Re: Simplification of definite integral?
- To: mathgroup at smc.vnet.net
- Subject: [mg40751] Re: Simplification of definite integral?
- From: "Carl K. Woll" <carlw at u.washington.edu>
- Date: Wed, 16 Apr 2003 01:35:50 -0400 (EDT)
- Organization: University of Washington
- References: <200304130617.CAA27308@smc.vnet.net> <b7dq00$67a$1@smc.vnet.net> <b7gf2c$dr1$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Wolfgang,
It might help to understand a little bit about what is going wrong when
using Integrate for definite integrals.
First, note that NIntegrate should always produce a correct answer. Also,
when using Integrate with indefinite integrals, Integrate always produces a
correct antiderivative, or to be more specific, a function which when
differentiated will produce the integrand. It is only when Integrate is used
with definite limits that problems can arise.
There are two possibilites here. Either Integrate can produce the
antiderivative of its integrand, or Integrate cannot figure out the
antiderivative of its integrand.When Integrate can produce the
antiderivative of its integrand, it will determine the definite integral by
simply plugging in the limits of integration. I don't know the algorithm
Mathematica uses when Integrate cannot produce the antiderivative of its
integrand, and so I will not consider this case any further.
The above discussion should give you a clue about where Mathematica is going
wrong. If Mathematica simply plugs in the limits of integration, and the
antiderivative contains branch cuts, there is no guarantee that the two
limits will lie on the same branch, and if they are not on the same branch
Integrate will give an incorrect answer. Consider the following simple but
contrived example:
Integrate[1/(x-1),{x,-I,I}]
-Log[-1 - I] + Log[-1 + I]
The integral of 1/(x-1) is of course Log[x-1] and so Mathematica found the
indefinite integral and plugged in the limits. However, Log has a branch
along the negative real axis, and integrating from -I to I will cross this
branch, and so the two limits are not on the same branch. Let's compare a
numerical approximation of the above integral to the answer provided by
NIntegrate:
Integrate[1/(x-1),{x,-I,I}]//N
NIntegrate[1/(x-1),{x,-I,I}]
0. + 4.71239 I
0. - 1.5708 I
We see that Integrate gives an incorrect answer, and the two answers differ
by 2Pi, as expected by the residue theorem.
In your example, Mathematica is indeed capable of finding the antiderivative
of your integrand. I won't bother copying Mathematica's output, but I will
point out that the answer contains two functions with branch cuts, Log and
CosIntegral. The Log terms don't present a problem here (as they vanish at
the limits), but the CosIntegral does, since Mathematica puts the branch cut
along the negative real axis and so having both CosIntegral[-2d+2x] and
CosIntegral[2d+2x] presents a problem for the lower limit of x=-Infinity.
Actually, this example is a little weird. Consider:
(Integrate[Sin[x-d]/(x-d) Sin[x+d]/(x+d),x]/.x->Infinity)-
(Integrate[Sin[x-d]/(x-d) Sin[x+d]/(x+d),x]/.x->-Infinity)
Pi Sin[2 d]
-----------
2 d
where I simply plug in the limits of integration into the indefinite
integral, and the correct answer pops out. I think what is happening here is
that Mathematica is going through a Limit process which undergoes a hiccup
at x->-Infinity.
At any rate, here are a couple more suggestions on how to get the correct
answer.
FullSimplify[Integrate[Sin[x-d]/(x-d)
Sin[x+e]/(x+e),{x,-Infinity,Infinity}]/.e->d]
Pi Cos[d] Sin[d]
----------------
d
<<Calculus`Limit`
Integrate[Sin[x-d]/(x-d) Sin[x+d]/(x+d),{x,-a,a}];
Limit[%,a->Infinity]
Pi Cos[d] Sin[d]
----------------
d
The second suggestion uses the Limit package, which you should avoid in
general. For David Cantrell's example we can do something similar (if you
have loaded the Limit package you should restart the kernel).
Integrate[Sin[x - 1]/(x - 1+d) Sin[x + 1]/(x + 1), {x,2,Infinity}]/.d->0
%//N
Cos[2] Log[3] Pi Sin[2] 2 Cos[2] CosIntegral[2] - 2 Cos[2]
CosIntegral[6] - 2 Sin[2] SinIntegral[2] - 2 Sin[2] SinIntegral[6]
------------- + ---------
+ --------------------------------------------------------------------------
---------------------------
4 4
8
-0.140037
In conclusion, be very careful of results returned by Integrate for definite
integrals. You should probably always do a numerical check on the answer
using NIntegrate. If the answer given by Integrate is incorrect, you can try
using options such as Principal Value, or you can try perturbing your
integral a little bit in the hope that the resulting antiderivative doesn't
have a branch cut along the path of integration. It would be nice if you had
the option of changing the location of the branch cuts of various functions.
Carl Woll
Physics Dept
U of Washington
"David W. Cantrell" <DWCantrell at sigmaxi.org> wrote in message
news:b7gf2c$dr1$1 at smc.vnet.net...
> Vladimir Bondarenko <vvb at mail.strace.net> wrote:
> > Sunday, April 13, 2003, 3:17:27 AM, "Dr. Wolfgang Hintze" <weh at snafu.de>
> > wrote:
> >
> > DWH> How do I get a satisfactory result from mathematica for this
> > function
> >
> > DWH> f[d]:=Integrate[Sin[x-d]/(x-d) Sin[x+d]/(x+d),{x,-Infinity,
> > Infinity}]
> >
> > DWH> I tried
> >
> > DWH> f[d]//ComplexExpand
> >
> > DWH> and several assumptions but I didn't succeed. Any hints?
> >
> > I am not sure of what is 'a satisfactory result'? Do you mean something
> > like this
> >
> > Integrate[Sin[x - d]/(x - d) Sin[x + d]/(x + d), {x, -Infinity,
> > Infinity}, Assumptions -> d > 0, PrincipalValue -> True]//TrigReduce
> >
> > (Pi*Sin[2*d])/(2*d)
>
> That is indeed satisfactory, at least to me. [However I'm surprised that
> PrincipalValue -> True was required; after all, the singularities are
> _removable_. Indeed, with the sinc function, the integrand could be easily
> rewritten without singularities.]
>
> But please do not overlook the fact that Mathematica can make mistakes
with
> integrals of this type even when no singularities are involved. As a
simple
> example, consider
>
> Integrate[Sin[x - 1]/(x - 1) Sin[x + 1]/(x + 1), {x, 2, Infinity}]
>
> The answer should clearly be purely real, yet Mathematica's answer is
> approximately -0.854198 - 0.326841*I. (Based on numerical investigations,
> I suspect the correct answer is approximately -0.14 .)
>
> Regards,
> David Cantrell
>
- References:
- Simplification of definite integral?
- From: "Dr. Wolfgang Hintze" <weh@snafu.de>
- Simplification of definite integral?