From: spellucci@mathematik.tu-darmstadt.de (Peter Spellucci)
Subject: Re: Integrals over infinite range
Date: 5 Mar 1999 15:15:46 GMT
Newsgroups: sci.math.num-analysis

In article <36DECB96.2806BDC6@ix.netcom.com>,
 Charles Bond <cbond@ix.netcom.com> writes:
|> In Quadpack, singular integrals with infinite range are transformed
|> so that the variable of integration goes from 'x' to '(1-x)/x' or
|> something
|> similar. This transform seems OK, generally. But can one argue that the
|> sensitivity of this transform  might be less than '1/(x^2)' or some
|> other
|> similar transform for some integrals? There doesn't seem to be much in
|> the way of quantitative comparisons of various transforms in the
|> literature
|> I've seen. Some writers suggest using tangent function or hyperbolic
|> tangents as transforms. any comments?
let us consider the simple infinite integral 

  \int_0^\infty exp(-x)dx = 1
with quadpack, using
   x+1=1/y   <-> x=1/y-1 <-> y=1/(1+x) 
zero goes to one and \infty to zero and we get
 \int_0^1  exp(1-1/y)(1/y^2)dy   
which is a regular integral at y=1 _and_  y=0. you can also use
   x^2+1=1/y <-> x=sqrt(1/y-1) <-> y=1/(x^2+1) 
which also sends zero to one and infty to zero. Now you get
  \int_0^1 exp( -sqrt(1/y -1) )* (1/(2*sqrt(1/y-1)) *(1/y^2) dy
and you are punished for not obeying the condition that the transformation
should have a nonzero derivative throughout by a newly singular integral
at  y=1 It is of course an integrabel singularity. using 
   y=tanh x    <-> x = ar tanh y = (1/2)(ln( (y+1)/(1-y) )
sends zero to zero and infinity  to one. now we have 
   \int_0^1  sqrt((1-y)/(1+y)) 1/(1-y^2) dy
and the integrand now has an integrable singularity at y=1.
clearly, the first transformation is superior but i have never seen a formal
proof that it is "best" in the sense never to introduce artificial 
singularities in the integrand .
hope this helps
peter
==============================================================================

From: "Hamer, Peter (EXCHANGE:HAL02:HK00)" <pgh@europem01.nt.com>
Subject: Re: Integrals over infinite range
Date: Mon, 08 Mar 1999 16:40:26 +0000
Newsgroups: sci.math.num-analysis

Charles,

I think that there are two issues. The optimum transform for a
particular class of problem, and making sure that you get a
sufficiently accurate answer with a `moderate' choice of
transform.

You are probably on your own for the first, as who knows what
your function is better than you do. Obviously the asymptotic
behaviour (power law, exponention, ...) is important. It may
be worth splitting into a finite region and an infinite tail
to avoid transformation woes `centre stage'.

The second seems to be adequately handled by techniques such as
the Romberg integration described in Numerical Recipes. Try 
increasingly fine meshes until extrapolation (eg Richardson's
defered approach to the limit) suggests that you have sufficient
accuracy. See http://beta.ulib.org/webRoot/Books/

Peter