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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