Integrating exp(-x^2)Date: 11/09/2005 at 21:43:59 From: Pablo Subject: integrating exp(-x^2) I know that integral exp(-x^2) can be evaluated from negative infinity to positive infinity and gives the result sqrt(pi), but it seems like all the ways of doing this involve making it a double integral and using polar coordinates, I was wondering if there is any way to do it without making it a double integral. I have tried shifting into polar coordinates with x = rcos(theta) and then doing another substitution of u = cos^2(theta) but it just gets really messy. Maybe it can't be done but I would like a better explanation for why you can shift it into a double integral, I mean it makes intuitive sense but I haven't seen any rigorous proof or reasoning as to why it's valid. Date: 11/11/2005 at 01:46:22 From: Doctor Minter Subject: Re: integrating exp(-x^2) Hi Pablo! I don't believe that there is any way to integrate this function without a change of variables, and thus a double-integral, but there is a way to do it with a simpler "u substitution." Start with defining I as the value of the integral of e^(-x^2) dx from negative infinity to infinity. Now square both sides of the equation, but instead of simply squaring the integral value, multiply it by the same integral, except replace x and dx with y and dy, respectively. Remember that for a definite integral, you get a value that is independent of the index (variable name) that you choose. So integrating e^(-x^2) dx is exactly the same as integrating e^(-y^2) dy over the same limits of integration. So now we have I^2 = {INT [e^(-x^2) dx]}*{INT [e^(-y^2) dy]}, both from -infinity to +infinity, where INT is a single integral. Which, by the properties of double integrals and the properties of exponential functions, is equivalent to I^2 = DOUBLE-INT {e^[-(x^2 + y^2)] dx dy, both from -infinity to +infinity, where DOUBLE-INT is a double integral. We now change to polar coordinates, where x^2 + y^2 = r^2, and dx dy = r dr dT, by the Jacobian Transformation, where T is substituted for the usual theta for convenience. Using these values in the double-integral formula, we now have I^2 = DOUBLE-INT [r e^(-r^2) dr dT]. The new limits of integration are: zero to infinity for dr, and zero to 2*Pi for theta. Since the integrand has no theta dependence, we can replace INT(dT) by the difference between its limits, which is 2*Pi. We can pull Pi out of the integral because it is a constant, but leave the 2 inside the dr integral for reasons that will be clear shortly. We are left with I^2 = Pi*INT[ 2r e^(-r^2) dr ]. Now for the u substitution. Let u = r^2, so du = 2r dr u = 0 when r = 0 and u = infinity when r = infinity, so the limits of integration do not change. Our formula is now I^2 = Pi*INT[ e^(-u) du ] Integrating this is straightforward, and we are left with I^2 = Pi*[-e^(-u)], evaluated from zero to infinity. This leaves us with I^2 = Pi*[0 - (-1)], which is also I^2 = Pi Taking the square root of both sides says that the value of the original integral is I = sqrt(Pi). I hope this has made things seem a little easier. Please feel free to write again if you need any further assistance, or if you have any other questions. Thanks for using Dr. Math! - Doctor Minter, The Math Forum http://mathforum.org/dr.math/ |
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