# Integrating a function of which poles appear on the branch cut

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• tworitdash
In summary, the conversation discusses a complicated function to integrate from -\infty to \infty, involving I0 and I2 functions containing Hankel functions of the second kind. The expressions for these functions are also given when \beta = \Omega, and they help avoid poles at those points. The square root term is evaluated using a branch cut and Hankel functions have a logarithmic discontinuity at the origin. Two integration paths are considered, with the first one starting from negative infinity on the real axis and the second one starting from negative imaginary infinity. The problem is that the poles at \pm \beta and \pm k lie on the branch cut, making it difficult to avoid them or apply Cauchy's theorem. The
tworitdash
I have a complicated function to integrate from $-\infty$ to $\infty$.

$$I = \int_{-\infty}^{\infty}\frac{(2k^2 - \Omega^2)(I_0^2(\Omega) + I_2(\Omega)^2) - \Omega^2 I_0(\Omega) I_2(\Omega)}{\sqrt{k^2 - \Omega^2}} \Omega d\Omega$$​
Where I0I0 and I2I2 are functions containing Hankel functions as below.

$$I_0(\Omega) = \frac{R}{\beta^2 - \Omega^2}\Big(aJ_1(\beta R)H_0^{(2)}(\Omega R) - \Omega J_0(\beta R) H_1^{(2)}(\Omega R)\Big)$$

$$I_2(\Omega) = \frac{R}{\beta^2 - \Omega^2}\Big(-aJ_1(aR)H_2^{(2)}(\Omega R) + \Omega J_2(aR) H_1^{(2)}(\Omega R)\Big)$$​
$H_n^{(2)}$ are Hankle functions of the second kind of order n. $J_n$ are the Bessel function of the first kind and order n. Along with this I also happen to have the expressions when $\beta = \Omega$ and they are,

$$I_0(\Omega) = \frac{1}{2} R^2 \Big( (H_0^{(2)}(\Omega R))^2 + H_1^{(2)}(\Omega R)H_2^{(2)}(\Omega R))\Big)$$

$$I_2(\Omega) = \frac{1}{2} R^2 \Big( (H_2^{(2)}(\Omega R))^2 - H_1^{(2)}(\Omega R)H_3^{(2)}(\Omega R))\Big)$$

The above 2 expressions are valid on only points where $|\Omega| = \beta$ These expressions actually help avoid the poles at those points in the actual expressions of $I_0$ and $I_2$. However, in my question I don't address the pole free expressions.

So, now the square root term is evaluated as $\sqrt{k^2 - \Omega^2} = -j \sqrt{-(k^2 - \Omega^2)}$. And Hankel functions have a logarithmic discontinuity at the origin. I extended it to the negative infinity for a branch cut. Therefore, the branch cuts of the square root and Hankel functions are shown in the figure here.

I have tried 2 integration paths. The first path starts from negative infinity (and a small delta below the real axis) on the real axis, goes just below the origin to avoid the branch cut of Hankel function, and moves to positive infinity on the real axis (and a small delta above the real axis). The delta is used to avoid the poles at $\pm \beta$ and at $\pm k$. The second integration path starts from negative imaginary infinity on the third quadrant and follows the same path about the origin but comes to again negative infinity on the imaginary axis in the fourth quadrant after covering the branch-cut of square root just after k on the real axis.In the problem $|\beta| < |k|$ always. Therefore, poles at $\pm \beta$ appear on the branch cut of the square root function. I am not using the pole free expressions for the integrated so the delta makes a huge impact on the integral. The poles lie on the branch cut, so I don't know how to avoid them or how to apply the Cauchy's theorem because one can't move around the pole jumping on the branch cut.

The integral is real when I take the integral only till k and it is purely imaginary when I take the integral from k to $\infty$. This happens because of the square root function. If I change the delta, the real part of the integral changes considerably which makes me think that the poles at $\pm \beta$ and $\pm k$ are responsible. Can I use the pole free expressions? I am afraid I can't because it is only valid at the pole points and not its locality. Please share your thoughts.

Last edited:
I figured out how to execute this without any warnings. I used pole free expressions at the locality of the pole. I took an offset of e-7 so that it doesn't come very close to the pole location. Furthermore, I didn't use the second integration path I have mentioned in the picture above and used the first integral path, but from 0 to $\infty$. For this reason, I changed all Hankle functions of the second kind to Bessel functions of the first kind. (because that was my original problem and keeping in mind that I can make a faster convergence, I used Hankel function from $-\infty$ to $\infty$)

## 1. What is a branch cut in the context of integrating a function?

A branch cut is a line or curve in the complex plane where a function is not well-defined. It is usually chosen to be along the negative real axis, but can vary depending on the function. Poles, or singularities, are points on the branch cut where the function becomes infinite.

## 2. Why do poles appear on the branch cut of a function?

Poles appear on the branch cut because they are points where the function is not well-defined. They can occur when the denominator of a rational function becomes zero, resulting in an infinite value for the function.

## 3. How do poles affect the integration of a function with a branch cut?

Poles can make the integration of a function with a branch cut more challenging because they can cause the integral to diverge. This means that the integral does not have a finite value and cannot be solved using traditional methods. Special techniques, such as contour integration, may be necessary to properly integrate a function with poles on the branch cut.

## 4. Can a function with poles on the branch cut still be integrated?

Yes, a function with poles on the branch cut can still be integrated. However, it may require more advanced techniques, such as contour integration, to properly evaluate the integral. It is important to carefully consider the location and type of poles when attempting to integrate a function with a branch cut.

## 5. How can I determine the location of poles on a branch cut?

The location of poles on a branch cut can be determined by analyzing the function's denominator. Poles occur when the denominator becomes zero, so solving for the roots of the denominator can give insight into where poles may appear on the branch cut. It is also important to consider the function's behavior near the branch cut to identify any potential poles.

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