证明：∫−∞∞e−x2dx=π\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt\pi∫−∞∞e−x2dx=π Easy integration question 7 专题 General / 综合 难度 L4 来源 QuantQuestion 题目详情 Please prove that the following relationship holds: ∫−∞+∞e−x2dx=π\int_{-\infty}^{+\infty}e^{-x^{2}}d x = \sqrt{\pi}∫−∞+∞e−x2dx=π 解析 令 I=∫−∞∞e−x2dxI=\int_{-\infty}^{\infty}e^{-x^2}dxI=∫−∞∞e−x2dx，则 I2=∫R2e−(x2+y2)dxdy.I^2=\int_{\mathbb{R}^2}e^{-(x^2+y^2)}dxdy.I2=∫R2e−(x2+y2)dxdy. 改用极坐标： I2=∫02π∫0∞e−r2r dr dθ=2π⋅[−12e−r2]0∞=π.I^2=\int_0^{2\pi}\int_0^{\infty}e^{-r^2}r\,dr\,d\theta =2\pi\cdot\left[-\frac12e^{-r^2}\right]_{0}^{\infty}=\pi.I2=∫02π∫0∞e−r2rdrdθ=2π⋅[−21e−r2]0∞=π. 所以 I=π\boxed{I=\sqrt\pi}I=π。