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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!

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    $\begingroup$ If your intuition is true then using the FindFit function in Mathematica I found $$I(t)\sim\frac{e^{2 t}}{t^{2.4}}$$ $\endgroup$
    – Raffaele
    Nov 2 '20 at 17:50
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    $\begingroup$ the integral seems to simplify to $\pi^2\int_0^1drI_1(rt)I_1((2-r)t)$ where $I_1$ is a Bessel function. Can yout take it from there? $\endgroup$
    – asgeige
    Nov 3 '20 at 8:32
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    $\begingroup$ Hi @asgeige , with modified Bessel of first kind, shouldn't it be $I(t) = \frac14 \pi^2 \int_0^t (I_0(s)-I_2(s)) (I_0(2t-s) - I_2(2t-s)) \mathrm d s = \pi^2 \int_0^t I_1(s) I_1(2t-s) \frac{1}{s(2t-s)} \mathrm d s $ ? It was the original form I was trying to work with, until I thought it might be easier to rewrite it in the form of the Watson Lemma, or Laplace method. $\endgroup$ Nov 3 '20 at 11:20
  • $\begingroup$ @user2673062 right, i was typing on the phone... $\endgroup$
    – asgeige
    Nov 3 '20 at 18:40
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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 $\mathcal{J_2(t)}\ll\mathcal{J_1(t)}$

$$ I(t)\approx C(t)\mathcal{J_1(t)} \approx \frac{C(t)}{2^{3/2}}\\ =\frac{e^{2t}\pi^{3/2}}{4t^{5/2}} $$


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 $I_0(z)-I_2(z)=I_1(z)/(2z)$

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  • $\begingroup$ Thank you for your answer, that was really helpful and I also learned a lot from the method you displayed! $\endgroup$ Nov 5 '20 at 16:12
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As remarked by @user2673062 result in the comments, the triple integral simplifies. By using the integral representation for the modified Bessel function $$I_{\nu}\left(z\right)=\frac{(\frac{1}{2}z)^{\nu}}{\pi^{\frac{1}{2}}\Gamma\left(\nu+\frac{1}{2}\right)}\int_{0}^{\pi}e^{\pm z\cos\theta}(\sin\theta)^{2\nu}\mathrm{d}\theta$$ for $\nu>-1/2$, we can perform the integrations over $x$ and $y$ with $\nu=1$ to obtain: \begin{align} 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\\ &=\frac{\pi^2}{t^2}\int_0^1\frac{I_1(tr)}r\frac{I_1\left(t(2-r)\right)}{2-r}\,dr \end{align} By changing $r\to 2-r$, this integral can be written as \begin{align}I(t)&= \frac{\pi^2}{2t^2}\int_0^2\frac{I_1(tr)}r\frac{I_1\left(t(2-r)\right)}{2-r}\,dr\\ &=\frac{\pi^2}{2t^2}\Phi(2) \end{align} where $$\Phi(\theta)=\int_0^\theta\frac{I_1(tr)}r\frac{I_1\left(t(\theta-r)\right)}{\theta-r}\,dr$$ can be considered as a convolution integral. Its Laplace transform is then \begin{align}\mathcal L[\Phi] &=\left\lbrace \mathcal L\left[\frac{I_1(tx)}x\right]\right\rbrace^2\\ &=\frac1{t^2}(u-\sqrt{u^2-t^2})^2 \end{align} By inversion, we find $$ \Phi(\theta)=\frac2\theta I_0(\theta t)-\frac4{\theta^2 t}I_1(\theta t)$$ These integral transforms were found using a CAS but could certainly be found from classical tables of Laplace transforms (Ederlyi TI, for example). Finally, $$ I(t)=\frac{\pi^2}{2t^2}\left[I_0(2 t)-\frac1{ t}I_1(2t)\right]$$ which seems to be numerically correct. Using the asymptotic expansions for the Bessel functions, we find $$ I(t)\sim \frac{\pi^{3/2}}{4}\frac{e^{2t}}{t^{5/2}}$$ which is not very far from @Raffaele estimation.

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    $\begingroup$ Thank you very much for your answer! Would have liked to be able to accept both answers because yours is actually really elegant and rigorous! Thanks a lot again for your help. $\endgroup$ Nov 5 '20 at 16:14

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