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MathGroup Archive 2006

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Re: How to make results from Integrate and NIntegrate agree

  • To: mathgroup at smc.vnet.net
  • Subject: [mg66693] Re: How to make results from Integrate and NIntegrate agree
  • From: "David W.Cantrell" <DWCantrell at sigmaxi.org>
  • Date: Sat, 27 May 2006 03:51:56 -0400 (EDT)
  • References: <e510la$8di$1@smc.vnet.net> <e56ejk$23m$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Paul Abbott <paul at physics.uwa.edu.au> wrote:
> In article <e510la$8di$1 at smc.vnet.net>, riazzi at hotmail.com wrote:
[snip]
> Perhaps you could tell us what the real problem is so that other
> approaches could be suggested?

I happen to know, from private email, that Will's real problem involves
integrating polynomials in x times the fractional part of 1/x.

> > So what I'm looking for is a way to make Integrate produce a
> > closed-form expression for my integrand that does not produce these
> > discontinuities,
>
> While I'm a big fan of obtaining closed-form expressions, their utility
> is variable. Sometimes they are too cumbersome to be useful -- so the
> application you have in mind should determine how much work you should
> devote to obtaining a closed-form expression.

I was able to provide Will with what he needed. I'll just copy here from my
email to him:

"In any event, let me
just give you what you need to get a continuous antiderivative for any
polynomial in x times {1/x}:  Since {t}, the fractional part of t, is
the same as t - Floor[t], all I should need to give you is a
continuous antiderivative for x^n*Floor[1/x]. It's

(x^(n + 1)*Floor[1/x] + Zeta[n + 1, Floor[1/x] + 1])/(n + 1)

for integer n >= 1. For n = 0, you already have the result from my
earlier response to you."

> Note that, since Fr is the fractional part of 1/x, you can compute the
> integral (in this simple example) using Piecewise as follows:
>
>   g[x_] = Piecewise[Table[{1/x - n, 1/(n + 1) <= x < 1/n]},{n, 5}]]
>
>   Integrate[g[x], x]
>
> but this will not work if you have an infinite number of discontinuities.
>
> > plus some insight as to why they appear. I'd guess
> > the latter has to do with branch cuts in the complex plane for the
> > functions involved, and where the solution might have to do with
> > picking a different contour or somesuch, but that's getting into areas
> > I've long forgotten or never learned.
>
> David Cantrell and Zdislav V. Kovarik have posted on this topic on
> sci.math. Essentially, computer algebra systems often fail to give
> continuous antiderivatives for piecewise integrands.

That's not the "bad" part, actually, IMO. They also sometimes fail to give
continuous antiderivatives for continuous integrands (defined in just one
piece). That, I think, is bad, at least when a continuous antiderivative
can be expressed easily in closed form. Here's a simple (and important type
of) example:

In[1]:= Integrate[1/(5 + 3*Cos[t]), t]
Out[1]= -1/2*ArcTan[2*Cot[t/2]]

The integrand is continuous on R. The above result, however, has jump
discontinuities at integer multiples of 2*Pi. Of course, that result is an
antiderivative over intervals which do not contain such multiples, but it
is not an antiderivative over R. But an antiderivative over R not only
exists (of course) but it can be expressed easily in closed form:

t/4 - ArcTan[Sin[t]/(3 + Cos[t])]/2

Regards,
David Cantrell


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