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Complex analysis - an integral with branch cuts

  • Thread starter Loro
  • Start date
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1
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]

3. 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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BiGyElLoWhAt

Gold Member
1,560
113
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]

3. 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.
 

statdad

Homework Helper
1,493
33
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?
 

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