I am attempting to figure out the first term of the asymptotic expansion at $t \to \infty$ of the following triple integral:
$$
I(t) = \int_0^1 \int_0^\pi \int_0^\pi e^{[r\cdot(\cos(x)-\cos(y))+2 \cos(y)]\cdot t} \sin^2(x) \sin^2(y) \mathrm d x \mathrm dy \mathrm dr
$$
Intuitively, I am expecting to find something like $I(t)\sim \gamma e^{2t} t^\alpha$ with some constants $\alpha, \gamma$, as the exponential is most certainly going to dominate for such values (for instance when $y=0, r=0$)
As for solving it, I am considering using Laplace's method – or at least a variant of it. Now, it's a bit puzzling because the inward term in the exponential function : $f(r,x,y) = r\cdot(\cos(x)-\cos(y))+2 \cos(y)$ has some issues: its maximum value 2 is reached for all values for which we have $y=0$ and either $x=0$ or $r=0$ – so that means potentially two segments on which the exponential is maximal (while the link provides inputs for a single point). Secondly, these points are on the border of the integration surface.
Would you know another variant of this method to solve this properly?
Thank you in advance!
Best Answer
Using the representation of the integral in terms of modified Bessel functions we have:
$$ I(t)=\frac{\pi^2}{t^2}\int_0^1dr \frac{I_1(tr)I_1(t(2-r))}{r(2-r)} $$
Observation: $t(2-r)\gg1$ for $r\in(0,1)$ so we won't do much harm if we just replace $$ I_1(t(2-r))\approx\frac{e^{2t-r}}{\sqrt{2t\pi(2-r)}}$$ (I am ignoring $O$'s for the moment but all statements can be made precise with a bit of extra work),
We get $$ I(t)=\underbrace{\frac{\pi^{3/2} e^{2t}}{\sqrt{2}t^{5/2}}}_{C(t)}\underbrace{\int_0^1dr \frac{I_1(tr)e^{-tr}}{r(2-r)^{3/2}}}_{\mathcal{J(t)}} $$
We now split this integral at $1/t \ll\delta \ll 1$: $$ \mathcal{J(t)}=\mathcal{J_1(t)}+\mathcal{J_2(t)} $$ We get (Lemma 1 plus large argument asymptotics): $$ \mathcal{J_1(t)}\approx \frac{1}{2^{3/2}}\int_0^{\delta} I_1(tr)e^{-tr}/r=\frac{1}{2^{3/2}}((1+O(\sqrt{\delta t}^{-1})) $$
for $\mathcal{J_2(t)}$ we can use again large argument asymptotics for $I_1(rt)$: $$\mathcal{J_2(t)}\approx\frac{1}{\sqrt{2\pi t}}\int_{\delta}^1\frac{dr}{r^{3/2}(2-r)^{3/2}}=\frac{\eta}{\sqrt{2\pi t}}-O(\delta^{5/2}) $$
where $\eta$ is a constant independet of $t$. Finally since $\mathcal{J_2(t)}\ll\mathcal{J_1(t)}$
Lemma 1: $$ \int \frac{dz}ze^{-z}I_1(z)=Const-e^{-z}(I_1(z)+I_0(z)) $$
Proof:
Differentiate w.r.t. $z$ and use $I_0'(z)=I_1(z), \,\, 2 I_1'(z)=I_0(z)+I_2(z)$ as well as $I_0(z)-I_2(z)=I_1(z)/(2z)$