Calculate $\int_{0}^{\infty} e^{(\frac{-x^2}{2a})}dx$

Question

Given that the derivative of $e^{(\frac{-x^2}{2a})}$ is $e^{(\frac{-x^2}{2a})}$ $ * \frac{-x}{a}$,
and that $\int_{0}^{\infty}x*e^{\frac{-x^2}{2a}} = 1$

I need to calculate $\int_{0}^{\infty} e^{(\frac{-x^2}{2a})}dx$.

I've been trying to do this using integration in parts, but I've failed in every attempt.

$\int_{0}^{\infty} \frac{x}{a}*e^{(\frac{-x^2}{2a})}dx = \frac{x}{a}|_{0}^{\infty}*\int_{0}^{\infty}e^{(\frac{-x^2}{2a})} - \frac{1}{a}\int_{0}^{\infty}e^{(\frac{-x^2}{2a})} $

$1 = \int_{0}^{\infty}e^{(\frac{-x^2}{2a})}(\frac{x}{a}|_{0}^{\infty}-\frac{1}{a})$

$\frac{1}{\frac{x}{a}|_{0}^{\infty}-\frac{1}{a}} = \int_{0}^{\infty}e^{(\frac{-x^2}{2a})}$

$0 = \int_{0}^{\infty}e^{(\frac{-x^2}{2a})}$

It can't be 0 since it's the gaussian integral.

What am I doing wrong?

Answer 1

It follows from the property of normal distribution that $\int_{-\infty}^{\infty} e^{-\frac{x^2}{2a}}dx =\sqrt{2\pi a}$. At the same time, since the integrand is an even function $$\int_{0}^{+\infty} e^{-\frac{x^2}{2a}}dx = \frac{1}{2} \int_{-\infty}^{\infty} e^{-\frac{x^2}{2a}}dx = \sqrt{\frac{\pi a}{2}}$$