Show that $\int\limits_{-\infty}^\infty e^{-\pi x^2}dx = 1$

Question

I want to show that $\int\limits_{-\infty}^\infty e^{-\pi x^2}dx = 1$.

By definition $$\int\limits_{-\infty}^\infty e^{-\pi x^2}dx = \lim\limits_{t\to\infty}\int\limits_{-t}^t e^{-\pi x^2}dx$$ and since the integrand $e^{-\pi x^2}$ is an even function $$\int\limits_{-\infty}^\infty e^{-\pi x^2}dx = \lim\limits_{t\to\infty}\int\limits_{-t}^t e^{-\pi x^2}dx = 2\lim\limits_{t\to\infty}\int\limits_0^t e^{-\pi x^2}dx$$ i.e. we can equivalently show that $\lim\limits_{t\to\infty}\int\limits_0^t e^{-\pi x^2}dx=\frac{1}{2}$.

Since the antiderivative of $e^{-x^2}$ is given by the error function we can't straightforwardly evaluate the integral, so I tried to use the power series expansion, hoping to be able to see that the resulting series will converge to $\frac{1}{2}$:

$$|\int\limits_0^t e^{-\pi x^2}dx-\frac{1}{2}| = |\int\limits_0^t\sum\limits_{n=0}^\infty\frac{\pi^n\cdot x^{2n}}{n!}dx - \frac{1}{2}| = |\sum\limits_{n=0}^\infty\frac{\pi^n\cdot t^{2n+1}}{n!\cdot(n+1)}-\frac{1}{2}|$$

However, I'm in a doubt that it converges and a quick check in Wolfram Mathematica shows indeed that with $t\to\infty$ the resulting series will diverge.

What am I doing wrong? Can anybody help me with a proof for this problem? Any help will be really appreciated.

Answer 1

How about doing a substitution, $x = \frac{1}{\sqrt{\pi}} u; dx = \frac{1}{\sqrt{\pi}} du$?

That'll convert your integral into an integral of $exp(-u^2)$, which is just the error function, whose value "at infinity" is well known.

Answer 2

Let $\pi x^2=t\implies dx=\dfrac{t^{-1/2}dt}{2\sqrt{\pi}}$ $$\int_{-\infty}^{\infty}e^{-\pi x^2} dx=2\int_{0}^{\infty}e^{-\pi x^2} dx$$ $$=2\int_{0}^{\infty}e^{-t}\dfrac{t^{-1/2}dt}{2\sqrt{\pi}}$$ $$=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}e^{-t}t^{-1/2} dt$$ Using Laplace transform: $L[t^n]=\int_0^{\infty}e^{-st}t^{n}dt=\dfrac{\Gamma(n+1)}{s^{n+1}}$, $$=\frac{1}{\sqrt{\pi}}L[t^{-1/2}]_{s=1}$$ $$=\frac{1}{\sqrt{\pi}}\left[\frac{\Gamma(-\frac12+1)}{s^{-\frac12+1}}\right]_{s=1}$$ $$=\frac{1}{\sqrt{\pi}}\left[\frac{\Gamma(\frac12)}{s^{\frac12}}\right]_{s=1}$$ $$=\frac{1}{\sqrt{\pi}}\left[\frac{\sqrt{\pi}}{1}\right]\quad (\because \Gamma(1/2)=\sqrt{\pi})$$ $$=1$$