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achille
Posts:
575
Registered:
2/10/09


Re: Difficult integral
Posted:
Jun 23, 2011 12:45 PM


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 20110622, G. A. Edgar <ed...@math.ohiostate.edu.invalid> wrote: > >> > In article > >> ><551e4221dcfa41079c80ed028627c...@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 nonnegative 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 halfplane, 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 offhand 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%28pi*a*n%29%2F%281%2B9*n^2*pi^2%29^2%2C{n%2C1%2CInfinity}]



