anyone feel like computing an integral using residues?
$$\int_0^{2\pi}\frac{d\theta}{a+\sin(\theta)},a\in \mathbb R, a1$$gimmick is to write
$$z=e^{i\theta}$$ and then use residues.

Question

anyone feel like computing an integral using residues?
$$\int_0^{2\pi}\frac{d\theta}{a+\sin(\theta)},a\in \mathbb R, a>1$$gimmick is to write 
$$z=e^{i\theta}$$ and then use residues.

anonymous · Answer

guess not.  i don't feel like it either, but i guess i am stuck with it

anonymous · Answer

ok it wasn't that bad after all

anonymous · Answer

think the answer is (2*pie/(a^2 -1)^0.5)  .. I used the following algorithm ..

algorithm is  sin(x) = (e^(ix) - e^(-ix))/2
                   hence we have to find poles for (2/((z^2 -1) +2*i*a*z))
                  poles would be z = i*(-a +- (a^2-1)^0.5)
only z =p1 =  i*(-a + (a^2 -1)^(0.5)) would lie in the region z = e^(ix) .. hence find the limit 
z -> p1  for (z - p1)*f(z) = (1/i*(a^2-1)^0.5)
now simply multiply by 2*pie*i to get 2*pie/(a^2-1)^0.5