Laurent series for pole to non integer power

  • Thread starter meldraft
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  • #1
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Main Question or Discussion Point

Hey all,

I am doing a Schwarz-Christoffel transformation and I am trying to calculate the integral analytically using the residue theorem.

My integral is the following:

[tex]\int^\zeta _{\zeta_0} (z+1)\frac{1}{(z+2.9)^{{b_1}/\pi}{(z-0.5)^{{b_2}/\pi}}}dz[/tex]

This has two poles at -2.9 and 0.5. [itex]b_1[/itex] and [itex]b_2[/itex] are not integers.

I want to do this integral for a contour that contains both poles. I know how to use the Laurent series to extract the [itex]a_{-1}[/itex] term (residue) needed for the residue theorem for integer powers (which is to take the limit of the derivative of the same power). Does anyone know how I can find the residue for a function where the poles are raised to a non-integer power?

Cheers

P.S. Lately my fraction lines appear in the web browser distorted, anyone knows what's up with that??
 

Answers and Replies

  • #2
When you've got non-integer powers in the denominator, it means you have branch cuts, hence your function is not continuous around z=-2.9 and z=0.5, hence not holomorphic, hence no poles.
 
  • #3
281
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Awww that sucks :cry: Is there no way to calculate this analytically?
 
  • #4
With the hypergeometric function you can.
 
  • #5
1,796
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Ok, that's interesting. Here's the antiderivative via Mathematica where I use just 2 and not 2.9 and the exponents are b and c:

[tex]
\begin{multline}\frac{1}{2 (-b+\pi )}\left(-\frac{1}{2}+z\right)^{-\frac{c}{\pi }} (2+z)^{-\frac{b}{\pi }} \left((-b+\pi ) (1-2 z)^{c/\pi } \left(1+\frac{z}{2}\right)^{b/\pi } z^2 \text{AppellF1}\left[2,\frac{c}{\pi },\frac{b}{\pi },3,2 z,-\frac{z}{2}\right]\\
+2 \pi \left(\frac{1}{5}-\frac{2 z}{5}\right)^{c/\pi } (2+z) \text{Hypergeometric2F1}\left[\frac{c}{\pi },\frac{-b+\pi }{\pi },1+\frac{-b+\pi }{\pi },\frac{2 (2+z)}{5}\right]\right)
\end{multline}
[/tex]

Ok, that antiderivative, call it [itex]M(x)[/itex] is full of multifunctions and in order to evaluate:

[tex]M(z)\biggr|_{z_1}^{z^2}[/tex]

you would have to take analytic extensions over each multifunction between the points z_1 and z_2. That's quite a challenge I think which means in order to do this one, you'd best work on some simpler ones where you have to analytically extend the antiderivative. Also, since the antiderivative is multivalued, so too will be the answer, one value for each sheet of each function you integrate over and if the exponents are irrational, the answer is infinitely-valued.

All in all, a nice problem to work on. Probably take me the entire semester. :)
 

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