Evaluating the integral $I(u,v,w)=\iint_{(0,\infty)^2}\sinh(upq) e^{-vq^2 - wp^2}pq(q^2-p^2)^{-1}dpdq$ I found in an article "Imperfect Bose Gas with Hard-Sphere Interaction", Phys. Rev. 105, 776–784 (1957) the following integral, but I don't know how to solve it. Any hints?
$$\int_0^\infty {\int_0^\infty  {\mathrm dp\mathrm dq\frac{\sinh(upq)}{q^2 - p^2}pq} } e^{-vq^2 - wp^2}  = \frac{\pi}{4}\frac{u(w - v)}{\left[(w + v)^2-u^2 \right]\left(4wv-u^2\right)^{1/2}}$$
for $u,v,w > 0$.
 A: First of all one can note that the integral converges and is a differentiable function of parameters for $4vw>u^2$. With the change of variables $p\to p' u^{1/2}$, $q\to q' u^{1/2}$ the general case be reduced to $u=1$. Now denoting the lhs
$$
f(v,w)=\int_0^\infty {\int_0^\infty  {\mathrm dp\mathrm dq\frac{\sinh(pq)}{q^2 - p^2}pq} } e^{-vq^2 - wp^2}
$$
we have 
$$
\frac{\partial}{\partial v}f(v,w)-\frac{\partial}{\partial w}f(v,w)=
\int_0^\infty {\int_0^\infty  {\mathrm dp\mathrm dq\sinh(pq)pq}} e^{-vq^2 - wp^2}
=\frac{\pi }{2 (4 v w-1)^{3/2}},
$$
the integral converging for $vw>1/4$. 
Since $f(w,v)=-f(v,w)$ we have $f(v,v)=0\;$. The solution of this Cauchy problem can be obtained in the standard way (rotating the coordinate system on $\pi/4$ etc.):
$$
f(v,w)=\frac{\pi  (w-v)}{4 \left((v+w)^2-1\right)\sqrt{4 v
   w-1} }\;.
$$
A: I will assume that $u, v, w > 0$ and $4vw > u^2$. Let $$
I := \int_{0}^{\infty} \int_{0}^{\infty} \frac{\sinh (upq)}{p^2 - q^2} \, pq \, e^{-vp^2} e^{-wq^2} \; dpdq.
$$ By polar coordinate transform, we obtain $$
\begin{eqnarray*}
I & = & \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{\sinh (r^2 u \cos \theta \sin \theta)}{r^2 \cos^2 \theta - r^2 \sin^2 \theta} \, r^2 \cos\theta \sin \theta \, e^{-vr^2 \cos^2 \theta} e^{-w r^2 \sin^2 \theta} \; r dr d\theta \\
& = & \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\tan \theta}{1 - \tan^2 \theta} \int_{0}^{\infty} \sinh(r^2 u \sin \theta \cos\theta) \, e^{-r^2 (v \cos^2 \theta + w \sin^2 \theta)} \; d(r^2) d\theta \\
& = & \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\tan \theta}{1 - \tan^2 \theta} \left( \frac{u \cos\theta \sin\theta}{(v \cos^2 \theta + w \sin^2 \theta)^2 - u^2 \cos^2 \theta \sin^2 \theta} \right) d\theta \\
& = & \frac{1}{4} \int_{-\infty}^{\infty} \frac{u t^2}{(1-t^2)\left( (v + w t^2)^2 - u^2 t^2 \right)}\; dt. \qquad (\text{where} \ t = \tan \theta)
\end{eqnarray*}
$$ Now the last integral can be attacked by standard contour integration techinque. In particular, let $$ f(z) = \frac{u}{4} \frac{z^2}{(1-z^2)\left( (v + w z^2)^2 - u^2 z^2 \right)} $$ be the integrand. Then considering appropriate upper-semicircular contour with vanishing dents at $\pm 1$, we obtain $$ \begin{align*} I = & \pi i \Bigg[ \mathrm{Res} \left\{ f, 1 \right\} + \mathrm{Res} \left\{ f, -1 \right\} \Bigg] \\ & + 2\pi i \Bigg[ \mathrm{Res} \left\{ f, \frac{u+i\sqrt{4vw-u^2}}{2w} \right\} + \mathrm{Res} \left\{ f, \frac{-u+i\sqrt{4vw-u^2}}{2w} \right\} \Bigg], \end{align*}$$
which yields the desired formula. (A tip : $\mathrm{Res} \{ f, 1 \} + \mathrm{Res} \{ f, -1 \} = 0$ because $\pm 1$ are simple poles of an even function $f$.)
p.s. While posting my solution, Andrew gave a nice solution.
A: Call $I(u,v,w)$ the integral to compute and note that this can be defined only when $4vw>u^2$. Using the definition of $\sinh$ and the parity of the function to be integrated one sees that
$$
4I(u,v,w)=\int_{-\infty}^\infty\int_{-\infty}^\infty{\mathrm dp\mathrm dq}\,\frac{e^{upq}}{q^2 - p^2}pq\, e^{-vq^2 - wp^2},
$$
that is,
$4I(u,v,w)=\partial_uJ(u,v,w)$ with
$$
J(u,v,w)=\iint{\mathrm dp\mathrm dq}\,\frac{e^{upq}}{q^2 - p^2}\, e^{-vq^2 - wp^2}.
$$
The function $J(u,\cdot,\cdot)$ is symmetric and
$$
\partial_wJ(u,v,w)-\partial_vJ(u,v,w)=\iint{\mathrm dp\mathrm dq}\,e^{upq}\,e^{-vq^2-wp^2}.
$$
The exponent in the exponential is a quadratic form in $(p,q)$ and one knows that
$$
\iint e^{-\frac12\xi^*C\xi}\,\text{d}\xi=2\pi\det(C)^{-1/2},
$$
hence
$$
\partial_wJ(u,v,w)-\partial_vJ(u,v,w)=\frac{2\pi}{\sqrt{4vw-u^2}}.
$$
This is enough to recover $J(u,v,w)$, hence $I(u,v,w)$. Since $J(u,\frac12(v+w),\frac12(v+w))=0$ by symmetry,  one gets $J(u,v,w)$ as an integral of $\partial_tJ(u,\frac12(v+w)-t,\frac12(v+w)+t)$, that is,
$$
J(u,v,w)=\int\limits_{0}^{(w-v)/2}\frac{2\pi \text{d}t}{\sqrt{4\left(\frac12(v+w)+t\right)\left(\frac12(v+w)-t\right)-u^2}},
$$
which is
$$
J(u,v,w)=\int\limits_{0}^{w-v}\frac{\pi \text{d}t}{\sqrt{s^2-t^2}},\quad
s^2=(v+w)^2-u^2.
$$
Hence,
$$
J(u,v,w)=\pi\text{Arcsin}\left(\frac{w-v}{s}\right).
$$
Differentiating this with respect to $u$ yields finally
$$
4I(u,v,w)=\frac{\pi(w-v)u}{((v+w)^2-u^2)\sqrt{4vw-u^2}}.
$$
