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ò-¥ ¥e-x2dx= pÖ2
By Andrew Hodges on Monday, November 26, 2001 - 10:04 pm:

How can you prove that ò-¥ ¥e-x2dx=pÖ2?

Thanks Andrew

By Dan Goodman on Monday, November 26, 2001 - 10:12 pm:

Two ways to do this. Either way you need to know about double integrals and the change of variable formula for them. The easy way is to look at this thread. The other way is to use this hint and see if you can do it yourself:

If I=ò-¥ ¥e-x2dx then what happens if you change I*I into a double integral and change to polar coordinates?

By David Loeffler on Tuesday, November 27, 2001 - 06:18 pm:

There is another very nice proof that cropped up on one of our Differential Equations problem sheets. Define f(x) = (ò0 xe-t^2dt)2 and g(x) = ò0 1e-x2(1+t2)/(1+t 2) dt.

Now calculate f'(x) and g'(x). We find that f'(x)+g'(x) = 0, so f(x)+g(x) = f(0)+g(0) = p/4 for all x. If we take the limit as x->¥, we find that g(x) -> 0, so f(x) -> p/4. Since f(x) is the square of the integral ò0 ¥e-t2dt, we see that this integral is sqrt(p)/2.

David

By Andrew Hodges on Tuesday, November 27, 2001 - 07:33 pm:

Could you explain to me what a double integral is, so I can attempt it for myself!

Thanks

Andrew

By Brad Rodgers on Friday, December 28, 2001 - 12:35 am:

It's probably too late, but try here. The basic idea is that a double integral is doing the operation of integration twice.