Calculate asymptotic value of an integral of exponential function

Question

In the existing literature [From the Schrödinger problem to the Monge–Kantorovich problem], below the Example 3.4, the author claimed that using Laplace method, we can obtain $$ \int_0^{\infty}e^{-z^p/p}\exp(\zeta z) dz\sim (2\pi(q-1))^{1/2}\zeta^{1-q/2}e^{\zeta^q/q} $$ as $\zeta\to\infty$, where $q$ is a parameter satisfying $1/p+1/q=1$. However, I don't know the detailed deviation step. Also, is it possible to calculate the asymptotic value for the integral $$ \int_0^{\infty}e^{-z^p/p}\exp(\zeta (z + cz^2)) dz, $$ where $c>0$ is some constant?

Answer 1

The strategy is as follows:

• put everything in the integrand in the exponent $e^{f(z)}$, with $f(z)=\zeta z-z^p/p$;
• calculate the saddle point, the (possibly complex) number $z_0$ where $f'(z)=0$; this gives $z_0=\zeta^{1/(p-1)}$;
• expand $f(z)$ to second order around $z_0$, discarding higher order terms; this gives the exponent $g(z)=f(z_0)-a(z-z_0)^2$ with $a=\tfrac{1}{2}(p-1)\zeta^{(p-2)/(p-1)}$;
• carry out the remaining Gaussian integral of $e^{g(z)}$ over $z$, along a "path of steepest descent", which gives the asymptotic expansion as $\zeta\rightarrow\infty$ of the integral $\int e^{f(z)}dz\rightarrow e^{f(z_0)}(\pi/a)^{1/2}$.

The same approach also works for the second integral, but then I am not able to give a closed-form expression for the saddle point at arbitrary $p$; for $p=2$ I arrive at $\exp[\zeta^2(2-4c\zeta)^{-1}]\bigl(2\pi/(1-2c\zeta)\bigr)^{1/2}$.