积分的计算

Let the integral be $I$. $I={\int }_{-\infty }^{\infty }{e}^{-\frac{y^2}{2}}dy$ Consider the square of the integral: $I^2=\left({\int }_{-\infty }^{\infty }{e}^{-\frac{x^2}{2}}dx\right)\left({\int }_{-\infty }^{\infty }{e}^{-\frac{y^2}{2}}dy\right)$ We can rewrite this as a double integral: $I^2={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-\frac{x^2}{2}}{e}^{-\frac{y^2}{2}}dxdy$ $I^2={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-\frac{x^2+y^2}{2}}dxdy$ Now, we convert to polar coordinates. Let $x=r\cos \theta$ and $y=r\sin \theta$. Then $x^2+y^2=r^2$, and the Jacobian of the transformation is $r$. The limits of integration become $0\le r<\infty$ and $0\le \theta \le 2\pi$. $I^2={\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-\frac{r^2}{2}}rdrd\theta$ We can separate the integrals: $I^2=\left({\int }_0^{2\pi }d\theta \right)\left({\int }_0^{\infty }r{e}^{-\frac{r^2}{2}}dr\right)$ The first integral is straightforward: ${\int }_0^{2\pi }d\theta =[\theta {]}_0^{2\pi }=2\pi$ For the second integral, we can use a substitution. Let $u=\frac{r^2}{2}$, so $du=rdr$. The limits of integration change from $0$ to $\infty$ for $r$ to $0$ to $\infty$ for $u$.

$$\begin{array}{rl}{\int }_0^{\infty }r{e}^{-\frac{r^2}{2}}dr& ={\int }_0^{\infty }{e}^{-u}du\\ & =[-e^{-u}{]}_0^{\infty }\\ & =-e^{-\infty }-(-e^{-0})\\ & =0-(-1)\\ & =1\end{array}$$

Now, we can combine the results: $I^2=(2\pi )(1)=2\pi$ Taking the square root of both sides, and noting that the integrand $e^{-y^2/2}$ is positive, so $I$ must be positive: $I=\sqrt{2\pi }$ Thus, ${\int }_{-\infty }^{\infty }{e}^{-\frac{y^2}{2}}dy=\sqrt{2\pi }$