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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.

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    I still know nothing about those functions, so I just hope you had fun =) I am not very much into integrals, mostly those I can't handle. =P But I see that I was out of context. Since you spoke of $-\infty$ to $\infty$ cases, perhaps an appropriate change of variable could be considered? Not at all a directed guess, just an idea. I have absolutely no reason or faith that this is the right direction though.2012-01-18

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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 $n\to-\infty$...

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    @AntonioVargas: just to give a bit more information. For the asymptotic expansion of an integral, you have to check either saddle points inside the integration region AND the boundary terms (you cannot deform your integration contour away from the initial and final point).2012-01-18