Gaussian integral with Feynman's technique for integration

Question

I tried to solve the gaussian integral with Feynman's technique for integration, and now I'm stuck. Here's what I've done: $$I(a)=\int_{-\infty}^\infty{e^{-ax^{2}}dx}=2\int_0^\infty{e^{-ax^{2}}dx}$$ $$I'(a)=2\int_0^\infty{\frac{\partial}{\partial x}e^{-ax^{2}}dx}=2\int_0^\infty{-x^2e^{-ax^{2}}dx}$$ (Here I used integration by parts, differenciating $x$, and integrating $-2xe^{-ax^2}$) $$I'(a)=\Bigl[x\frac{1}{a}e^{-ax^2}\Bigr]_0^\infty-\frac{1}{a}\int_0^\infty{e^{-ax^2}dx}$$ The limit evaluates to zero, and the lower bound is also 0, so only the integral remains, which is the half of the original integral: $$I'(a)=-\frac{1}{2a}I(a)$$ $$\frac{I'(a)}{I(a)}=-\frac{1}{2a}$$ $$\int{\frac{I'(a)}{I(a)}da}=-\frac{1}{2}\int{\frac{da}{a}}$$ $$\ln(I(a))=-\frac{1}{2}\ln(a)+c_1$$ $$I(a)=a^{-\frac{1}{2}}c_2=\frac{c_2}{\sqrt{a}}$$ And here I am stuck, because I don't know how to figure out what the value of $c_2$ is, and the only initial value I know is as $a\to \infty$, but that isn't useful. If someone could nudge me in the right direction, I would appreciate that.

Answer 1

If your heart's set on a solution using Feynman's trick, note$$\int_0^\infty re^{-ar^2}dr=\frac{1}{2a}\implies\int_0^\infty r^3e^{-ar^2}dr=\frac{1}{2a^2}.$$So$$-I(a)I^\prime(a)=\int_{\Bbb R^2}x^2e^{-ar^2}dxdy=\int_0^{2\pi}\cos^2\theta d\theta\int_0^\infty r^3e^{-ar^2}dr=\frac{\pi}{2a^2}.$$Integrating, $I^2(a)=\frac{\pi}{a}+C$. Since $I>0$ and $\lim_{a\to\infty}I(a)=0$, $I=\sqrt{\frac{\pi}{a}}$.

Answer 2

You proved that $I(a)=\frac{I(1)}{\sqrt{a}}$ (notice that if you substitude $u=\sqrt{a}x$, it is way faster). There are many ways to find $I(1)$, for instance with polar coordinates : $$ I(1)^2=\int_{-\infty}^{+\infty }\int_{-\infty}^{+\infty}e^{-(x^2+y^2)}dxdy=\int_0^{+\infty}\int_0^{2\pi}e^{-r^2}rdr d\theta=2\pi\int_0^{+\infty}re^{-r^2}dr=\pi$$ so that $I(1)=\sqrt{\pi}$.