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.

$$u=\frac{(x-a)}{\sqrt{2a}}\Longrightarrow du=\frac{1}{\sqrt {2a}}dx\Longrightarrow$$
$$\int\limits_0^\infty e^{\frac{-(x-a)^2}{2a}}\,dx=\int\limits_{-\sqrt\frac{a}{2}}^\infty e^{-u^2}du\sqrt{2a}$$
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.
