If $1 = c . \int_{- \infty}^{\infty} e^{-x^2}dx$ , then what is c?

Question

I basically wonder if $\displaystyle \int_{- \infty}^{\infty} e^{-x^2}dx$ is convergent or divergent, and if convergent, what is it equal to?

$c=\frac{1}{\sqrt{\pi}}$ This is a classical result – Jonathan Davidson Jul 12 '17 at 06:12 — Jonathan Davidson, Jul 12 '17 at 06:12
https://en.wikipedia.org/wiki/Gaussian_integral – charmd Jul 12 '17 at 06:18 — charmd, Jul 12 '17 at 06:18

score 1 · Accepted Answer · answered Jul 12 '17 at 06:31

Take a look at what the integral provides in a slightly more general case: \begin{align} \int_{- \infty}^{\infty} e^{- a x^2} \, dx &= 2 \, \int_{0}^{\infty} e^{- a x^2} \, dx \\ &= \frac{1}{\sqrt{a}} \, \int_{0}^{\infty} e^{- t} \, t^{- \frac{1}{2}} \, dt \hspace{5mm} \text{where} \, t = a \, x^2\\ &= \sqrt{\frac{\pi}{a}} \end{align} Now if it is required that $$1 = c \, \int_{- \infty}^{\infty} e^{- a \, x^2} \, dx$$ then it is determined that $$c = \sqrt{\frac{a}{\pi}}$$.

score 0 · Answer 2 · answered Jul 12 '17 at 06:19

0

The pdf of normal distribution with mean $0$ and variance $\frac{1}{\sqrt{2}}$ is $f(x) = \frac{1}{\sqrt{2\pi}\frac{1}{\sqrt{2}}}e^{-x^2}$.

Since, $\int_{-\infty}^{\infty}f(x)dx = 1 \implies \frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-x^2}dx = 1$

answered Jul 12 '17 at 06:19

Dhruv Kohli

5,216

If $1 = c . \int_{- \infty}^{\infty} e^{-x^2}dx$ , then what is c?

2 Answers2