习题

习题 Gauss 积分

计算

$$ \bbox[5pt]{ \int_{\mathbb{R}}{e^{-x^2}}{dx} = \sqrt{\pi} } $$

Proof.

$$ \bbox[5pt]{ \begin{aligned} & \left(\int_{\mathbb{R}}{e^{-x^2}}{dx}\right)^2 \\ = & \int_{\mathbb{R}}{e^{-x^2}}{dx}\int_{\mathbb{R}}{e^{-y^2}}{dy} \\ = & \iint_{\mathbb{R}^2}{e^{-(x^2+y^2)}}{dxdy} \qquad & (\text{定理* Fubini定理}) \\ = & \int_{0}^{2\pi}{\int_{0}^{+\infty}{e^{-r^2}r}{dr}}{d\theta} \qquad & (\text{定理* 极坐标换元}) \\ = & \frac{1}{2}\int_{0}^{2\pi}{\int_{0}^{+\infty}{e^{-s}}{ds}}{d\theta} \qquad & (\text{定理* 凑微分}) \\ = & \frac{1}{2}\int_{0}^{2\pi}{1}{d\theta} \\ = & \pi \end{aligned} } $$

综上

$$ \bbox[5pt]{ \int_{\mathbb{R}}{e^{-x^2}}{dx} = \sqrt{\pi} } $$