1. Move the constants a and b out of the integral by a substitution of
variables.
2. Now you've reduced the problem to calculating integrals of the form

I[n] = Int[0,INFINITY] r^n exp(-r^2) dr.

Partial integration now gives you the recursion relation I[n] = (n-1)/2 *
I[n-2], valid for n > 0.
3. Now you've reduced the problem to calculating I[0] and I[1]. This is most
easily calculated by studying the double integrals

I[0]^2 = Int[0,INFINITY] Int[0,INFINITY] exp(-x^2-y^2) dx dy
= Int[0,2*pi] Int[0,INFINITY] r exp(-r^2) dr dtheta = pi

and

I[1] = Int[0,INFINITY] r exp(-r^2) dr = 1/2.
4. Thus, we have I[n] = 1/2 * Gamma((n+1)/2).

Erik

----- Original Message -----
From: "ZHANG Yan" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, May 24, 2004 2:13 PM
Subject: [edstat] How to compute the integral?Thanks


> Suppose that n is positive integer. Suppose that a and b are positive
> real number.
>
> How to simplify/compute the following integral.
>
> Int[0, INFINITE] x^n exp^( - (x-a)^2/ (2b^2)) dx
>
> In this expression, "Int[0 , INFINITE]" represents the integral from
> zero to infinite. x is the variable.
>
> Thanks for your suggestions.Any suggestions are greatly appreciated.
>
> --------------
> ZHANG Yan
> .
> .
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