$\int^{\infty}_{-\infty} \exp \{-\frac{1}{2} y^2\} \; dy$

Question

$$\int^{\infty}_{-\infty} \exp \{-\frac{1}{2} y^2\} \; dy$$

I tried letting $u = -\frac{1}{2} y^2$ then $dy = - \frac{1}{\color{red}y} du$... but theres still a $y$ term in $dy$?

Answer 1

Hint: If $I$ is the desired integral, then

$$I^2 = \int_{-\infty}^{\infty} e^{-x^2/2} dx \int_{-\infty}^{\infty} e^{-y^2/2} dy = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-\frac{1}{2}(x^2 + y^2)} dy dx$$

Now convert to polar coordinates.