Define $$ f(n, k)= \int_0^\infty e^{-(n+1)y} (1+y)^n \int_y^\infty e^{-(kn+1)x} (1+x)^{kn} \mathrm dx \mathrm dy, $$ where $k > 0$ is a fixed integer. I would like to estimate the growth $f(n,k)$ as $n \to \infty$.

In the case that $k=1$, we have $$ \begin{align*} f(n, 1) & = \int_0^\infty e^{-(n+1)y} (1+y)^n \int_y^\infty e^{-nx} (1+x)^{n} \mathrm dx \, \mathrm dy \\ & = \int_0^\infty e^{-(n+1)x} (1+x)^n \int_0^x e^{-ny} (1+y)^{n} \mathrm dy \, \mathrm dx \\ & = \int_0^\infty e^{-(n+1)x} (1+x)^n \left( \int_0^\infty e^{-ny} (1+y)^{n} \mathrm dy - \int_x^\infty e^{-ny} (1+y)^{n} \mathrm dy \right) \mathrm dx \\ & = \left(\int_0^\infty e^{-(n+1)x} (1+x)^n \mathrm dx \right)^2 - f(n,1). \end{align*} $$ So we have $$ \begin{align*} f(n,1) & = \frac 1 2 \left(\int_0^\infty e^{-(n+1)x} (1+x)^n \mathrm dx \right)^2 \\ & = \frac 1 2 \left( \left(\frac{e}{n+1}\right)^{n+1} \Gamma(n+1, n+1) \right)^2 \\ & \sim \frac 1 2 \frac {\pi}{2n}, \end{align*} $$ where $\Gamma(a,z)$ is the incomplete Gamma function.

But is there any way to find the first order approximation this double integral for genera $k$?


1 Answer 1


First, change variables $x-y=z$ $$ f(n, k)= \int_0^\infty e^{-(n+1)y} (1+y)^n \int_y^\infty e^{-(kn+1)x} (1+x)^{kn} \mathrm{d}x \mathrm {d}y= f(n, k)= \int_0^\infty \mathrm{d}y e^{-(n+1)y} (1+y)^n \int_0^\infty e^{-(kn+1)(y+z)} (1+y+z)^{kn} \mathrm{d}z, $$ and then rewrite as $$ f(n, k)= \iint_0^\infty \mathrm{d}y\mathrm{d}z\ e^{-2y-z} e^{-n \phi_k(y,z)}\ , $$ where $$ \phi_k(y,z)=y-\log(1+y)+k(y+z)-k\log(1+y+z)\ . $$ The largest contribution to the integral for large $n$ comes from the region where $\phi_k(y,z)$ is as small as possible. Finding the minimum $$ \frac{\partial}{\partial y}\phi_k(y,z)=1-\frac{1}{1+y}+k-k\frac{1}{1+y+z}=0 $$ $$ \frac{\partial}{\partial z}\phi_k(y,z)=k-k\frac{1}{1+y+z}=0 $$ we obtain that the minimum is achieved at $(y^\star,z^\star)=(0,0)$, which corresponds to a boundary point of the integration domain. Expanding $\phi_k(y,z)$ around the point $(y^\star,z^\star)=(0,0)$ up to second order, the function $f(n,k)$ can be approximated as $$ f(n,k)\sim\iint_0^\infty \mathrm{d}x_1\mathrm{d}x_2 \exp\left[-\frac{n}{2}\mathbf{x}^T A\mathbf{x}\right] $$ where $A$ is the Hessian matrix evaluated at $(x^\star,y^\star)$, i.e. $$ A=\begin{pmatrix} \frac{\partial^2 \phi_k(y,z)}{\partial y^2} &\frac{\partial^2 \phi_k(y,z)}{\partial y\partial z}\\\frac{\partial^2 \phi_k(y,z)}{\partial z\partial y} & \frac{\partial^2 \phi_k(y,z)}{\partial z^2} \end{pmatrix}\Big|_{(x,y)=(0,0)} =\begin{pmatrix} 1+k & k\\ k & k\end{pmatrix}\ , $$ and $\mathbf{x}=(x_1,x_2)^T$. Solving the integral, we obtain $$ f(n,k)\sim\frac{\arctan\left(\frac{1}{\sqrt{k}}\right)}{n\sqrt{k}}\ . $$ Note that for $k=1$, we recover $f(n,1)\sim \pi/(4n)$.

  • $\begingroup$ Thanks a lot! I got stuck on this for a while. $\endgroup$
    – ablmf
    Feb 24, 2017 at 10:23

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