Finding asymptotic growth using the Laplace Method

Question

I am asked to find the the asymptotic growth of the integral $$I(k,m)=\int_0^1{x^{nk}}{{(1-x)}^{nm}}dx$$ with $k,m>0$ fixed. How do I approach this solution? I know it means finding the growth as $n \to \infty$, but I am clueless on what to do. Does it have something to do with Stirling's approximation?

Answer 1

To actually derive the behavior for large $n$, we begin by rewriting in Laplace's integral form \begin{align} I(k,m)& = \int_0^1{x^{nk}}{{(1-x)}^{nm}}dx \\ & = \int_0^1 e^{n[k \ln x + m \ln(1-x)]} \ dx. \end{align} Define $\phi(k,m,x)$ such that the integral is $e^{n\phi}$. This is maximal when $\partial \phi/\partial x = 0$ and is given by $x_c = \frac{k}{k+m}$. You can check this is maximal. Next we return to the integral and expand $\phi$ to second order. \begin{align} I(k,m)& = \int_0^1{x^{nk}}{{(1-x)}^{nm}}dx \\ & = \int_0^1 e^{n[k \ln x + m \ln(1-x)]} \ dx \\ & \approx \int_{x_c - \epsilon}^{x_c + \epsilon} e^{n[\phi(x_c) - \frac{1}{2}|\phi''(x_c)|(x-x_c)^2]} \ dx \\ & = e^{n \phi(x_c)} \int_{-\epsilon}^{\epsilon} e^{-n|\phi''(x_c)|\frac{1}{2}u^2]} \ du \\ & = e^{n\phi(x_c)} \int_{-\sqrt{|\phi''(x_c)|n}}^{\sqrt{|\phi''(x_c)|n}} e^{-\frac{1}{2}s^2} \ \frac{ds}{\sqrt{|\phi''(x_c)|n}} \\ & \approx \frac{e^{-n\phi(x_c)}}{\sqrt{|\phi''(x_c)|n}} \int_{-\infty}^\infty e^{-\frac{1}{2}s^2} \ ds \\ & = e^{-n\phi(x_c)}\sqrt{\frac{2\pi}{|\phi''(x_c)|n}} \end{align} You can insert the values for $\phi$ evaluated at $x_c$ to check against the asymptotics.

Answer 2

$\newcommand{\B}{\mathcal{B}}$This is the Beta function.

We have that:

$$I(k,m,n)=\B(nk+1,nm+1)=\frac{\Gamma(nk+1)\Gamma(nm+1)}{\Gamma(n(k+m)+2)}$$

Now you can use a Stirling approximation, as $\Gamma(x+1)=x!$, loosely speaking, and the Stirling series indeed provides an asymptotic formula for $\Gamma$.

$$\begin{align}I(k,m,n)&\sim\sqrt{2\pi}\frac{(nk)^{nk}(nm)^{nm}\cdot e^{n(k+m)+1}}{e^{n(k+m)}\cdot(n(k+m)+1)^{n(k+m)+1}}\cdot\sqrt{\frac{(nk)(nm)}{n(k+m)+1}}\\&=e\sqrt{\frac{2\pi km}{k+m+\frac{1}{n}}}\cdot\frac{n^{n(k+m)+1/2}k^{nk}m^{nm}}{(n(k+m)+1)^{n(k+m)+1}}\\&=e\sqrt{\frac{2\pi\cdot km}{n(k+m)+1}}\cdot\frac{(k^km^m)^n}{\left(k+m+\frac{1}{n}\right)^{n(k+m)+1}}\\&=e\sqrt{\frac{2\pi\cdot km}{n(k+m)+1}}\cdot\left(\frac{k^k\cdot m^m}{\left(k+m+\frac{1}{n}\right)^{k+m+1/n}}\right)^n\end{align}$$

This approximation is strikingly accurate even for small $n$ due to the decaying nature of $I$. If you so wished you could have included more terms from the Stirling series for $\Gamma$.