at the moment I am trying to reproduce the results of a paper.
There, it turns out that a specific physical problem is mapped onto an integral to be calculated:
$I(\Theta; a, b) = 2\cdot\int_0^\infty \frac{a-u^2}{\left( u^2+\frac{a^2}{b-a}\right) \left(u^2+\frac{b^2}{b-a} \right) \sqrt{a - u^2 -\frac{a}{1-a/b}\sin^2(\Theta) } }\mathrm {d}u \equiv \int_0^\infty f(u)du$
where I took the liberty to replace $\epsilon_d\rightarrow a$ and $\epsilon_m \rightarrow -b$ in contrast to the paper and one can assume that both
$a,b > 0$ and $b > a$.
Somehow, Mathematica manages to be able to calculate the numerical values of this integral with some warning messages due to the pole at $u_p = \sqrt{a -\frac{a}{1-a/b}\sin^2(\Theta)}$
Also, Mathematica can calculate the indefinite integral, both for $\Theta = 0$ (which is a special case to compare results) and in the general case. Nevertheless, I am not able to use the result since it is indefinite in all cases for $u\rightarrow \infty$.
So, I am asking for some advice to calculate the integral at hand with respect to given constant $a$ and $b$.
The special case of $\Theta = 0$ might already be worth to take a look since it is much easier to calculate it than for the general case.
In the meantime I tried something like a Cauchy principal value integration around $u_p$ using the parametrization $u_\delta (\varphi) = u_p -\delta e^{\mathrm{i}\varphi}$ along a half circle $C_\delta$ interpreting $u^2$ as $\bar{u}u$. Then,
$I_\delta = \int_0^\pi f(u_\delta(\varphi))\delta d\varphi$ is the integral around $C_\delta$ which turned out to vanish for $\delta\rightarrow 0$ such that the whole integral should be given in terms of a principal value one. Noteworthy, I am not sure if my result is correct.
Please, if my question is not stated correctly, or anything is obvious don't hesitate to give me some advice.
Thank you in advance.
Sincerely
Robert