Estimating the integral $\int_0^1 (1-t^2)^{-1/2} e^{-nt} \,dt$ for large $n$.

I would like to find the asymptotic behavior of the integral

$$\int_0^1 (1-t^2)^{-1/2} e^{-nt} \,dt$$

for large $n$. It seems reasonably obvious that the integral goes to zero. At least it is bounded; the integral is between $0$ and

$$\int_0^1 (1-t^2)^{-1/2} \,dt = \pi/2.$$

I am just learning asymptotic methods and I'm having trouble even approaching this. I thought that Laplace's method might be appropriate but only the case of $\int_{-\infty}^{\infty}$ is discussed in the books I have.

Full, detailed steps would be greatly appreciated. My goal is to try to estimate a slightly more complicated integral.

Solution 1:

For large $n$ the integral is dominated at points where $\text{Re} t$ is smallest. In your case this is at $t=0$. Thus to get the correct asymptotic expansion (up to exponential accuracy), you need to expand the integrand around $t=0$: $$\int_0^1 \!dt\,\frac{e^{-nt}}{\sqrt{1-t^2}} = \int_0^1 \!dt\,e^{-nt} \left[ 1 + \frac{t^2}2 + O(t^4) \right].$$ Next step is to note that we introduce only exponential small errors (in $e^{-n}$) by extending the integral up to $t=\infty$. Thus, we have $$\int_0^1 \!dt\,\frac{e^{-nt}}{\sqrt{1-t^2}} \sim \int_0^\infty \!dt\,e^{-nt} \left[ 1 + \frac{t^2}2 + O(t^4)\right]= n^{-1} + n^{-3} + O(n^{-5}). $$

This asymptotic expansion (because it only involves integer powers in $n$) you could as well have obtained by successive integrating (integrating $e^{-nt}$ and differentiating the rest).

Much more interesting is the asymptotic expansion for $n\to-\infty$...