# Integration of 1/(square root of a third order polynomial) on the complex plane

I have to compute the integral

$$\oint_{a,b} \frac{d \lambda}{\sqrt{(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)}},$$

Here is the picture of the integration contours and cuts:

The "saw-like" lines are the cuts, while a and b are the integration contours. I have to compute the integral on this 2 contours, and I feel completely lost, especially for the a cycle where the contour does not seem to cross the cut. With my basic knowledges I know how to compute some easy integrals like a square root of z on the unit circle, but I wonder if I can do the same kind of computation here. Any suggestion?

• See residues theorem Commented Jul 26, 2016 at 14:37
• It evokes me the Carlson form of an elliptic integral. It would suffice to take a transform that sends $-\lambda_0,-\lambda_1, -\lambda_2$ onto, for example $x=0,y=1,z=2$ resp. ? Commented Jul 26, 2016 at 14:49

For the path named $(a)$ in your drawing, preserving the value of the integral you may shift the contour. You could e.g. shrink it towards the cut between $\lambda_1$ and $\lambda_2$ or you may blow it up to the one shown in my figure. I'll discuss the latter option. There is a small half-circle around $\lambda_3$ of radius $r>0$, a large (almost complete) circle of radius $R$ and two segments along the cut from $\lambda_3$ to $-\infty$ . Letting $r\rightarrow 0$ and $R\rightarrow +\infty$ the circular parts go to zero (not too difficult to see) and the only remaining part is along the cut. The integral above the cut then gives the same contribution as the lower (the directions are opposite but you get a minus sign from the determination of the square root which changes sign when going through the cut). The path, say below the cut, is given by $z=\lambda_3-t$ with $t$ going from $0$ to $+\infty$. Set $\lambda_{1,2}-\lambda_3=a\pm i b$. The integral may then be expressed as follows:

$$2i \int_0^{+\infty} \frac{dt}{\sqrt{t\ ((t-a)^2+b^2)}}$$

The $i$ outside comes from $\sqrt{z-\lambda_3}=\sqrt{-t}=-i \sqrt{t}$, $t>0$, but the sign is arbitrary, or rather, should be determined by the value of the square root at some fixed chosen point in the complex plane. To my knowledge there is no closed form solution other than this 'abstract' elliptic integral.

Later edit:

Regarding the loop $(b)$: You may construct a Riemann surface by taking two copies of your cut plane and glue them together across the cuts. The integrand becomes globally holomorphic (or extends to such) on the resulting Riemann surface (topologically a torus). Loop $(b)$ is then glued together from one path on each surface. In order to calculate the integral I believe that it is better to make a cut from $\lambda_2$ to $\lambda_3$ and from $\lambda_1$ to $\infty$. You may then play the same game as in the previous case. The function to integrate will, however, this time not be real.

The two paths (a) and (b) corresponds to two generators for the torus and yields two ${\Bbb R}$ independent complex numbers $z_a$ and $z_b$. Any other loop will give rise to an integer combination of the two values, i.e. a lattice $\Lambda = {\Bbb Z} z_a + {\Bbb Z} z_b \subset {\Bbb C}$. The quotient ${\Bbb C}/\Lambda$ is then the above mentioned torus.

• Thanks for your reply, I think I see what you mean. Sorry for my poor understanding of complex integration, but is seems that the contours are more complicated. Here is the picture of the contours and cuts: img15.hostingpics.net/pics/851878Capturedcran20160726171104.png The 3 singularities of the polynomial appears, and the "saw-like" lines are the cuts introduced by the author of the article. I have to compute the integral along the a and b cycles. Do you think it's still possible to do what you said? Commented Jul 26, 2016 at 15:41
• From the picture it appears that $\lambda_1$ and $\lambda_2$ are complex conjugates? And that $\lambda_2$ is real and negative? Do you really need numbers or more like a relationship between the two integrals? (which seems feasible) Commented Jul 26, 2016 at 15:54
• Sorry for the second comment, but I edited my question to be more clear. Commented Jul 26, 2016 at 15:54
• Yes, this two are complex conjugates. I just need to evaluate the value of the integral on this 2 contours. Numerically, I have no idea if it is possible. Commented Jul 26, 2016 at 16:21
• Your explanation for the b contour seems clear to me, thanks a lot. But I have a question for the a contour: Let's imagine 2 cuts on the real axis, one on $(z_1,\infty)$, another on $(-\infty,z_2)$. Now, integrate something like $\sqrt{(-z+z_1)(z+z_2)}dz$ on a contour which crosses the 2 cuts. I guess it is possible to do that? Doesn't it look like the integration on the a contour? what is wrong here? Commented Jul 27, 2016 at 14:30