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Re: Integral
Posted:
Jul 22, 2011 3:00 PM


On 22.07.2011 16:01, Axel Vogt wrote: > On 21.07.2011 02:42, rancidmoth@yahoo.com wrote:
>> 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) ... > > For a=1/2/2, b=1/2 I get 0.765149580251623 for the integral and > 0.765143786388934 for your formula, using Maple with 15 Digits. > So it seems that it is not correct for a,b<1 as well.
Though I have no idea, why a closed form should exist I played with it, considering a series in b=0:
Even derivatives vanish, since the function is odd and the odd ones in b=0 seem to be
Int(sin(a*cos(t))*sin(t)^4/cos(t),t = 0 .. 2*Pi)
which is just twice that integral over 0 .. Pi.
There one can invert cos and t =arccos(x) leads to
Int(sin(a*x)*(1x^2)^(k1/2)/x,x = 1 .. 1)
which Maple can solve in terms of special function (including BesselJ).
There I stopped, as I have no idea, why the resulting series in b should give something helpful.
May numerical is would be a the better way ...



