Contour integration with branch cut This is an exercise in a course on complex analysis I am taking:
Determine the function $f$ using complex contour integration:
$$\lim_{R\to\infty}\frac{1}{2\pi i}\int_{c-iR}^{c+iR}\frac{\exp(tz)}{(z-i)^{\frac{1}{2}}(z+i)^{\frac{1}{2}}} dz$$
Where $c>0$ and the branch cut for $z^\frac{1}{2}$ is to be chosen on $\{z;\Re z=0, \Im z \leq0\}$.
Make a distinction between:
$$t>0, \quad t=0, \quad t<0$$
I think I showed that for $t<0$, $f(t)=0$ by using Jordan's Lemma. For $t=0$ I think the answer must be $f(0)=\frac{1}{2}$. For $t>0$ however, I have no idea what contour I have to define, nor how I have to calculate the residues in $i$ and $-i$.

For $t\equiv-\tau<0$, consider a half-circle of radius $M$ centred at $c$  and lying on the right of its diameter that goes from $c-iM$ to $c+iM$. By Cauchy's theorem
$$
\int_{c-iM}^{x+iM}\frac{e^{tz}}{\sqrt{1+z^2}}dz
=\int_{-\pi/2}^{+i\pi/2}\frac{e^{-\tau(c+Me^{i\varphi})}}{\sqrt{1+(c+Me^{i\varphi})^2}}iMe^{i\varphi}d\varphi;
$$
the right-hand side is bounded in absolute value by an argument in the style of Jordan's lemma, and hence goes to $0$ as $M\to\infty$.
So, indeed:
$$\boxed{
\int_{c-i\infty}^{c+i\infty}\frac{e^{tz}}{\sqrt{1+z^2}}dz=0,\text{ for }t<0.}
$$
For $t=0$ the integral diverges logarithmically, but we can compute its Cauchy principal value:
$$
\int_{c-iM}^{x+iM}\frac{1}{\sqrt{1+z^2}}dz=\left[ \sinh^{-1}z \right]_{c-iM}^{c+iM}=\log\frac{cM^{-1}+i+\sqrt{(cM^{-1}+i)^2+M^{-2}}}{cM^{-1}-i+\sqrt{(cM^{-1}-i)^2+M^{-2}}}
$$
and for $M\to\infty$
$$
\boxed{
PV\int_{c-i\infty}^{c+i\infty}\frac{1}{\sqrt{1+z^2}}dz=i\pi.
}
$$
Finally, for $t>0$, consider the contour below:

It is easy to see that the integrals along the horizontal segments vanish in the $M\to\infty$ limit, as well as the integral along the arc on the left (the latter, again by Jordan's lemma). Even the integrals on the small arcs give no contribution as the contour approaches the branch cuts.
The only contribution comes from the branch discontinuities:
$$
\int_{c-i\infty}^{c+i\infty}\frac{e^{tz}}{\sqrt{1+z^2}}dz=
4i \int_{1}^{+\infty}\frac{{\sin (ty)}}{\sqrt{y^2-1}}dy.
$$
Now, letting $y=\cosh \psi$, we have
$$
4i\Im \int_1^{+\infty}\frac{e^{ity}}{\sqrt{y^2-1}}dy=
2i\Im \int_{-\infty}^{+\infty}e^{it\cosh\psi}d\psi=
2i\Im \left(i\pi H_0^{(1)}(t)\right)=i2\pi J_0(t),
$$
thanks to the integral representation of cylindrical Bessel functions.
In this step, the analytic continuation $t\mapsto t+i\delta$, for a small $\delta>0$ which is then sent to $0$, has been employed.
So, finally
$$
\boxed{
\int_{c-i\infty}^{c+i\infty}\frac{e^{tz}}{\sqrt{1+z^2}}dz
=i2\pi J_0(t), \text{ for }t>0.
}
$$
To sum up
$$
f(t)=
\begin{cases}
0 &\text{if }t<0\\
1/2 &\text{if }t=0\text{ (in the }PV\text{ sense)}\\
J_0(t)&\text{if }t>0.
\end{cases}
$$
