Re: Simplification of definite integral?
- To: mathgroup at smc.vnet.net
- Subject: [mg40735] Re: Simplification of definite integral?
- From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
- Date: Tue, 15 Apr 2003 04:00:06 -0400 (EDT)
- References: <200304130617.CAA27308@smc.vnet.net> <b7dq00$67a$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
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?