Why do poles occur at specific values of q in algebraic function integration?

[Master J]

Say I am integrating some algebraic function with respect to a variable q. There is also an exp( iq ) in the integral as a factor.

There are poles when q takes a certian value as it is in the denominator as q - A, so pole at q = A.I know the poles are in the upper left plane, and lower right , but I can't see why this is. Could someone explain this ?

In addition, there are finite limits to the integral. Can one still use the residue method on this?

Cheers!

PS

If someone would like to also show me how to integrate it, I would be ever so grateful:

Integral of dq. exp ( iqd) / cos(qd) - E between pi/d and -pi/d

 

[Master J]

\int^{d/pi}_{-d/pi}exp( iqd) / cos(qd) - E dq

 

[homology]

\int^{d/pi}_{-d/pi}exp( iqd) / cos(qd) - E dq

\int^{d/\pi}_{-d/\pi} \frac{e^{iqd}}{\cos(qd)-E}dq

click on the formula and you can see how its tex'd

 

[Master J]

The wonders of Latex! :-) Thanks for putting that straight.


Now that the integral is clear, can anyone share with me HOW one does it??

 

[homology]

yeah sorry. My residue knowledge is a little rusty, I was paying attention to follow along when someone else answers :p

 

[Master J]

No problem. Hopefully someone here can enlighten me? I haven't done much Complex Analysis, I've just recently encountered it in a course on the application of Green's Functions to physics, so I am at a loss here!

 

[mathwonk]

I am a little puzzled. Since cosine is periodic, it seems to equal E an infinite number of times, and where this happens should depend on E.

If you want to look at your integral as along an interval on the x axis, and as part of some closed path in the plane, it does not lend itself to the usual method of residues since the poles seemingly keep occurring more and more often as you try to go to infinity. but maybe you can take a vertical rectangle, based on your interval and maybe there will only be a finite number of poles in there.

I would look in some complex book like churchill.