Hy, I would like some help with my integral : $ \oint_B\frac{1}{z^2}\sqrt{(z-a)(z+\bar{a})(z-\frac{1}{a})(z+\frac{1}{\bar{a}})} $
Where B is the contour that connect the two branch cuts associated to my square root. And moreover, I know that the points z=1 and z=-1 belong to my two branch cuts.
I naively tried to compute half of the integral by choosing one branch of my multi-valued function : ( by putting $z=e^{-i\theta}$ from 0 to $\pi$):
\begin{eqnarray} \int^{\pi}_{0}d\theta\sqrt{(2\cos{\theta}-A)(2\cos{\theta}+\bar{A})} \end{eqnarray} Where $A=a+\frac{1}{a}$
Then by changing the variable $u=\sin{\frac{\theta}{2}}$ :
\begin{eqnarray} = 4\int^{1}_{0}du \sqrt{\frac{(1-2u^2-\frac{A}{2})(1-2u^2+\frac{\bar{A}}{2})}{1-u^2}} \end{eqnarray}
Then $v=1-2u^2$ : \begin{eqnarray} =-4\int^{-1}_1dv\frac{1}{4\sqrt{(\frac{1-v}{2})(\frac{1+v}{2})}}\sqrt{(v-\frac{A}{2})(v+\frac{\bar{A}}{2})} \end{eqnarray} \begin{eqnarray} = -2\int^{-1}_1dv\sqrt{\frac{(v-\alpha)(v+\bar{\alpha})}{(1-v)(1+v)}} \end{eqnarray} where $\alpha=\frac{A}{2}$
But since $\alpha$ is complex, I didn't find a way to express to espress it using elliptic functions. I have two questions :
- Does any one know how to compute this kind of integral, that connects two branch cuts ? I'm a physicist, I'm not quite familiar with Riemann surfaces etc ..
- If my naive approach is correct, does anyone know how to simplify my last expression ?
Any help or even a reference would be very helpful !
Here is the kind of cycle I want to compute : B-cycle integral
Thanks