Large $n$ behavior of $\int_0^{\infty}\int_0^{\infty}e^{-x} e^{-y}y^n \cosh\left(2c\sqrt{xy}\right){dx}{dy}$

Question

I stumbled on a double integral $$\int_0^{\infty}\int_0^{\infty}e^{-x} e^{-y}y^n \cosh\left(2c\sqrt{xy}\right){dx}{dy},\quad 0<c<1$$ and wanted to find the exponential term in its asymptotic as $n\rightarrow \infty$ using Laplace's method. First, I made the substitution $x = nt, y =nv$, so we get $$n^{n+2}\int_0^{\infty}\int_0^{\infty}e^{n(-t-v+\ln v)}\cosh(2c\sqrt{tv}\cdot n) {dt}{dv}$$ For large $n$ we can use that $\cosh\left(2c\sqrt{xy}\cdot n\right)\sim \frac{1}{2}\exp\left(2c\sqrt{xy}\cdot n\right)$ and the integral becomes $$\frac{1}{2}n^{n+2}\int_0^{\infty}\int_0^{\infty}e^{n(-t-v+\ln v+2c\sqrt{tv})} {dt}{dv}$$

Now, the maximum of the function in exponent is reached at $$t\rightarrow\frac{c^2}{1-c^2},v\rightarrow\frac{1}{1-c^2}$$ And so the exponential term in the asymptotic should be $$\left(\frac{1}{1-c^2}\right)^n e^{-n}$$ But this doesn't seem to match the numerical evaluations, especially for small values of $c$. What am I doing wrong?

Answer 1

Too long for a comment

Thought the asymtotics was already obtained by @Eric Towers and @Gary, let's do it by Laplace'methode.

$$I(n,c)=\frac{n^{n+2}}2\int_0^\infty\int_0^\infty e^{n(-t-v+\ln v+2c\sqrt{tv})} dtdv=\frac{n^{n+2}}2\int_0^\infty\int_0^\infty e^{-nf(u,v)} dtdv$$ The maximum point was already found: $$t_0=\frac{c^2}{1-c^2}; \,v_0=\frac1{1-c^2}; \,\frac{t_0}{v_0}=c^2;\,t_0v_0=\frac{c^2}{(1-c^2)^2}$$ then $$f''_{tt}(t_0, v_0)=\frac{1-c^2}{2c^2};\,f''_{vv}(t_0,v_0)=(1-c^2)(1-\frac{c^2}2);\,f''_{vt}(t_0,v_0)=\frac{1-c^2}4$$ hence $$I(n,c)\sim\frac{n^{n+2}}2e^{-n}\left(\frac1{1-c^2}\right)^n\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1-c^2}{4c^2}n(t^2-2c^2vt+(2-c^2)c^2v^2)}dvdt\tag{1}$$ The quadratic form is positive definite; its determinant $\det||\,\,||=c^2(2-c^2)-c^4=2c^2(1-c^2)$

and $$I(n,c)\sim\frac{n^{n+2}}2e^{-n}\left(\frac1{1-c^2}\right)^n\frac{4c^2}{n(1-c^2)}\frac\pi{\sqrt{\det||\,\,||}}=\frac{\sqrt2\pi c}{(1-c^2)^\frac32}n^{n+1}e^{-n}\left(\frac1{1-c^2}\right)^n$$

Answer 2

The dominant contribution for large $n$ arises from large values of $y$. Therefore, we compute the large-$y$ asymptotics of the $x$-integral: \begin{align*} \int_0^{ + \infty } {{\rm e}^{ - x} \cosh (2c\sqrt {xy} ){\rm d}x} & = \frac{1}{2}\int_0^{ + \infty } {\exp ( - x + 2c\sqrt {xy} ){\rm d}x} +o(1) \\ & = y\exp (c^2 y)\int_0^{ + \infty } {t\exp ( - y(t^2 - 2ct + c^2 )){\rm d}t} +o(1) \\ & = c\sqrt {\pi y} \exp (c^2 y) + \mathcal{O}(1) \end{align*} as $y\to+\infty$. Substituting this into the original integral gives the asymptotics $$ \int_0^{ + \infty }\!\!\! {\int_0^{ + \infty } {{\rm e}^{ - x} {\rm e}^{ - y} y^n \cosh (2c\sqrt {xy} ){\rm d}x} {\rm d}y} = c\sqrt \pi \frac{{\Gamma \!\left( {n + \frac{3}{2}} \right)}}{{(1 - c^2 )^{n + 3/2} }} + \mathcal{O}(\Gamma (n + 1)), $$ as $n\to+\infty$. Expanding the gamma function for large $n$ yields the asymptotic expansion \begin{multline*} \int_0^{ + \infty }\!\!\! {\int_0^{ + \infty } {{\rm e}^{ - x} {\rm e}^{ - y} y^n \cosh (2c\sqrt {xy} ){\rm d}x} {\rm d}y} \\ \sim \sqrt 2 \pi \frac{c}{{(1 - c^2 )^{3/2} }}n\left( {\frac{n}{{(1 - c^2 ){\rm e}}}} \right)^n \left( {1 + \frac{{11}}{{24n}} - \frac{{23}}{{1152n^2 }} + \ldots } \right), \end{multline*} as $n\to+\infty$.