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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.

Antonio Vargas
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    FWIW, Wolfram Alpha gives an asymptotic expansion for large $n$: http://www.wolframalpha.com/input/?i=expansion+of+Integrate%5B%281-t%5E2%29%5E%28-1%2F2%29Exp%5B-nt%5D%2C+%7Bt%2C+0%2C+1%7D%5D+near+n%3Dinfinity – Mariano Suárez-Álvarez Jan 18 '12 at 03:31
  • @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 04:12
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    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 04:13
  • @Patrick, those integrals are Beta functions:

    $$\int_0^1 (1-t^2)^{-1/2} t^k ,dt = \frac{1}{2}B!\left(\frac{1}{2},\frac{k+1}{2}\right).$$

    But I'm not sure you can switch the sum and the integral, mostly because the resulting series

    $$\frac{\pi}{2}-\frac{n B(1/2,1)}{2}+\frac{n^2 B(1/2,3/2)}{4}-\frac{n^3 B(1/2,2)}{12} + \cdots$$

    diverges as $n \to \infty$.

     – Antonio Vargas Jan 18 '12 at 04:42
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    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 04:45
  • @Patrick: You're right, I spoke too soon. According to WA, the series converges to

    $$\frac{\pi}{2}(I_0(n)-L_0(n)+1),$$

    where $I_0$ is the modified Bessel function of the first kind of order $0$, and $L_0$ is the modified Struvel function of order $0$ (that one's certainly new to me!). The graph of this seems to indicate convergence as $n\to\infty$.

    Link to WA: http://preview.tinyurl.com/6shhvc5 (I hope tinyurl isn't frowned upon here.)

    However, estimating this by looking up the relevant expansions wouldn't help me learn about handling integrals, which was my goal for this question.

     – Antonio Vargas Jan 18 '12 at 04:59
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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.

     – Patrick Da Silva Jan 18 '12 at 05:06

1 Answers1

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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 $n\to-\infty$...

Fabian
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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
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    @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
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    @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
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    Fabian, thank you so much for your help :) – Antonio Vargas Jan 18 '12 at 17:19
  • @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). – Fabian Jan 18 '12 at 17:20