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How can I show that $$\int_0^\infty\exp\left(-{(x-a)^2\over 2a}\right)dx$$ can be approximated by $$\sqrt{2\pi a} \,\,\,e^{-a}$$ when $a\to \infty$?


It looks suspiciously similar to $$e^{-a}\int_{-\infty}^\infty \exp\left(-{t^2\over 2a}\right)dt$$ (given the result) but I can't see how to convert the integral to this form, esp the limits seem a bit problematic...

Please help, thanks.

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2 Answers 2

$$u=\frac{(x-a)}{\sqrt{2a}}\Longrightarrow du=\frac{1}{\sqrt {2a}}dx\Longrightarrow$$

$$\int\limits_0^\infty e^{\frac{-(x-a)^2}{2a}}\,dx=\int\limits_{-\sqrt\frac{a}{2}}^\infty e^{-u^2}du\sqrt{2a}$$

But

$$\lim_{a\to\infty}\int\limits_{-\sqrt\frac{a}{2}}^\infty e^{-u^2}du=\sqrt \pi$$

so the above limits seems to be $\,\infty\,$...Check if this helps somehow, or whether you have written the correct expressions...or, of course, whether I'm wrong.

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I don't think this is true. If $a$ is huge and positive the integral will be about $\sqrt{2 \pi a}$ because we can take the lower limit to $-\infty$ without changing much. Are you sure of the $e^{-a}$ term? Without it, you can bound the error made by making the lower limit $-\infty$ because $0$ is $\sqrt a$ standard deviations below the mean.

Added: The $(x-a)^2$ in the numerator shifts the peak of the Gaussian to $+a$. The $a$ in the denominator increases the standard deviation to $\sqrt a$. You can see that by making the change of variable $t=\frac {x-a}{\sqrt a}$. Your integral becomes $\sqrt a\int_{-a}^\infty e^{-\frac {t^2}2}dt$. Now as $a \to \infty$, the lower limit goes far left, so there isn't much area there. The error in changing to lower limit to $-\infty$ becomes zero. You can make this explicit by looking at the normal distribution. You are $\sqrt a$ standard deviations below the mean. This gives the result that the integral gets close to $\sqrt {2 \pi a}$

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Thank you, Ross. It is quite possible that there is a typo on the sheet I have -- it happens sometimes. Would you mind explaining the error and lower bounds etc.? –  Ella Feb 1 '13 at 21:24
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