Find the value of the integral $I=\int_{-\infty}^\infty e^\frac{-y^2}{2}~dy$ .

Question

$$I=\int_{-\infty}^\infty e^\frac{-y^2}{2}~dy$$

Let $\dfrac{y^2}{2}=z$ ,

$y~dy=dz$

$$I=\int_{-\infty}^\infty\frac{e^{-z}}{\sqrt{2z}}dz$$

Answer 1

Let $x^2 = y^2/2$, so $\mathrm{d}y = \sqrt{2}\,\mathrm{d}x$ then $$ \int_{-\infty}^\infty e^{-y^2/2} \mathrm{d}y = \sqrt{2}\int_{-\infty}^\infty e^{-x^2} \mathrm{d}x $$ Can you recognize the last integral as the Gaussian Integral ?

Answer 2

$$I^2=\int_{\mathbb R}e^{-x^2/2}dx\cdot \int_{\mathbb R}e^{-y^2/2}dy=\int_{\mathbb R^2}e^{(-x^2-y^2)/2}dxdy=\\ =\lim_{n\to\infty}\int_{\{x^2+y^2\leq n^2\}}e^{(-x^2-y^2)/2}dxdy=\lim_{n\to+\infty}\int_0^{2\pi}\int_0^ne^{-r^2/2}rdrd\varphi=\\ =2\pi\lim_{n\to+\infty}(-e^{-n^2/2}+e^0)=2\pi.$$ Hence $I=\sqrt{2\pi}.$

Answer 3

Well, if you know something about distributions then you'll see that $\exp(-y^2/2)/\sqrt{2\pi}$ is the density of the normal distribution which integrates to one, so your answer is $\sqrt{2\pi}$.