Q:

What is $\int_{0}^1 x^{- m x} dx$ when $m$ is large?

I am looking for an estimate to the following definite integral:
$$I(m) = \int_{0}^1 x^{-m x} dx$$
as a function of $m$, asymptotically when $m$ is large.
I found out that the maximum value of the integrand is attained when $x=1/e$, and it is $M(m) := exp({m\over e})$.  So $J(m) := I(m) / M(m)$ is finite. Moreover, by plotting the integrand of $J(m)$ it seems that $J(m) \to 0$ as $m\to \infty$, so:
$$
I(m) \in o\left(exp({m\over e})\right)
$$
What is a more accurate estimate of $I(m)$?

A:

This is a laplace method integral. You can write the integrand as $e^{-mg(x)}$ for $g(x)=x\log x.$ Then you Taylor expand $g(x)$ about its local minimum at $x=1/e$ as $$g(x)\approx -1/e +\frac{e}{2}(x-1/e)^2.$$ Finally changing the limits of the integral to $\pm\infty$ only introduces exponential error terms. So you have a Gaussian integral to do and you get $$ I(m) = \sqrt{\frac{2\pi }{em}} e^{m/e}(1+ O(1/\sqrt m))$$

