Originally posted by: chuckywang
Originally posted by: Narmer
The boundary is from negative to positive infinity. I know the answer is the square root of pi. I also know that it's good to switch to polar coordinate. But I cannot seem to get the right function to plug in the (new) boundaries. I haven't done calculus since high school and cannot remember this.
Let I = integral e^(-x^2) dx from -inf to inf.
Therefore, I^2 = integral e^(-x^2) dx from -inf to inf * integral e^(-y^2) dy from -inf to inf
= double integral of e(-(x^2+y^2)) from -inf to -inf dx and -inf to inf dy
Converting to polar coordinates, we get: integral of r*e^(-r^2) from -inf to inf dr and 0 to 2pi d(theta). Integrating that we get that I^2 = pi, so I = sqrt (pi).
Hope that helps.