# Show $\int_0^\infty xe^{-ax^2}dx=\frac{1}{2a}$ [closed]

I tried integrating by parts but I always arrive at an expression other than $\frac{1}{2a}$ which contains $\sqrt{\frac{\pi}{a}}$ from the Gaussian integral $\int_0^\infty e^{-ax^2}dx=\frac12 \sqrt{\frac{\pi}{a}}$. Is there some kind of trick to evaluating this integral?

• Let $-ax^2=u$.. May 3, 2017 at 19:08

Hint: let $u=ax^2$. Then $1/2du=axdx$.

• $1/2du=axdx$ I think. May 3, 2017 at 19:13
• @MyGlasses, thanks, fixed.
– Paul
May 3, 2017 at 19:14

$\int_0^{+\infty}xe^{-ax^2}dx=\frac{1}{-2a}\int_0^{+\infty}-2axe^{-ax^2}dx=\frac{1}{-2a}[e^{-ax^2}]_0^{+\infty}=\frac{1}{2a}$

Substitute $x^2=y$. Then $$\int_0^\infty xe^{-ax^2}\,dx=\frac12\int_0^\infty e^{-ay}\,dy.$$

This integral is very simple and can be explained and understood in a better way just using simple properties of Gamma function Put $$ax^2 =z$$ then use the definition of Gamma function also $$\Gamma \frac{1}{2}=\sqrt{\pi}$$

• No need to switch to polar coordinates, switch only if you know the topic with complete understanding May 3, 2017 at 19:09
• This has nothing to do with the problem being asked.
– Paul
May 3, 2017 at 19:10
• I am just giving him information so that he could solve the question on his own May 3, 2017 at 19:11
• But your information doesn't help. You didn't read the problem carefully.
– Paul
May 3, 2017 at 19:12
• Gamma function is not required. This is a simple substitution integral.
– user307169
May 3, 2017 at 19:20