Prove $\int e^{x^{2}} = \sqrt{\pi}$

Question

I have to prove $\int e^{x^{2}} = \sqrt{\pi}$.

I aready know that $\int e^{x^{2}} = \frac{\sqrt{\pi} * erfi(x)}{2}$. What i cannot understand is how $\frac{erfi(x)}{2}$ get cancelled.

Thank you.

Answer 1

$erf$ is a name we sometimes give to the primitive (anti-derivative) of the function $x \mapsto e^{-x^2}$, which cannot be expressed in the usual way. The reason we give it a name is because we know a few interesting properties about it. But in any case, since we don't know its expression, it's not gonna help us compute the integral of $e^{-x^2}$. You have to proceed otherwise...

Here's a hint: think about the expression $\left(\displaystyle\int_{-\infty}^{+\infty} e^{-x^2} \mathrm{d}x\right)^2$, and try to turn it into a simple integral, instead of a product of two integrals...

Answer 2

I'm assuming you meant ${\int e^{-x^2}dx}$ (correct me if I'm wrong). When you talk about the expression $$ \int e^{-x^2}dx $$ This is not going to just be ${\sqrt{\pi}}$. This is an anti-derivative - that is, $$ \frac{d}{dx} \int e^{-x^2}dx = e^{-x^2} $$ But clearly $$ \frac{d}{dx}\sqrt{\pi}=0 $$ What your professor is probably meaning to ask you is to show $$ \int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi} $$ which is true. There is surprisingly more than one way to calculate this integral - polar coordinates, feynmans trick.... so it's difficult to give you the right guidance towards an answer that complements the material you have covered. Please edit the question and add more context

Answer 3

you have a simple way of showing this with double integrals. Let : $$I=\int_{-\infty}^\infty e^{-x^2}dx $$ Now we get: $$I^2=\int_{-\infty}^\infty\int_{\infty}^\infty e^{-x^2}e^{-y^2}dxdy $$ Now you can do a polar change of variable, with $r^2=x^2+y^2 $, and $dxdy=rdrd\theta$

$$I^2=\int_{r=0}^{\infty}\int_{\theta=0}^{2\pi} re^{-r^2}drd\theta=\pi\int_0^\infty 2re^{-r^2}dr\\ =\pi\left[-e^{-r^2} \right]_0^\infty=\pi$$

Hence the result : $$\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi} $$