An approximation of an integral

Question

Is there any good way to approximate following integral?
$$\int_0^{0.5}\frac{x^2}{\sqrt{2\pi}\sigma}\cdot \exp\left(-\frac{(x^2-\mu)^2}{2\sigma^2}\right)\mathrm dx$$
$\mu$ is between $0$ and $0.25$, the problem is in $\sigma$ which is always positive, but it can be arbitrarily small.
I was trying to expand it using Taylor series, but terms looks more or less this $\pm a_n\cdot\frac{x^{2n+3}}{\sigma^{2n}}$ and that can be arbitrarily large, so the error is significant.

Answer 1

A standard way to get a good approximation for integrals that "look" Gaussian is to evaluate the Taylor series of the logarithms of their integrands through second order, expanding around the point of maximum value thus (continuing with @Ross Millikan's substitution):

$$\eqalign{ &\log\left(\sqrt{y}\cdot \exp\left(-\frac{(y-\mu )^2}{2\sigma ^2}\right)\right) \cr = &\frac{-\mu ^2-\sigma ^2+\mu \sqrt{\mu ^2+2 \sigma ^2}+2 \sigma ^2 \log\left[\frac{1}{2} \left(\mu +\sqrt{\mu ^2+2 \sigma ^2}\right)\right]}{4 \sigma ^2} \cr + &\left(-\frac{1}{2 \sigma ^2}-\frac{1}{\left(\mu +\sqrt{\mu ^2+2 \sigma ^2}\right)^2}\right) \left(y-\frac{1}{2} \left(\mu +\sqrt{\mu ^2+2 \sigma ^2}\right)\right)^2 \cr + &O\left[y-\frac{1}{2} \left(\mu +\sqrt{\mu ^2+2 \sigma ^2}\right)\right]^3 \cr \equiv &\log(C) - (y - \nu)^2/(2\tau^2)\text{,} }$$

say, with the parameters $C$, $\nu$, and $\tau$ depending on $\mu$ and $\sigma$ as you can see. The resulting integral now is a Gaussian, which can be computed (or approximated or looked up) in the usual ways. The approximation is superb for small $\sigma$ or large $\mu$ and still ok otherwise.

The plot shows the original integrand in red (dashed), this approximation in blue, and the simpler approximation afforded by replacing $\sqrt{y} \to \sqrt{\mu}$ in gold for $\sigma = \mu = 1/20$.

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(Added)

Mathematica tells us the integral, when taken to $\infty$, can be expressed as a linear combination of modified Bessel Functions $I_\nu$ of orders $\nu = -1/4, 1/4, 3/4, 5/4$ with common argument $\mu^2/(4 \sigma^2)$. From the Taylor expansion we can see that when both $\mu$ and $\sigma$ are small w.r.t. $1/2$--specifically, $(1/4-\mu)/\sigma \gg 3$, the error made by including the entire right tail will be very small. (With a little algebra and some simple estimates we can even get good explicit bounds on the error as a function of $\mu$ and $\sigma$.) There are many ways to compute or approximate Bessel functions, including polynomial approximations. From looking at graphs of the integrand, it appears that the cases where the Bessel function approximation works extremely well more or less complement the cases where the preceding "saddlepoint approximation" works extremely well.

Answer 2

If you write y=x^2 and pull the constants out you have $$\frac{1}{2\sqrt{2\pi}\sigma}\int_0^{0.25}\sqrt{y}\cdot \exp(-\frac{(y-\mu )^2}{2\sigma ^2})dy$$ If $\sigma$ is very small, the contribution will all come from a small area in $y$ around $\mu$. So you can set $\sqrt{y}=\sqrt{\mu}$ and use your error function tables for a close approximation. A quick search didn't turn up moments of $\sqrt{y}$ against the normal distribution, but maybe they are out there.

Answer 3

Not an answer, but might still be helpful. Using the variable substitution that Ross mentions in his answer, we can treat a simpler case $\mu=0$ more easily.

For the following integral, Wolfram Alpha tells us that (I hope I did not make a transcription error here):

$$\int_0^\infty \sqrt{y}e^{-y^2/(2\sigma^2)}dy = \frac{\sigma^{3/2}\Gamma(3/4)}{2^{1/4}},$$

But your problem goes from $0$ to $0.25$, so some approximations are needed.

Link-text: http://www.wolframalpha.com/input/?i=Integrate[Sqrt[x]+Exp[-x^2%2F%282+s^2%29]%2C{x%2C0%2Cinf}]

Answer 4

How about some good old-fashioned trapezoid rule?