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I have an integral of the following form $$ \int^w_0 \frac{bz}{(z-a)(z+a)}\textrm{d}z,\quad z\in\mathbb{C} $$ which goes from the origin to the point $w$ in the complex-plane. I'm feeling a bit shaky about how to solve this integral. I'm thinking about parameterise $z$, but I don't really know if that will help me. I'm also aware of the two poles that I have at $z=\pm a$.

A clue of how to get started would be of great appreciation.

Background: I have a scalar potential given by $\phi$ that's a function of two complex variables $\alpha$ and $\beta$. Such a potential defines a generalised force $$\vec{\Phi}(\alpha,\beta)=-\vec{\nabla}\phi(\alpha,\beta)$$ For the function $\phi$ to be a well-behaved potential one must require that the crossed derivatives of the force components are the same $$ \frac{\partial}{\partial\alpha}\Phi_\beta= \frac{\partial}{\partial\beta}\Phi_\alpha $$ In my case the force vector is given by $$ \vec{\Phi}(\alpha,\beta)= 2 \begin{pmatrix} \dfrac{k\beta\alpha^2+(i\kappa/4-k)\alpha-\beta\varepsilon_p}{k\alpha^2-\varepsilon_p}\\ \dfrac{-k\alpha\beta^2+(i\kappa/4+k)\beta+\alpha\varepsilon^*_p}{-k\beta^2-\varepsilon^*_p} \end{pmatrix} $$ and the scalar potential can be found from $$ \phi(\alpha,\beta)=\phi(0,0)-\underbrace{\int^{(\alpha,0)}_{(0,0)}\Phi_\alpha(\alpha',0)d\alpha'}_{\textrm{I}} -\int^{(\alpha,\beta)}_{(\alpha,0)}\Phi_\alpha(\alpha,\beta')d\beta' $$ Focusing on the integral I we have $$ \int^{(\alpha,0)}_{(0,0)}\Phi_\alpha(\alpha',0)d\alpha' =2\int^{(\alpha,0)}_{(0,0)}\frac{(i\kappa/4-k)\alpha'}{k\alpha^{\prime 2}-\varepsilon_p}d\alpha' $$ which leads back to my original problem if we call $b=(i\kappa/4-k)$, $a=\sqrt{\varepsilon/k}$ and $z=\alpha'$

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  • $\begingroup$ I think so you can use Cauchy Residue Theorem $\endgroup$ – Abhinav Jha Feb 20 '18 at 9:15
  • $\begingroup$ @AbhinavJha I though so to, but don't I need a closed contour for that? $\endgroup$ – Turbotanten Feb 20 '18 at 9:15
  • $\begingroup$ If you have it, where's the question? What are you planning to do with it? $\endgroup$ – Professor Vector Feb 20 '18 at 9:16
  • $\begingroup$ @ProfessorVector What do you mean if I have it? The integral? I'm solving a difficult physics problem and I need to solve this integral to find the solution to my physics problem. $\endgroup$ – Turbotanten Feb 20 '18 at 9:18
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    $\begingroup$ @ProfessorVector That's excellent $\endgroup$ – Turbotanten Feb 20 '18 at 9:30
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Hint:

$$\frac1{z-a}+\frac1{z+a}=\frac{2z}{(z-a)(z+a)}$$ leads to the antiderivative

$$\log(z-a)+\log(z+a)=\log(x^2-y^2-a^2+i2xy).$$

Now you have to discuss the handling of the singularities.

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  • $\begingroup$ You just blew my mind. Thank you so much. This is a perfect hint! $\endgroup$ – Turbotanten Feb 20 '18 at 10:08
  • $\begingroup$ The antiderivative is elementary. The problem is that $\log(z^2-a^2)$ is multivalued. That's because the value of the integral depends on the integration path: if a pole (or two) lies between two paths, the value of the integral will be different. Usually, that's taken care of by introducing branch cuts, and forbidding the integration path to cross that cut. If that's meaningful for your physical problem, you'll have to decide. $\endgroup$ – Professor Vector Feb 20 '18 at 10:40
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You apply CR Theorem only for closed curve intergals else you need to do the integral directly provided no poles on your path.

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