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I need to evaluate the following integral: \begin{equation} \int_{z_1}^{z_2} d z \sqrt{g(z)}, \end{equation} where the function $g(z)$ is given by \begin{equation} g(z)=-\left(\alpha-\frac{\beta}{z^2}+\frac{\gamma}{z^4}\right)=\frac{\alpha\,(z^2-z_1^2)(z_2^2-z^2)}{z^4}, \end{equation} and $z_1$ and $z_2$ are two zeros of $g(z)$ with positive real parts.

When $z_1$ and $z_2$ are real numbers, by the change of variable $z^2=z_2^2-(z_2^2-z_1^2)\sin^2(\theta)$ I write the integral in the form of elliptic functions. The result is:
\begin{equation} 2\sqrt{\alpha} z_2 \left((1-\frac{m^2}{2})K(m)-E(m)\right), \end{equation} where $m:=(z_2^2-z_!^2)/z_2^2$ and $K(m)$ and $E(m)$ are respectively the complete elliptic integrals of the first and second kind.

The problem arises when $z_1=z_2^*$ are complex conjugate roots. In this case, in the above integral the path lies along the imaginary axis and the branch of $\sqrt{g(z)}$ is real and positive. I don't know how to perform the integration in this case. But, surprisingly, for some values of $z_1=z_2^*$, the numerical value obtained by Mathematica is exactly identical with the result obtained in the case of real zeros of $g(z)$.

My questions are:

(1) Which appropriate contour I should take to perform the integral in the case $z_1=z_2^*$?

(2) Is there any analytical result for the integral in this case? If yes, how can I find it?

(3) Two special choices of the parameters may be helpful. $(\alpha, \beta, \gamma)=(1.5, 0.75, 0.25)$ and $(\alpha, \beta, \gamma)=(4, -0.05, 0.1225)$.

Any help will be appreciated.

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  • $\begingroup$ Do you know how to show that $f(z)=( (z+1) (z-1))^{1/2}$ is analytic on $\Bbb{C} - [-1,1]$ ? Your function is $g(z)^{1/2}=\frac{c}{z^2} f(z/z_1)(f(z/z_2)$ which is analytic on $U = \Bbb{C}- [-z_1,z_1]-[-z_2,z_2]$. Then $G(z)=\int_0^z g(s)^{1/2}ds$ is analytic on simply connected open sets contained on $U$ and its other branches are $G(z)+u_1n_1+u_2n_2$ where $u_j = \int_{\gamma_j}g(z)^{1/2}dz$ with $\gamma_j$ the closed loop enclosing $[-z_j,z_j]$ $\endgroup$ – reuns Sep 12 at 11:39
  • $\begingroup$ @reuns Thanks for your comment. I know the branch points and cuts of $f(z)=\sqrt{z^2-1}$ and $g(z)^{1/2}$. My first question was bad. I don't know how to take the appropriate branch cut and contour to perform the integral. I had to ask this one: "how can I take the appropriate contour of integration?" I edited that. $\endgroup$ – Nzamani Sep 12 at 20:29
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Let $\alpha = 1$. I assume that you mean that it's possible to choose a branch of $\sqrt {g(z)}$ in such a way that $I = \int_{z_1}^{z_2} \sqrt {g(z)} \, dz$ is real when $z_1 = z_2^*$; $\sqrt {g(z)}$ itself isn't real on $[z_1, z_2]$.

Since the possible values of $m$ are such that $|m - 1| = 1 \land m \neq 0$ (in particular, $m = 2$ when $\arg z_2 = \pi/4$), we need to construct a branch of $$\left( 1 - \frac m 2 \right) K(m) - E(m)$$ (not $1 - m^2/2$) which is continuous at those points. For this, we can use the algebraic relation between $E$ and $K$ and take $$K(m) = m^{-1/2} (K(m^{-1}) + i K(1 - m^{-1})),$$ then the arguments do not lie on $[1, \infty)$. We obtain $$I = z_2 \frac {(2 E(1 - m) - m K(1 - m)) (K(m^{-1}) + i K(1 - m^{-1})) - \pi m^{1/2}} {m^{1/2} K(1 - m)}, \\ m = 1 - \left( \frac {z_2^*} {z_2} \right)^{\! 2}, \quad 0 < \arg z_2 < \frac \pi 2,$$ where $E(m)$ and $K(m)$ are analytic continuations from $(-\infty, 1)$ to $\mathbb C \setminus [1, \infty)$, which I assume is what is desired for numerical calculations.

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  • $\begingroup$ @Mxim, Thanks a lot for your answer. I'm working in physics, not math. I couldn't understand whether it is a complexification of the result obtained for real $z_1$ and $z_2$ or an analytic continuation. Why did you work with $(1-m/2)$ instead of $(1-m^2/2)$? Also, would you please let me know which handbook you used for the connection relations between $K$ and $E$? Thanks again. $\endgroup$ – Nzamani Sep 16 at 12:41
  • $\begingroup$ @Maxim, I cannot get the $I$ formula Can you just explaining a little more, please $\endgroup$ – user62498 Sep 16 at 13:16
  • $\begingroup$ I was referring to Legendre's relation. $(1 - m^2/2) K(m) - E(m)$ is simply not correct for real $z_1, z_2$. I'm not sure what your definition of complexification is, but the point is that we want to take the result obtained for real $z_1, z_2$ and extend it analytically to the relevant domain of $m$ values. $\endgroup$ – Maxim Sep 16 at 13:43

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