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Re: Simplification of definite integral?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg40701] Re: Simplification of definite integral?
  • From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
  • Date: Mon, 14 Apr 2003 04:01:39 -0400 (EDT)
  • References: <b7avdo$qlg$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Dr. Wolfgang Hintze" <weh at snafu.de> wrote:
> How do I get a satisfactory result from mathematica for this function
>
> f[d]:=Integrate[Sin[x-d]/(x-d) Sin[x+d]/(x+d),{x,-Infinity,Infinity}]
>
> I tried
>
> f[d]//ComplexExpand
>
> and several assumptions but I didn't succeed. Any hints?

Note that you already know that the answer is pi*sinc(2*d),
where sinc(x) denotes the sine cardinal function,
i.e., sinc(x) = 1        for x = 0
              = sin(x)/x for nonzero x.
[BTW, I know that you already know the answer because I gave that answer
in a thread in alt.math.recreational at the end of February and you then
supplied a proof based, IIRC, on series expansion.]

So how do we get a satisfactory answer from Mathematica for that integral?
We don't, as far as I can tell; Mathematica's answer is wrong. To be more
specific, if it would be reasonable to assume that your d is positive, then

Integrate[Sin[x-d]/(x-d) Sin[x+d]/(x+d),{x,-Infinity,Infinity},
Assumptions-> d>0]; FullSimplify[%, d>0]

gives a simple result, but it is incorrect, having a spurious imaginary
part. The real part of that result, however, is the correct answer.

But it should also be noted that, using a specific value of d, Mathematica
can even get the real part wrong. For example, with d = 1,

Integrate[Sin[x-1]/(x-1) Sin[x+1]/(x+1),{x,-Infinity,Infinity}]

gives the purely imaginary result -1/2*I*Pi*Cos[2],

rather than the correct pi*sin(2)/2.

David Cantrell


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