Re: SIGMA 1/(n^2 + 1)
From: Zdislav V. Kovarik (kovarik_at_mcmaster.ca)
Date: 10/18/04
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Date: Mon, 18 Oct 2004 14:06:17 -0400
On Sun, 17 Oct 2004, David C. Ullrich wrote:
> On Sat, 16 Oct 2004 20:11:10 +0000 (UTC), trimble1@optonline.net (Todd
> Trimble) wrote:
>
> >On 16 Oct 2004, Eamon Warnock wrote:
> >>Is it possible to evaluate SIGMA 1/(n^2 + 1) from 0 to infinity without
> >>using contour integrals?
> >>
> >
> >Here's a hint:
> >
> > pi*cot(pi*x) = 1/x - 2x*Sum_{n >= 1} 1/(n^2 - x^2)
>
> Of course this raises the question of whether you can prove
> that formula without using contour integrals - the standard,
> farily routine proof is by the resisue theorem.
>
> (The answer is yes, for example it's not too hard to show
> that the difference of the two quantities is a bounded entire
> function, hence constant; then looking carefully at what
> happens for "x = iy", y -> infinity shows the constant is 0.
> Of course it's really not clear that this counts; "without
> contour integrals" could well really mean "without complex
> analysis"...)
>
>
> ************************
>
> David C. Ullrich
In fact, I saw the "real variable" trick before the complex variable one:
it assumes that you can rely on Fourier expansions of well-behaved
2*pi-periodic functions (e.g. continuous, piecewise continuously
differentiable).
With b>0, expand cosh(b*x) on (-pi,pi] into a cosine series and
observe what happens at x=pi. As a bonus, you can sum some other
series by setting x=something else.
Cheers, ZVK(Slavek).
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