∫ 0 ∞ x exp ( − 3 2 x 2 ) d x = ∫ 0 − ∞ − 1 3 d v d x exp v d x = − 1 3 ∫ 0 − ∞ exp v d v = − 1 3 ∫ 0 − ∞ exp v d v = − 1 3 [ exp v ] 0 − ∞ = 1 3 {\displaystyle \int _{0}^{\infty }x\exp {\left(-{\tfrac {3}{2}}x^{2}\right)}\,dx=\int _{0}^{-\infty }-{\frac {1}{3}}{dv \over dx}\exp {v}\,dx=-{\frac {1}{3}}\int _{0}^{-\infty }\exp {v}\,dv=-{\frac {1}{3}}\int _{0}^{-\infty }\exp {v}\,dv=-{\frac {1}{3}}\left[\exp {v}\right]_{0}^{-\infty }={\frac {1}{3}}}
∫ 0 ∞ x 3 exp ( − 3 2 x 2 ) d x = {\displaystyle \int _{0}^{\infty }x^{3}\exp {\left(-{\tfrac {3}{2}}x^{2}\right)}\,dx=}
= [ − 1 3 x 2 exp ( − 3 2 x 2 ) ] 0 ∞ − ∫ 0 ∞ − 2 3 x exp ( − 3 2 x 2 ) d x = 0 + 2 3 ⋅ 1 3 {\displaystyle =\left[-{\frac {1}{3}}x^{2}\exp {\left(-{\tfrac {3}{2}}x^{2}\right)}\right]_{0}^{\infty }-\int _{0}^{\infty }-{\frac {2}{3}}x\exp {\left(-{\tfrac {3}{2}}x^{2}\right)}\,dx=0+{\frac {2}{3}}\cdot {\frac {1}{3}}}
![{\displaystyle \int _{0}^{\infty }x\exp {\left(-{\tfrac {3}{2}}x^{2}\right)}\,dx=\int _{0}^{-\infty }-{\frac {1}{3}}{dv \over dx}\exp {v}\,dx=-{\frac {1}{3}}\int _{0}^{-\infty }\exp {v}\,dv=-{\frac {1}{3}}\int _{0}^{-\infty }\exp {v}\,dv=-{\frac {1}{3}}\left[\exp {v}\right]_{0}^{-\infty }={\frac {1}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34cb08a4f9267f99982c719626e8a9dcbaaf1b73)
and find:
then sub in:![{\displaystyle =\left[-{\frac {1}{3}}x^{2}\exp {\left(-{\tfrac {3}{2}}x^{2}\right)}\right]_{0}^{\infty }-\int _{0}^{\infty }-{\frac {2}{3}}x\exp {\left(-{\tfrac {3}{2}}x^{2}\right)}\,dx=0+{\frac {2}{3}}\cdot {\frac {1}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1c67cebd9f9e95bba01975c3c629f4172d8d16a)
