From: hrubin@odds.stat.purdue.edu (Herman Rubin)
Subject: Re: Another tricky integration
Date: 14 Apr 2001 08:59:13 -0500
Newsgroups: sci.math
Summary: Expressing functions defined as integrals in terms of ODEs

In article <9b8642$7uu$1@pegasus.csx.cam.ac.uk>,
Ziggi <ziggi@mindpower.swinternet.co.uk> wrote:
>Hello again.  Could someone help me with this integration:

>Int (from -infinity to +infinity) of [exp^(ikx)]/(k^2+a^2) dk

>a is a constant.  I've tried and tried, but the solution eludes me.  It
>would be really helpful if you could tell me HOW to do the integral, as
>opposed to the just answer.  Thanks in advance.

>Ziggi

The usual method, which has been posted here, is to
use contour integration.  However, it CAN be done by
undergraduate methods.  I will leave out some details,
so you can work it out yourself.

First, one needs x real, and hence can use the symmetry
to get 

	f(x) = 2*\int_0 cos(kx)/(k^2+a^2) dk.

Assume x > 0.  Integrate this by parts, integrating the
cosine.  Multiply by x, and then differentiate with respect
to x.  This gives you an expression, which is a linear
combination of f(x) and another function h(x).  Moving the
f(x) to the other side, differentiate again.  This gives a
recognizable second order differential equation on the open
interval (0, infinity).  Solve it, and then use the
continuity of f and its behavior at 0 and infinity to find
the solution.

I have never seen this in the literature, but it looks like
18th century.
-- 
This address is for information only.  I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
hrubin@stat.purdue.edu         Phone: (765)494-6054   FAX: (765)494-0558