Integrating the exponential $\exp\left(-{(x-a)^2\over 2a}\right)$ 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.
 A: $$u=\frac{(x-a)}{\sqrt{2a}}\Longrightarrow du=\frac{1}{\sqrt {2a}}dx\Longrightarrow$$
$$\int\limits_0^\infty e^{\frac{-(x-a)^2}{2a}}\,dx=\int\limits_{-\sqrt\frac{a}{2}}^\infty e^{-u^2}du\sqrt{2a}$$
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.
A: 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 $\sqrt {2 \pi a}$
