Complex analysis - an integral with branch cuts

In summary: If so, try differentiable functions and see if that helps. If the integral does not converge, that might be an indication that you need to modify your equation in some way. In summary, I think it might help to try splitting the numerator and denominator into fractions and seeing if that helps.
  • #1
Loro
80
1

Homework Statement



Hi, I need to calculate the following integral:

[itex]\int_{-\infty}^{+\infty}dx \frac{(\pi+\sqrt{x^2+m^2})^2(1+\cos x)}{(x^2-\pi^2)^2\sqrt{x^2+m^2}}[/itex]

The Attempt at a Solution



I tried complexifying it:

[itex]\oint dz \frac{(\pi+\sqrt{z^2+m^2})^2(1+e^{iz})}{(z^2-\pi^2)^2\sqrt{z^2+m^2}}[/itex]

And having this over the following contour (sorry for the quality of the image):

https://www.dropbox.com/s/t7ioou1kjs3y7ej/paint.jpg?dl=0

The red dots are the poles at: [itex]\pm\pi[/itex] and [itex]\pm im[/itex].

But is it a valid contour in this case, or should I pay extra attention to the branch cuts of the sqrt in the denominator? If so, how do I choose the contour properly?
 

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  • #2
Loro said:

Homework Statement



Hi, I need to calculate the following integral:

[itex]\int_{-\infty}^{+\infty}dx \frac{(\pi+\sqrt{x^2+m^2})^2(1+\cos x)}{(x^2-\pi^2)^2\sqrt{x^2+m^2}}[/itex]

The Attempt at a Solution



I tried complexifying it:

[itex]\oint dz \frac{(\pi+\sqrt{z^2+m^2})^2(1+e^{iz})}{(z^2-\pi^2)^2\sqrt{z^2+m^2}}[/itex]

And having this over the following contour (sorry for the quality of the image):

https://www.dropbox.com/s/t7ioou1kjs3y7ej/paint.jpg?dl=0

The red dots are the poles at: [itex]\pm\pi[/itex] and [itex]\pm im[/itex].

But is it a valid contour in this case, or should I pay extra attention to the branch cuts of the sqrt in the denominator? If so, how do I choose the contour properly?

Well let me start off by saying that improper integrals are by no means my specialty, but I'll try to offer some assistance if that's alright.

I don't know if I would "complexify " the integral like that, as I'm not sure that really simplifies it.

What I am noticing however, is that you have similar terms in the numerator and denominator, and that makes me think expanding your numerator and splitting it up into 2 (or more) fractions might be helpful, as you might get some useful cancellations.

If that didn't work for me, I might try rewriting this whole thing (or maybe the new split up fraction) by splitting the denominator via partial fractions.

Have you tried any of these things? If so I'd be curious to see what it looked like.
 
  • #3
With a little grunting you can get
[tex]
\begin{align*}
\dfrac{(\pi + \sqrt{x^2 + m^2})^2 (1 + \cos(x))}{(x^2 - \pi^2)^2 \sqrt{x^2+m^2}} & = %
\dfrac{(\pi^2 + 2 \pi \sqrt{x^2 + m^2} + (x^2 + m^2))(1+\cos(x))} {%
(x^2-\pi^2)^2 \sqrt{x^2 + m^2}} \\
& = T1 + T2 + T3 \text{ where}\\
T1 & = \dfrac{\pi^2 (1+\cos(x))}{(x^2-\pi^2)^2 \sqrt{x^2 + m^2}} \\
T2 & = \dfrac{2 \pi(1+\cos(x))}{(x^2-\pi^2)^2} \\
T3 & = \dfrac{\sqrt{x^2 + m^2}(1+\cos(x))}{(x^2-\pi^2)^2}
\end{align*}
[/tex]

You can also note that your integrand is even so you can try integrating from 0 to ∞ and see if that helps.

Are you sure this integral converges?
 

1. What is complex analysis?

Complex analysis is a branch of mathematics that deals with functions of complex numbers. It involves the study of complex-valued functions, their properties, and their behavior. It is a powerful tool in solving problems in various fields, including physics, engineering, and economics.

2. What is an integral with branch cuts?

An integral with branch cuts is an integral that involves a complex-valued function with a branch cut. A branch cut is a curve on the complex plane where a function is not continuous. It is used to deal with multivalued functions, where there are multiple possible values for a given input.

3. Why are branch cuts important in complex analysis?

Branch cuts are important in complex analysis because they allow us to evaluate integrals involving multivalued functions. Without branch cuts, we would not be able to integrate these functions properly and obtain correct results. Branch cuts also help us to define functions that are not continuous on the complex plane.

4. How do we handle branch cuts in complex analysis?

In complex analysis, branch cuts are handled by choosing a branch of the multivalued function and then defining it on the complex plane. This branch is chosen in a way that avoids the branch cut, and the function is defined everywhere else on the complex plane. This allows us to evaluate integrals involving the function while avoiding any discontinuities.

5. What are some applications of complex analysis with branch cuts?

Complex analysis with branch cuts has many applications in different fields of science and engineering. It is used in solving problems related to fluid mechanics, electromagnetism, and quantum mechanics. It also has applications in signal processing, control theory, and image processing. Additionally, complex analysis with branch cuts is used in the development of algorithms for numerical computations and simulations.

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