[mathjax]
問題
$\displaystyle\int_{0}^{\infty} x^2 e^{-ax^2}dx$
解答
$\displaystyle\int_{0}^{\infty} x^2 e^{-ax^2}dx$
$=\displaystyle\int_{0}^{\infty} x (xe^{-ax^2}) dx$
※ここがポイント！
$=\left[x\left(-\displaystyle\frac{1}{2a}e^{-ax^2}\right)\right]_{0}^{\infty}+\displaystyle\int_{0}^{\infty} \displaystyle\frac{1}{2a}e^{-ax^2} dx$
※部分積分
$=\displaystyle\frac{1}{2a}\displaystyle\int_{0}^{\infty} e^{-ax^2} dx$
$=\displaystyle\frac{1}{2a}\displaystyle\frac{1}{2}\sqrt{\displaystyle\frac{\pi}{a}}$
$=\displaystyle\frac{\sqrt{\pi}}{4\sqrt{a^3}}$
