Okay, today I feel like proving it!
Let "I" equal the integral from -infinity to infinity of e^(-.5t^2) with respect to x. Then,
I^2 = int_x=(-∞,∞) [e^(-.5 x^2) dx] * int_y=(-∞,∞) [e^(-.5 y^2) dy] (1)
= int_(x,y)=R2 [e^(-.5 (x^2+y^2)) dx dy] (2)
= int_r=(0,∞) int_θ=[0,2π) [e^(-.5 r^2) r dr dθ] (3)
= 2π [-e^(-.5 ∞^2) + e^(-.5*0^2)] (4)
I^2 = 2π (5)
I = sqrt(2π) (6)
I just squared the integral by multiplying it by itself in step one. The dummy variable does not matter at all, so I used both x and y to make it an expression with two variables. In step 2, I combined the integrals into a single double integral over the real plane (R2). I then converted the expression into polar coordinates, making the substitutions x^2+y^2=r^2 and dx dy=dA=r dr dθ. The limits of integration become θ=[0,2π) and r=(0,∞) because this also represents integrating over the whole plane, minus the origin (excluding this point does not affect the value for the integral). The expression is evaluated as usual, and the result is that I^2=2π. This suggests that I=sqrt(2π) or I=-sqrt(2π). The positive value is chosen because the original function e^(-.5x^2) is always positive for real x.
Q.E.D.
