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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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FWIW, Wolfram Alpha gives an asymptotic expansion for large $n$:… – Mariano Suárez-Alvarez Jan 18 '12 at 3:31
@Mariano, am I interpreting that correctly as $O(e^n n^{-3/2})$? I think that may be a bit off, considering the integral is bounded between $0$ and $\pi/2$. – Antonio Vargas Jan 18 '12 at 4:12
Have you tried using an expansion for $e^{-nt}$ in series and then switch the sum and the integral? If you have a formula for $$ \int_0^1 \frac{t^k}{\sqrt{1-t^2}} \, dt $$ perhaps it would become useful. It's just a suggestion at the moment because I'm tired and your question make me think of this. – Patrick Da Silva Jan 18 '12 at 4:13
Heh, funny you found the Beta function thing. Fancy. Are you sure of the divergence? I know nothing about beta functions beside their definition. – Patrick Da Silva Jan 18 '12 at 4:45
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. – Patrick Da Silva Jan 18 '12 at 5:06
up vote 6 down vote accepted

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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Thanks Fabian. Could you go into detail on how to justify that the integral is dominated by small values of $t$? – Antonio Vargas Jan 18 '12 at 16:43
@Antonio Vargas: for $n$ (very) large $e^{-nt_1} \gg e^{-n t_2}$ when $t_1<t_2$ (and this remains even true when multiplied by some algebraic function). – Fabian Jan 18 '12 at 16:51
I think the part I'm getting hung up on is how to estimate the contribution of the pole at t=1 and then conclude that it is negligible. For example, I tried showing that I may extend the integral to $\infty$ with small error, but I can only show that $$\left|\int_{1+\delta}^{\infty} (1-t^2)^{-1/2} e^{-nt} \,dt \right| \leq \frac{1}{\sqrt{\delta}} \cdot \frac{1}{n e^n}$$ and hence that $$\int_{1+\frac{1}{n}}^{\infty} (1-t^2)^{-1/2} e^{-nt} \,dt = O(n^{-1/2} e^{-n}).$$ Is this enough to conclude that the contribution of $\int_1^{\infty}$ is negligible? – Antonio Vargas Jan 18 '12 at 17:14
@AntonioVargas: this is exactly my statement. The other contributions are exponentially small. The contributions from $t\approx 0$ give $S_0=\sum_m a_m n^{-m}$ and the contribution from $t\approx 1$ give $S_1=e^{-n}\sum_m a_m n^{-m}$. For $n\to\infty$, $S_1 \ll S_0$ (exponents kill everything algebraic). – Fabian Jan 18 '12 at 17:17
Fabian, thank you so much for your help :) – Antonio Vargas Jan 18 '12 at 17:19

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