skipjack | February 18th, 2018 07:30 PM | Using $x = e^{-u^2}$, where $u\geqslant0$, so that $dx = -2ue^{-u^2}du$, one gets
$\displaystyle \int_\infty^0\! -2u^2e^{-3u^2/2}du = -(2/3)\!\int_0^\infty\! -3u^2e^{-3u^2/2}du$.
Using integration by parts, with $v = e^{-(3/2)u^2}$, so that $dv = -3ue^{-(3/2)u^2}du$, one gets
$\displaystyle -(2/3)\left[ue^{-(3/2)u^2}\right]_0^\infty + (2/3)\!\int_0^\infty\! e^{-(3/2)u^2}du = (2/3)\!\int_0^\infty\! e^{-(3/2)u^2}du$.
If $u = \sqrt{\frac23}t$, so that $du = \sqrt{\frac23}dt$, one gets $\frac23\sqrt{\frac23}$ $\displaystyle \!\int_0^\infty\! e^{-t^2}dt$.
It's well known that $\displaystyle \!\int_0^\infty\! e^{-t^2}dt = \small\frac{\sqrt\pi}{2}$. |