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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.

Hasek
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    Exercise 7.19 of Apostol's Analysis book (2nd edition) will guide you to prove of the famous identity along the first lines of your problem (i.e, by using only Calculus on the real line). – Mittens Jun 01 '20 at 03:07
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    Are you aware of that "trick" where you convert it to a 2D integral – Calvin Khor Jun 01 '20 at 03:09
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    The Exercise 7.19 mentioned above: Define $$ f(x)=\left(\int_{0}^{x} e^{-t^{2}} d t\right)^{2}, \quad g(x)=\int_{0}^{1} \frac{e^{-x^{2}\left(t^{2}+1\right)}}{t^{2}+1} d t $$ a) Show that $g^{\prime}(x)+f^{\prime}(x)=0$ for all $x$ and deduce that $g(x)+f(x)=\pi / 4$

    b) Use (a) to prove that $$ \lim {x \rightarrow \infty} \int{0}^{x} e^{-t^{2}} d t=\frac{1}{2} \sqrt{\pi} $$

     – Calvin Khor Jun 01 '20 at 03:11
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    There's a number of different proofs here - https://kconrad.math.uconn.edu/blurbs/analysis/gaussianintegral.pdf Maybe someone with better MSE-oriented googlefu can find a duplicate target – Calvin Khor Jun 01 '20 at 03:13
  • While your effort is commendable, it is really weird that you don't look for the easy proof of the known integral of the standard normal distribution function, since you do in fact know that it is called the error function. Even the wikipedia page you linked to mentions the Gaussian integral! – user21820 Jul 26 '20 at 12:17

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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.

John Hughes
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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: $L[t^n]=\int_0^{\infty}e^{-st}t^{n}dt=\dfrac{\Gamma(n+1)}{s^{n+1}}$, $$=\frac{1}{\sqrt{\pi}}L[t^{-1/2}]_{s=1}$$ $$=\frac{1}{\sqrt{\pi}}\left[\frac{\Gamma(-\frac12+1)}{s^{-\frac12+1}}\right]_{s=1}$$ $$=\frac{1}{\sqrt{\pi}}\left[\frac{\Gamma(\frac12)}{s^{\frac12}}\right]_{s=1}$$ $$=\frac{1}{\sqrt{\pi}}\left[\frac{\sqrt{\pi}}{1}\right]\quad (\because \Gamma(1/2)=\sqrt{\pi})$$ $$=1$$

Harish Chandra Rajpoot
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