Show that the normal PDF satisfies the normalization

QUESTION:

Show that the normal PDF satisfies the normalization property. Hint: 2 The integral Jx e-x /2 dx is equal to the square root of 1 = 100 -x 2 2 /2 e e x - y /2 d d y, -x -00 and the latter integral can be evaluated by transforming to polar coordinates. Solution. We note that = - e -r /2r dr dO 1 1 271" 1 = 2 2 1T" 0 0 = 1, where for the third equality, we use a transformation into polar coordinates, and for the fifth equality, we use the change of variables u = r2 / 2. Thus, we have 1= 1_ -x2/ 2 d - e x - 1 , - 00 ..j'i; because the integral is positive. Using the change of variables u = (x - J-l)/u, it follows that fx (x) dx = e(x-/L) / 217 dx = __ e-u /2du = 1. I x I X 1 2 2 lOG 1 2 -:x; - = v'2ii u _ = ..j'i;