Gaussian integral asymptotics

Question

I am trying to derive the asymptotics of $$\int_{2\sqrt{m}}^{\infty}e^{-\frac{x^2}{4}}x^mdx$$ as $m\to\infty$ with no success. I tried integrating by parts, but could get no nice expression. Any help would be very appreciated.

Answer 1

If you make the change of variables $x = 2\sqrt{m} y$ then the integral becomes

$$ \int_{2\sqrt{m}}^\infty e^{-x^2/4} x^m\,dx = \left(2\sqrt{m}\right)^{m+1} \int_1^\infty \exp\left[m\left(\log(y)-y^2\right)\right]\,dy. $$

Over the interval $[1,\infty)$ the exponent $\log(y)-y^2$ has a maximum at $y = 1$. According to the Laplace method we may obtain an asymptotic for the integral by expanding the exponent in Taylor series around this point and using the first two terms. So, since

$$ \log(y)-y^2 = -1 - (y-1) + O(y-1)^2, $$

we replace $\log(y) - y^2$ with $-1-(y-1) = -y$ and find that

$$ \int_1^\infty \exp\left[m\left(\log(y)-y^2\right)\right]\,dy \sim \int_1^\infty \exp(-my)\,dy = m^{-1}e^{-m} $$

as $m \to \infty$. Thus

$$ \int_{2\sqrt{m}}^\infty e^{-x^2/4} x^m\,dx \sim \left(2\sqrt{m}\right)^{m+1} m^{-1} e^{-m} = 2 \left(\frac{2}{e}\right)^m m^{(m-1)/2} $$

as $m \to \infty$.