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

 Messages: [ Previous | Next ]
 Axel Vogt Posts: 1,068 Registered: 5/5/07
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)*(1-x^2)^(k-1/2)/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 ...

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