I am attempting to figure out the first term of the asymptotic expansion at $t \to \infty$ of the following double integral: $$ I(\tau) = \iint_{[\sqrt{\tau},\tau]^2} \frac{e^{2(u+v)} I_1(2 (2\tau - u - v))}{(2\tau - u - v)\sqrt{uv}} d u d v $$
My hypothesis is that: $I(t) \sim e^{4t} \sqrt \frac{\pi}{2t}$
I believe if I could make the assumption 1 that the bessel function $I_1$ can be replaced by its asymptotic form $\frac{e^{z}}{\sqrt{2 \pi z}}$, then it seems quite doable. Indeed Mathematica seems to provide me with an answer for the input:
Integrate[1/(Sqrt[x y] (2 t - x - y)^(3/2)), {y, Sqrt[t], t}, {x, Sqrt[t], t}, Assumptions -> t > 1]
output: $$ \frac{2 \sqrt{2} \left(\tan ^{-1}\left(\frac{\sqrt{2} \sqrt{t-\sqrt{t}}}{\sqrt[4]{t}}\right)-\tan ^{-1}\left(\frac{1}{\sqrt{2} \sqrt{\sqrt{t}-1}}\right)+\cot ^{-1}\left(2 \sqrt{t-\sqrt{t}}\right)\right)}{\sqrt{t}} $$ Therefore, I can run "Series" on top of that and get it is equivalent to $\frac{\sqrt{2} \pi }{\sqrt{t}}$
Actually, it seems I could even make the assumption 2 that it is also equivalent to taking the integration domain to $[0,\tau]^2$ for which I also get the same result with:
Integrate[1/(Sqrt[x y] (2 t - x - y)^(3/2)), {y, 0, t}, {x, 0, t}, Assumptions -> t > 1]
Would you know if there is any justification to make assumption 1 and potentially assumption 2 ? Or would you see another way to tackle such a problem ?
Thank you in advance!
EDIT: A slight improvement to try to reduce the problem to a simple integral. With change of variable $x = \tau - u$, $y = \tau - v$ we find: \begin{equation} e^{-4 \tau} I(\tau) = \iint_{[0,\tau - \sqrt \tau]^2} e^{-2(x+y)} \frac{I_1(2(x+y))}{(x+y)\sqrt{(\tau - x)(\tau - y)}} d u d v \end{equation} Further, with $s=x+y$ we have: \begin{eqnarray} e^{-4 \tau} I(\tau) & = & \int_{0}^{\tau - \sqrt \tau} \int_{x}^{x+\tau-\sqrt \tau} \frac{e^{-2s} I_1(2s)}{s \sqrt{(\tau -x )(\tau + x - s)}} d s d x\\ & = & \int_{0}^{\tau - \sqrt \tau} \int_0^{s} \frac{e^{-2s} I_1(2s)}{s \sqrt{(\tau -x )(\tau + x - s)}} d x d s \\ & + & \int_{\tau - \sqrt \tau}^{2(\tau - \sqrt \tau)} \int_{s-(\tau - \sqrt \tau)}^{\tau - \sqrt \tau} \frac{e^{-2s} I_1(2s)}{s \sqrt{(\tau -x )(\tau + x - s)}} d x d s \\ \end{eqnarray} Now we can "easily" integrate the subintegrals, if I am not wrong we have: \begin{equation} \int \frac{d x}{\sqrt{(\tau -x )(\tau + x - s)}} = 2 \sinh^{-1}\left( \sqrt \frac{\tau - x}{2\tau - s} \right) \end{equation} But not sure how to deal with simple integrals with $I_1$ and $\sinh^{-1}$.