Find that Residue!

Calculus Level 5

$\oint_C\Gamma(z) \, dz$

Find the integral above, where $C$ is the circle $x^2+y^2=8$ , oriented counter-clockwise.

Notation : $\Gamma(\cdot)$ denotes the Gamma function .

2\pi i

-\pi i

\pi i

-2\pi i

-1 0

1 solution

Peter Macgregor
Mar 22, 2016

The wonderful residue theorem from complex analysis says that the value of a contour integral in the complex plane is

$2 \pi i \sum{\text{residues within the contour}}$

According to WolframMathWorld the gamma function has residues only at the non-positive integers, where they take the values $(-1)^{n}/n!$ .

Since our contour is a circle centred on the origin with radius $\sqrt{8}$ , and since $\sqrt{8}<3$ we need consider only the residues at 0,-1 and -2.

And so the integral is

$2 \pi i \left(\frac{(-1)^0}{0!} + \frac{(-1)^1}{1!} + \frac{(-1)^2}{2!} \right) = 2 \pi i \left(1-1+\frac{1}{2}\right) = \pi i$

Yes exactly! Very nice! (+1) The poles of the Gamma function are well explained here

Otto Bretscher - 5 years, 2 months ago

Sir, I know The Basics of line Integral.Form my point of n]view, $\Gamma(z)$ is A single variable function of z. Now it moves on the points on the circle,Sir,can you please tell the physicality of the integral,Than I can mange the Algebra

A Former Brilliant Member - 5 years, 2 months ago

This is a pretty elaborate theory; you will need to study some complex analysis.

The big picture is this (I'm omitting some technical details): If you have a simple closed curve $C$ in the complex plane and a function $f(z)$ that is (complex) differentiable in the region enclosed by $C$ , then $\oint_{C}f(z)dz=0$ by Stokes' theorem.

Things get more complicated when $f(z)$ fails to be differentiable at some point $a_1,a_2,...,a_n$ inside $C$ ; in our case those are the points $0,-1,-2$ . That's where the residues come in. See here .

See how much sense this makes, and feel free to ask more questions!

Otto Bretscher - 5 years, 2 months ago

Find that Residue!

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