$.  When $k \lt 0$, the integral on the left is zero because there are no poles or branch points of the integrand in the region $\operatorname{Im}{z} \lt a$.  
When $a \gt 0$, reverse the sign of $k$ in the above results.  Putting this altogether, we finally have
$$\int_{i a -\infty}^{i a + \infty} dt \frac{e^{i k t}}{\sqrt{t^2-b^2}} = i 2 \pi \operatorname{sgn}{(a)} J_0(b k) H\left [-k \operatorname{sgn}{(a)} \right ]$$
where $H$ is the Heaviside step function.
N.B. Proof of the Bessel integral may be found here.  Also, I realize that $C$ as defined is not exactly closed.  The pieces that would connect $C_1$ and $C_2$ vanish as $R \to \infty$, so I did not bother to include them.  I will expound on this in more detail in a future post, as I have yet to see a clear demonstration of this simple, yet essential fact, in the textbooks.

