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?
Answer
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.
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