Integrals involving exponential functions and the gamma function I'm having trouble evaluating this integral
$$\int_0^\infty {e^{-ax^2}} \,dx $$
My guess is that it would evaluate into something like
$$\int_0^\infty \frac 12e^{-s}s^{\frac 12} \ldots \,dx = \frac {\Gamma\left(\frac 12\right)}{\frac{a^{\frac 12}}{2}}$$ 
When you do a substitution $ \sqrt{s}= \sqrt{a}x $ so that $ s = ax^2 $. I'm having trouble convincing myself though that $ \frac {d}{ds}\sqrt{s} = \left(\ldots a^\frac 12\right) $ which would satisfy the answer that I provided.
Am I doing something wrong or is my guess wrong?
 A: The trick (at least one of them) is to write:
$$I = \int_0^\infty {e^{-ax^2}} dx$$ and since the variable doesn't matter,
$$I = \int_0^\infty {e^{-ay^2}} dy$$
So that 
$$I^2=\int_0^\infty {e^{-ax^2}} dx\int_0^\infty {e^{-ay^2}} dy=\int_0^\infty\int_0^\infty {e^{-a(x^2+y^2)}}dx dy$$.
We then switch to circular polar coordinates so that $dxdy=r drd\theta$ and $x^2+y^2=r^2$ and then $I^2$ becomes, with appropriate limits to cover only the first quadrant:
$$I^2 = \int_0^\infty\int_0^{\pi/2} e^{-ar^2} r dr d\theta = {\pi\over 2}\int_0^\infty e^{-ar^2} r dr$$ and then use the substitution $u=r^2$ and $du=2r dr$:
$$I^2 = {\pi\over4} \int_0^\infty e^{-au}du = {\pi\over4} \left({1\over a}\right)$$
So finally, 
$$I = \int_0^\infty {e^{-ax^2}} dx = \sqrt{\pi\over 4a}$$
A: If you want to use the Gamma function the substitution is $a x^2 = t$, so “$dx = \frac{1}{2\sqrt{a}}t^{-1/2}dt$".
Then the integral appears as,
$$\frac{1}{2\sqrt{a}} \int_0^\infty dt\,t^{-1/2}e^{-t} = \frac{1}{2\sqrt{a}}\Gamma(1/2)\ .$$
That's all.
