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

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 achille Posts: 575 Registered: 2/10/09
Re: Difficult integral
Posted: Jun 24, 2011 2:34 AM

On Jun 24, 12:45 am, achille <achille_...@yahoo.com.hk> wrote:
> On Jun 23, 10:09 pm, David C. Ullrich <ullr...@math.okstate.edu>
> wrote:
>
>
>

> > 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
> > >> > 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

>
> > >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/2
>   - 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}]

The correct expression for the integral should be:

pi/2
- 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

According to wolframalpha, the infinite sum in 3rd line can be
rewritten as a finite sum of Lerch trasncedent in exp(-pi a).

URL: http://www.wolframalpha.com/input/?i=Sum[exp%28-pi*a*n%29%2F%281-9*pi^2*n^2%29^2%2C{n%2C1%2CInfinity}]

Date Subject Author
6/22/11 gmarkowsky@gmail.com
6/22/11 G. A. Edgar
6/22/11 gmarkowsky@gmail.com
6/22/11 Herman Rubin
6/22/11 Christopher J. Henrich
6/23/11 gmarkowsky@gmail.com
6/23/11 David C. Ullrich
6/23/11 achille
6/24/11 achille