Question

How do you find the Improper integral `e^[(-x^2)/2]` from `x=-∞` to `x=∞`?

Answer 1

`sqrt(2pi)`

Explanation:

Calling `I =int_oo^oo e^{-x^2/2}dx` we know that

`I^2 = (int_oo^oo e^{-x^2/2}dx)(int_oo^oo e^{-y^2/2}dy)` but the integrals are independent so

`I^2 = int_oo^oo int_oo^oo e^{-(x^2+y^2)/2}dx dy`

Changing to polar coordinates

`rho^2 = x^2+y^2`
`dx dy equiv rho d rho d theta`

To cover the whole plane in polar coordinates we have

`I^2= int_{rho=0}^{rho=oo}int_{theta=0}^{theta=2pi}e^{-rho^2/2} rho d rho d theta`

`I^2 = (int_{rho=0}^{rho=oo}rho e^{-rho^2/2} d rho)(int_{theta=0}^{theta=2pi} d theta) = 1 xx 2pi`

Then `I = sqrt(2pi)`

Note:

`d/(drho)e^{-rho^2/2} = -rho e^{-rho^2/2}`