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Topic: Integral
Replies: 5   Last Post: Jul 22, 2011 3:00 PM

 Messages: [ Previous | Next ]
 Dann Corbit Posts: 1,424 Registered: 12/8/04
Re: Integral
Posted: Jul 21, 2011 7:53 PM

In article <3ise275f397jphilkph0rpnhb99k3gv46l@4ax.com>,
rancidmoth@yahoo.com says...
>
> Hello all,
>
> I'm trying to evaluate the following integral
>
> S(a,b) = integrate(0,2pi) sin(a*cos(t))*sin(b*sin(t)) /
> (sin(t)*cos(t)) dt
>
> I get S(a,b) = 16pi*a*b*J(2,sqrt(a^2+b^2))/(a^2+b^2)
>
> where J(k,z) is the bessel function of the first kind, order k.
> However funny numerics started occuring and i think i have traced it
> back to this integral. It would appear (in mathematica and maple)
> that numerically this appears reasonable for a,b<1. but for a,b>1 the
> numerical integration and my result diverge quite significantly...too
> much to perhaps be numerical error in the integration.
>
> My steps are as follows:
>
> cos(a*z+b/z) = -J(0,2sqrt(ab)) + sum(k=0,oo) (-1)^k ((a*z)^(2k) +
> (b/z)^(2k))*J(2k,2sqrt(ab))/(sqrt(ab))^(2k)
>
> I derived this by using binomial theorem in the series for cos - also
> numerically it appears bang on. Using this, convering the integral to
> that over the unit circle, yeilds my result.
>
> Alternatively one may use similar laurent series for sin(z+1/z),
> multiply them together, get the residue and get the same result.
>
> what have i missed?

http://integrals.wolfram.com/index.jsp?expr=sin%28a*cos%28t%29%29*sin%
28b*sin%28t%29%29+%2F%28sin%28t%29*cos%28t%29%29+&random=false

result:
x Csc[t] Sec[t] Sin[a Cos[t]] Sin[b Sin[t]]

Date Subject Author
7/20/11 Rancid Moth
7/21/11 Dann Corbit
7/22/11 Rancid Moth
7/22/11 Axel Vogt
7/22/11 Sam2718
7/22/11 Axel Vogt