題目

Problem

Find the integral

\int_0^\infty x^n e^{-x^2}\,dx.

(a) $\dfrac12\Gamma\!\left(\dfrac{n+1}{2}\right)$

(b) $\dfrac12\Gamma\!\left(\dfrac{n}{2}\right)$

(d) $\Gamma\!\left(\dfrac{n}{2}\right)$

(e) None of the above.

解答

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令 $u=x^2$ ，則 $\mathrm{d}x=\dfrac{1}{2}u^{-1/2}\,\mathrm{d}u$ ，所以

\begin{align*} \int_0^\infty x^ne^{-x^2}\,\mathrm{d}x =&\,\frac12\int_0^\infty u^{(n-1)/2}e^{-u}\,\mathrm{d}u\\[4mm] =&\,\boxed{\frac12\Gamma\left(\frac{n+1}{2}\right)}. \end{align*}

答案為 $\boxed{\text{(a)}}$ 。