Find the value of $a$ if $\int_0^\infty \frac{2x}{a}e^{\Large\frac{-x^2}{a}}\ dx=1 $

Question

Find the value of $a$ if $$\int_0^\infty \frac{2x}{a}e^{\Large\frac{-x^2}{a}}\ dx=1 $$ I tried to use integration by parts but I didn't get a good response.

Answer 1

Hint :

Let $u=\dfrac{x^2}{a}\;\Rightarrow\;du=\dfrac{2x}{a}\ dx$, then the integral turns out to be $$ \int_0^\infty e^{-u}\ du=1. $$

Answer 2

here's a hint: $xdx = d(\frac{x^2}{2})$

Answer 3

Let $Y$ be a Half-normal random variable (http://en.wikipedia.org/wiki/Half-normal_distribution). The expected value of the Half-normal distribution is given by

$E[Y]=\int_{0}^{\infty}\frac{y}{\sigma}\sqrt{\frac{2}{\pi}}exp(-\frac{y}{2\sigma^{2}})dy=\sigma\sqrt{\frac{2}{\pi}}$. This implies that

$\int_{0}^{\infty} y~exp(-\frac{y}{2\sigma^{2}})dy=\sigma^{2}$.

Since it is given that $\int_{0}^{\infty}2 \frac{y}{\sigma^{2}}exp(-\frac{y}{2\sigma^{2}})dy=1,$ using the above result we note that $\sigma^{2}=\frac{1}{2}$

Answer 4

You know that, $\int_{0}^\infty e^{-ax^2} \, \mathrm dx = \sqrt{\pi}/2a$. Take derivative with respect to $a$, you will get the answer. Then compare both the sides.