Re: Integral of [sin(x)/x]^2

• Previous message: Patrick Hamlyn: "Re: MuPAD bugs: series can return 6 terms only"
• In reply to: Robert Israel: "Re: Integral of [sin(x)/x]^2"
• Messages sorted by: [ date ] [ thread ]

Date: Wed, 23 Feb 2005 20:35:28 -0500

Robert Israel wrote:
>
> In article <atsh2f15fkv9@legacy>, casey <ccb147@psu.edu> wrote:
> >What about the same integral with the term exp(ixt) in it? (for all
> >real-valued t)
>
> Do you mean the integral of (exp(ixt)/x)^2 from x=-infinity to infinity?
> That diverges at 0 (if t <> 0).
> I hope you mean the integral of ((1-exp(ixt))/x)^2.
> Maple 9.5 says that is 0, and I believe it's right.
>
> Robert Israel israel@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia Vancouver, BC, Canada

I think he means f(t) = \int {exp(ixt) (sin(x) / x)^2 } . That is, he wants the
Fourier transform of (sin(x) / x)^2 .

That's a fairly easy one to do since the imaginary part vanishes
identically because of symmetry. Anyway, he can evaluate the contour
integral over the same contour.

I expect the result is something like

        \pi ( |t+2| /2 + |t-2| /2 - |t| ) /2 .

-- 
Julian V. Noble
Professor Emeritus of Physics
jvn@lessspamformother.virginia.edu
    ^^^^^^^^^^^^^^^^^^
http://galileo.phys.virginia.edu/~jvn/
   "For there was never yet philosopher that could endure the 
    toothache patiently."  
        -- Wm. Shakespeare, Much Ado about Nothing. Act v. Sc. 1.

• Previous message: Patrick Hamlyn: "Re: MuPAD bugs: series can return 6 terms only"
• In reply to: Robert Israel: "Re: Integral of [sin(x)/x]^2"
• Messages sorted by: [ date ] [ thread ]