Trying to Integrate f(x) = exp(-ax^2)Date: 11/23/2001 at 09:07:34 From: Craig Robinson Subject: Integration How would you go about integrating a function of the form f(x) = exp(-ax^2)? Date: 11/23/2001 at 09:32:53 From: Doctor Jubal Subject: Re: Integration Hi Craig, Thanks for writing to Dr. Math. The function f(x) = exp(-ax^2) doesn't have an antiderivative that can be expressed in terms of other commonly used functions. What this means is you can't analytically integrate it. Yet, the function pops up a lot in many places where it would be nice to be able to integrate it. So mathematicians have come up with several workarounds. Your options are: (i) Hope your limits of integration are zero and infinity. For these limits of integration, the integral does have an anlytical solution. The integral of exp(-ax^2) from 0 to infinity is (1/2)sqrt(pi/a). If you'd like to see a proof of this, feel free to write back. (ii) Solve it numerically. So your limits of integration aren't zero and infinity. Use Simpson's Rule or the numerical integration method of your choice to approximate the integral. (iii) Use the error function. Mathematicians like this integral a lot. They like it so much, in fact, that they defined a function just so this integral would have an "analytical" solution. The error function is defined 2 /x erf(x) = --------- | exp[-t^2] dt sqrt(pi) /0 and /x sqrt(pi/a) | exp[-at^2] dt = ---------- erf(x) /0 2 The error function has been evaluated numerically at many, many points, and the results tabulated in mathematical handbooks. Essentially this is just doing (ii), but taking advantage of the fact that someone else has probably already done the work for you. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/ Date: 11/23/2001 at 09:36:37 From: Craig Robinson Subject: Re: Integration Hi, Could I see a proof of the integral of f(x) = exp(-ax^2) between 0 and infinity, please? Thanks. Date: 11/23/2001 at 12:12:37 From: Doctor Jubal Subject: Re: Integration Hi Craig, thanks for writing back. We want to evaluate /inf | exp(-ax^2) dx /0 However, using any of our usual arsenal of integration tricks to produce an antiderivative of this will fail. When I was an undergraduate, I spent five hours one evening attempting various ways to integrate this by parts and trying various substitutions. Little did I know there was no antiderivative. The solution came to me over breakfast, within half an hour of my waking up, thus illustrating that in mathematics, there is no substitute for a good night's sleep. But I digress. Let's call the value of the definite integral value I. For reasons that will become apparent shortly, it's a lot easier to evaluate I^2. /inf /inf I^2 = | exp(-ax^2) dx | exp(-ax^2) dx /0 /0 It's rather irrelevant what symbols we use for the variables of integration, so I'm going to use x in the first integral and y in the second integral. This is okay because the two integrals are completely independent of each other. /inf /inf I^2 = | exp(-ax^2) dx | exp(-ay^2) dy /0 /0 And because the two integrals are independent of each other, we can combine them into a single double integral. /inf /inf I^2 = | | exp(-ax^2) exp(-ay^2) dx dy /0 /0 /inf /inf = | | exp[-a(x^2 + y^2)] dx dy /0 /0 Now so far, it looks as if all we've managed to do is make a mess of things. This integral doesn't look any easier to deal with than what we started with, and we have an extra variable to boot. But we have managed to create an x^2 + y^2 term, which means this double integral may be more manageable in polar coordinates. The region of integration is the first quadrant, so we're integrating over r from 0 to infinity and theta from 0 to pi/2. I'll use Q for theta. Remember that in polar coordinates, r^2 = x^2 + y^2, and dx dy = r dr dQ. /(pi/2) /inf I^2 = | | exp(-ar^2) r dr dQ /Q=0 /r=0 /(pi/2) /inf = | dQ | exp(-ar^2) r dr /Q=0 /r=0 This integral can be solved analytically, but I'll leave this as something for you to work out on your own. (Hint: try the substitution u^2 = ar^2). If there's any part of this that doesn't make sense to you, feel free to write back. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/ |
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