Recognizing Integral as $J$-Bessel Function In Iwaniec-Kowalski, in the derivation of the Fourier expansion of the Poincare series at the end of the computation, they arrive at the following integral:
$$
\int_{-\infty+iy}^{\infty+iy}(cv)^{-k}e\!\left(\frac{-m}{c^2v}-nv\right)dv
$$
where the integral is along the contour of $\Im(v)=y$ and $e(z)=e^{2\pi iz}$. I previously saw that this integral was talked about in this post; however, there was no answer there. The author says

Then putting together the different terms, and recognizing the Bessel function in the integral (see $\textbf{[GR,8.315.1,8.412.2]})$, we obtain the lemma.

Now the reference is to the Table of Integrals, Series, and Products, and the two equations that are being referenced are
\begin{equation}
\frac{1}{\Gamma(z)}=\frac{i}{2\pi}\int_C(-t)^{-z}e^{-t}dt
\end{equation}
where $C$ is the Hankel contour, and
\begin{equation}
J_\nu(z)=\frac{z^\nu}{2^{\nu+1}\pi i}\int_{-\infty}^{(0^+)}t^{-\nu-1}\exp\left(t-\frac{z^2}{4t}\right)dt
\end{equation}
I have no idea how we are supposed to get the $J$-Bessel function out of this. I will note that in the final Fourier expansion there is a $J_{k-1}(\frac{4\pi\sqrt{mn}}{c})$. I just have no idea how to see that this is true. I have tried to complete the square in the exponent, I have tried expanding out a power-series and just integrating term by term to try and get the power-series of the $J$-Bessel function, but it seems nothing that I try works. Thus, if someone can show me how to get a $J$-Bessel function out of this integral, I would be eternally grateful.
 A: It is essentially just a change of variables.
Our target is
$$ \tag{T}\label{1}
J_{k - 1}\Bigl( \frac{4 \pi \sqrt{m n}}{c} \Bigr) = \frac{(4 \pi \sqrt{m n})^{k - 1}}{2^k \pi i c^{k - 1}} \int_{-\infty}^{(0+)} t^{-k} \exp\Bigr( t - \frac{4 \pi^2 m n}{c^2 t} \Bigr) \, d t.
$$
I will leave it to you to keep track of the contours in the integrals as we go through the motions, but here's the idea. In the original integral
$$\int_{-\infty+iy}^{\infty+iy}(cv)^{-k}e\!\left(\frac{-m}{c^2v}-nv\right)\, dv$$
let us first do the change of variables $v = t/n$ (with $d v = d t / n$) so as to make the lonely $t$ in $\eqref{1}$ appear. We get
$$(c / n)^{-k} n^{-1} \int t^{-k}e\!\left(\frac{-m n}{c^2 t}-t\right)\, dt. $$
Since $\eqref{1}$ uses $\exp$ instead of $e(x) = \exp(2 \pi i x)$, take $v = 2 \pi i t$ (with $d v = 2 \pi i \, d t$) and we get
$$ \begin{align*} &\frac{(c / n)^{-k} n^{-1}}{2 \pi i} \int_{}^{} (v / 2 \pi i)^{-k} \exp\Bigl( \frac{- (2 \pi i)^2 m n}{c^2 v} - v \Bigr) \, d v \\
&= \frac{(2 \pi i n)^{k - 1}}{c^k} \int_{}^{} v^{-k} \exp\Bigr( \frac{4 \pi^2 m n}{c^2 v} - v \Bigr) \, d v.
\end{align*}$$
This is now very close, it just remains to switch $v = -t$ with $d v = - d t$ and we end up with
$$ \frac{(- 2 \pi i n)^{k - 1}}{c^k} \int t^{-k} \exp\Bigl ( t - \frac{4 \pi^2 m n}{c^2 t} \Bigr) \, d t. $$
There's obviously still a fair bit of bookkeeping to keep track of to finish off Lemma 14.2 in Iwaniec-Kowalski, but this is essentially where the Bessel function comes from.
