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Contour Integration
By Brad Rodgers (P1930) on Saturday, March 24, 2001 - 03:33 am:

What is contour integration? When would it be used?

Thanks,

Brad

By Michael Doré (Md285) on Saturday, March 24, 2001 - 03:44 am:

See Contour Integral.

By Kerwin Hui (Kwkh2) on Saturday, March 24, 2001 - 09:57 am:

Brad,

To say what contour integration means, we first have to define path integration. We can parameterise a path G from A to B in the complex plane as, say, z(t) from t=a to t=b. Then, we define the path integral of a complex function f(z) along G as

òGf(z)dz = òa bf(z(t))×z'(t)dt

[I assume you know how to do that integral - it is simply

òabf(z(t))×z'(t)dt = òa bRe(f(z(t))×z'(t))dt + iòabIm(f(z(t))×z'(t))dt]

We can define contour integration in a similar manner. A contour is simply a closed curve in the complex plane, and we can evaluate the contour integral as a sum of 2 path integrals.


Some 'nasty' integrals can be evaluated easily by choosing appropriate contours and applying Cauchy's Residue Theorem, for example

ò-¥ ¥1/(1+x4)dx

This integral can be evaluated by elementary means(by partial fractions) or by considering the contour integral along the semicircle with base {(r,0):-r£r£r}.

See Michael's link for a demonstration of using this technique.

Kerwin

By Brad Rodgers (P1930) on Monday, March 26, 2001 - 11:27 pm:

Could someone go through what process occured from "Can you work out the value of this integral by contour integration (you have already worked out where the poles are - just work out the residues too)?" (I've no idea what this even means) to where we've found what the contour integral evaluates to?

Thanks,

Brad

By Kerwin Hui (Kwkh2) on Tuesday, March 27, 2001 - 01:03 am:

For any complex function f(z), we can find a Laurent series about an arbitrary point z:

f(z)=...+a2(z-z)-2+a1(z-z)-1+a0+a1(z- z)+...

generally speaking, this series converges for z in some annulus centred at z. The residue of f(z) at z=z is a-1, written Res(f, z).

A number z is a pole of order k (integer) if (z-z)kf(z) is analytic at z but (z-z)k-1f(z) is not. If there is no such k then z is called an essential singularity.

Cauchy's Residue Theorem states that, for a simple closed path C encircling poles z1, ..., zN

ó
õ

C 
 f(z) dz = 2pi N
å
i=1 
 Res(f,zi)
where C is taken to transverse anticlockwise.

Back to the example, we have f(z)=z2/cosh z. The only pole inside the contour is at z=½pi. This is a simple pole (pole of order 1), and the residue can be evaluated easily, hence the integral.

Kerwin