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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]]



