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

Question

Cameron Alexander · Accepted Answer

#sqrt(2pi)#

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}#

Luke Alvarez · Answer

undefined The improper integral of $ e^{-\frac{x^2}{2}} $ from $ x = -\infty $ to $ x = \infty $ cannot be evaluated using elementary functions. This integral represents the area under the standard normal distribution curve, and it is a well-known result in calculus that it cannot be expressed in terms of elementary functions. However, it is equal to $ \sqrt{2\pi} $, which is often derived using advanced techniques such as contour integration in complex analysis.