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Re: Re: Simplification of definite integral?

• To: mathgroup at smc.vnet.net
• Subject: [mg40726] Re: [mg40701] Re: Simplification of definite integral?
• From: Dr Bob <majort at cox-internet.com>
• Date: Tue, 15 Apr 2003 03:58:44 -0400 (EDT)
• References: <b7avdo$qlg$1@smc.vnet.net> <200304140801.EAA06363@smc.vnet.net>
• Reply-to: majort at cox-internet.com
• Sender: owner-wri-mathgroup at wolfram.com

This looks reasonable:

Integrate[f[x, d], {x, -Infinity, Infinity}, PrincipalValue -> True]
   // Simplify[#, d > 0] &

(Pi*Cos[d]*Sin[d])/d

but this doesn't:

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

(Pi*(2*I*Cos[2*d] + Sin[2*d]))/(2*d)

(Mathematica 4.2.1)

Bobby

On Mon, 14 Apr 2003 04:01:39 -0400 (EDT), David W. Cantrell 
<DWCantrell at sigmaxi.org> wrote:

> "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
>
>



-- 
majort at cox-internet.com
Bobby R. Treat

• References:
  • Re: Simplification of definite integral?
    • From: "David W. Cantrell" <DWCantrell@sigmaxi.org>