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 $u,v,w > 0$.

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    $\begingroup$ Aren't there supposed to be restrictions on $u,v,$ and $w$? $\endgroup$ – J. M. isn't a mathematician Sep 11 '11 at 12:44
  • $\begingroup$ i think the only restrictions are that the quantity in the square root is positive and the denominator don't vanish! $\endgroup$ – Pablo Sep 11 '11 at 12:51
  • $\begingroup$ If $v$ or $w$ are negative, for instance... $\endgroup$ – J. M. isn't a mathematician Sep 11 '11 at 12:52
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    $\begingroup$ u,v and w are defined in the article, they are positive! $\endgroup$ – Pablo Sep 11 '11 at 12:57
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    $\begingroup$ ...and those are the restrictions I was talking about. I've edited in your correction on the rooted portion. $\endgroup$ – J. M. isn't a mathematician Sep 11 '11 at 13:25

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.):
$$ f(v,w)=\frac{\pi (w-v)}{4 \left((v+w)^2-1\right)\sqrt{4 v w-1} }\;. $$

  • $\begingroup$ Two steps baffle me in your solution: the computation of the last double integral and this rotating the coordinate system thing. Do these correspond more or less to what I explain in mine? $\endgroup$ – Did Sep 11 '11 at 16:29
  • $\begingroup$ I don't know how to handle this integral. I think he means that $v' = \frac{{\sqrt 2 }}{2}v - \frac{{\sqrt 2 }}{2}w$ and $w' = \frac{{\sqrt 2 }}{2}v + \frac{{\sqrt 2 }}{2}w$. The derivate is of v' now. If you do the calculations you get the correct answer. $\endgroup$ – Pablo Sep 11 '11 at 17:15
  • $\begingroup$ @Pablo, The main cause of my bafflement is the direct evaluation of the double integral. Maybe using a symbolic computation program? $\endgroup$ – Did Sep 11 '11 at 17:35
  • $\begingroup$ @Didier Yes, rotating the coordinate system on $\pi/4$ are similar to the calculations in your answer. Namely, it is the coordinate change $\xi=(v+w)/\sqrt2\;$, $\eta=(v-w)/\sqrt2\;$ which simplifies calculations somewhat. $\endgroup$ – Andrew Sep 11 '11 at 17:40
  • $\begingroup$ @Didier As for the integral $$I=\int_0^\infty {\int_0^\infty {\mathrm dp\mathrm dq\sinh(pq)pq}} e^{-vq^2 - wp^2}$$ rewriting it as $$\frac{1}{4wv}\int_0^\infty {\int_0^\infty {\mathrm dp\mathrm dq\sinh(pq)}} \frac{\partial^2}{\partial p\partial q}e^{-vq^2 - wp^2}$$ and integrating twice by parts gives $$I=\frac{I}{4wv}+\frac{1}{4wv} \int_0^\infty {\int_0^\infty {\mathrm dp\mathrm dq\cosh(pq)}} e^{-vq^2 - wp^2}.$$ The last integral can be expressed as 1/4 of an integral over the plane and is equal to $\frac{\pi }{2 \sqrt{4 v w-1}}$. It leads to the answer above. $\endgroup$ – Andrew Sep 11 '11 at 17:41

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 $$ 4I(u,v,w)=\frac{\pi(w-v)u}{((v+w)^2-u^2)\sqrt{4vw-u^2}}. $$

  • $\begingroup$ Can you explain me the formula of quadratic form?If there are other threads about this thing please send me to these. $\endgroup$ – Pablo Sep 11 '11 at 17:12
  • $\begingroup$ This is because the function to be integrated, divided by the RHS, is a probability distribution function (whose integral is 1, by definition), see here and the link therein to multivariate normal distributions. $\endgroup$ – Did Sep 11 '11 at 17:28
  • $\begingroup$ This is one nice solution. Thank you sir. $\endgroup$ – hbp Oct 21 '15 at 3:22

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.

  • $\begingroup$ I did the calculations with derive!It's all right! $\endgroup$ – Pablo Sep 11 '11 at 17:19

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