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Topic: Difficult integral
Replies: 8   Last Post: Jun 24, 2011 2:34 AM

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Posts: 575
Registered: 2/10/09
Re: Difficult integral
Posted: Jun 23, 2011 12:45 PM
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On Jun 23, 10:09 pm, David C. Ullrich <ullr...@math.okstate.edu>
> On Wed, 22 Jun 2011 19:22:02 -0400, Christopher Henrich
> <chenr...@monmouth.com> wrote:

> >In article <slrnj04diq.74a.hru...@skew.stat.purdue.edu>,
> > Herman Rubin <hru...@skew.stat.purdue.edu> wrote:

> >> On 2011-06-22, G. A. Edgar <ed...@math.ohio-state.edu.invalid> wrote:
> >> > In article
> >> ><551e4221-dcfa-4107-9c80-ed028627c...@r27g2000prr.googlegroups.com>,
> >> > TefJlives <gmarkow...@gmail.com> wrote:

> >> >> Anyone with any ideas on this one? Mathematica and Matlab can't do it.
> >> >> \int_0^\infinity sin(a y) coth(y)/((1+9y^2)^2)dy
> >> >> Here coth is the hyperbolic cotangent, and a is a positive parameter.
> >> >> It's giving me fits.

> >> >> Greg
> >> > Is there any reason to think it has a simpler form?
> >> I do not know if this is simpler, but it is another
> >> solution.  Instead of sin(ay), write exp(iay), and
> >> use the imaginary part.

> >> The expression then has poles at i/3 and at (n+.5)i*pi,
> >> n a non-negative integer.  The function is well enough
> >> behaved that the integral is 2i*pi times the sum of the
> >> residues at the infinite sequence of poles, which convenges
> >> at a rate similar to a geometric series.

> >> Using the sin instead of the complex exponential with
> >> cause the function to grow badly in the upper half plane.

> >The poles of coth are at ni*pi. In the integral as given, the pole at
> >y=0 is cancelled out by the sin(ay) factor. What I think you can do
> >about this* is take the following steps:

> >1. Change the original integral to an integral from -\infinity to
> >\infinity. (The integrand is an "even" function, so this step merely
> >requires you not to forget a factor of 1/2 .)

> >2. Deform the contour to dodge above 0 in the complex plane.
> >3. Separate sin(ay) into a linear combination of exp(iay) and exp(-iay).
> >4. For the part with exp(iay) , use the poles in the upper half plane.
> >5. For the part with exp(-iy), use the poles in the lower half plane,
> >but DO NOT FORGET the pole at y=0. (Remember that you dodged *above* 0
> >in step 2.

> Equivalently, change to an integral over the whole line and say
> sin(t) is the imaginary part of exp(it) - 1. Now you can just consider
> the upper half-plane, no dodging the origin necessary.
> Otoh one _does_ need to choose the contour carefully so as to stay
> within a region where coth is bounded...
> I'm curious how this comes out (not curious enough to actually
> try it) - I didn't bother suggesting this the other day because
> I assumed that the answer would be an infinite series that
> we'd be unable to evaluate in closed form.

> >6. Put it all together; the appearance of the answer is as if you
> >replaced sin(ay) by exp(iay) and used all the poles where Im(y) >= 0,
> >but mysteriously took only 1/2 the residue at y=0. Baffle your
> >colleagues by off-hand references to the Cauchy Principal Value, or the
> >"partie finie" of Hadamard.


If I didn't make any mistake, the integral is:

- pi/36 e^(-a/3) ((a+3) cot(1/3)+csc(1/3)^2)
+ pi \sum_{n=1}^{oo} e^(-pi a n)/(1+ 9 pi^2 n^2)^2

and according to wolframalpha, the infinite sum in last
line can be rewritten as a finite sum of 3F2 and 2F1
hypergeometric functions in exp(-pi a).

URL: http://www.wolframalpha.com/input/?i=Sum[exp%28-pi*a*n%29%2F%281%2B9*n^2*pi^2%29^2%2C{n%2C1%2CInfinity}]

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