Show $\int_0^\infty xe^{-ax^2}dx=\frac{1}{2a}$ 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?
Hint: let $u=ax^2$.  Then $1/2du=axdx$.
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 $$\Gamma \frac{1}{2}=\sqrt{\pi}$$