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I am trying to obtain an analytic estimate of this integral:

$\int_0^1\frac{1}{\sqrt{x}}\exp\left(-a(x-x_0)^2\right) dx$,

where $a\gg1$, $x_0\in[0,1]$. Saddle-point approximation doesn't work due to infinite derivative of $1/\sqrt{x}$ at 0. Any tips on how to get a handle on this will be much appreciated.

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Might the substitution x=t^2 help? – Michael Renardy May 11 '11 at 22:41
That substitution doesn't help with respect to saddle-point approximation. – Mikhail Kagalenko May 11 '11 at 23:07
up vote 1 down vote accepted

Well, now you tell us that $x_0$ depends on $a$. If $x_0$ is of order $1/\sqrt{a}$, set $x_0=y_0/\sqrt{a}$, $x=y/\sqrt{a}$, and your integral becomes $$a^{-1/4}\int_0^{\sqrt{a}}{1\over\sqrt{y}}\exp(-(y-y_0)^2)\,dy.$$ Up to exponentially small error, you can replace the upper bound of $\sqrt{a}$ by $\infty$.

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No, $x_0$ doe not depend on $a$. I mean that I need the value of the integral for all $x_0\in [0,1]$, including the region $x_0\sim 1/\sqrt{a}$. Sorry if my comment was unclear. – Mikhail Kagalenko May 11 '11 at 23:51
If $x_0$ does not depend on $a$, then either $x_0 = 0$ (which does require a different analysis) or $x_0 > 0$, in which case $a x_0^2 \to \infty$ as $a \to \infty$. If you want something for $x_0 \sim 1/\sqrt{a}$, that says $x_0$ does depend on $a$. – Robert Israel May 12 '11 at 1:43
For $x_0 = 0$, i.e. $y_0 = 0$, Maple says $\int_0^\infty \frac{1}{\sqrt{y}} {\rm exp}(-y^2)\, dy = \frac{\pi \sqrt{2}}{2 \Gamma(3/4)}$ so your integral is asymptotic to $\frac{\pi \sqrt{2}}{2 \Gamma(3/4)} a^{-1/4}$. – Robert Israel May 12 '11 at 2:12
If $x_0/\sqrt{a}$ is bounded, the calculation above works. If $x_0/\sqrt{a}$ is large, the saddle point method works. – Michael Renardy May 12 '11 at 2:16
I am interested in this integral as a function of $x_0$ for fixed large, but finite $a$. The interval of values $x_0\in[0, \sqrt{a}]$ is of particular interest. – Mikhail Kagalenko May 12 '11 at 10:03

Saddle-point approximation should work fine, if $x_0 > 0$. If the singularity at 0 bothers you, integrate instead from $x_0/2$ to 1: the integral from $0$ to $x_0/2$ is $O(exp(-a x_0^2/4))$.

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This doesn't produce the correct behaviour in the region $x_0\sim 1/\sqrt{a}$, which I do need. – Mikhail Kagalenko May 11 '11 at 23:25

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