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?

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