Understanding Contour Integrals with Examples

In mathematics, specifically in complex analysis, a contour integral is a line integral in the complex plane. This article aims to elucidate the concept of a contour integral, particularly focusing on the evaluation of integrals using Cauchy's Integral Formula and Cauchy's Residue Theorem. The primary goal is to provide a clear understanding of how these theorems can be used to solve complex integrals.

What is a Contour Integral?

A contour integral is a type of line integral taken along a specific curve called a contour. Contours are often closed paths in the complex plane. The integrand is a function of a complex variable, and the integral is taken along the specified contour. The contour integral is defined as:

( oint_C f(z) dz )

where (C) is a closed curve in the complex plane, and (f(z)) is a function of the complex variable (z).

Evaluating Contour Integrals Using Cauchy's Integral Formula

Consider the specific contour integral defined as:

( oint_C frac{1 - cos{z}}{z - 1 (z - 1)^2} dz )

The given contour integral assumes a counterclockwise orientation and an integrand that is analytic everywhere inside and on the contour (C: z 4), except when (z pm 1) from setting the denominator equal to zero. Since both singularities lie inside (C), we can decompose the contour into two smaller contours (C_1) and (C_2) that contain only (z -1) and (z 1) respectively, such that the integrand is otherwise analytic inside and on these respective contours.

Applying Cauchy's Integral Formula to evaluate each one, we have:

( oint_C frac{1 - cos{z}}{z - 1 (z - 1)^2} dz oint_{C_1} frac{1 - cos{z}}{z - 1 (z - 1)^2} dz oint_{C_2} frac{1 - cos{z}}{z - 1 (z - 1)^2} dz 0 )

This simplifies due to the cancellation of residues, leading to the final result of 0.

Evaluating Contour Integrals Using the Cauchy's Residue Theorem

The same integral can be evaluated using the Cauchy's Residue Theorem, which states that for a simple closed curve, the value of the integral is equal to the sum of residues of the integrand computed at all poles inside the contour, scaled by (2pi i).

For the function ( f(z) frac{1 - cos{z}}{z - 1 (z - 1)^2} ), the poles are at (z -1) and (z 1), both being simple poles.

Calculating Residue at (z -1):

( text{Res}(f, z-1) lim_{z to -1} (z 1) frac{1 - cos{z}}{z - 1 (z - 1)^2} - frac{1}{2} (1 - cos(-1)) )

Calculating Residue at (z 1):

( text{Res}(f, z1) lim_{z to 1} (z - 1) frac{1 - cos{z}}{z - 1 (z - 1)^2} frac{1}{2} (1 - cos(1)) )

Given the symmetry of the cosine function, we have:

( cos(-z) cos(z) )

Thus, the residues sum up to zero, leading to:

( oint_{z4} frac{1 - cos{z}}{z - 1 (z - 1)^2} dz 0 )

Therefore, the value of the contour integral is zero.

Conclusion

Contour integrals are a powerful tool in complex analysis, and their evaluation using Cauchy's Integral Formula and the Cauchy's Residue Theorem provide efficient methods to solve complex problems. Understanding these concepts is crucial for students and researchers in the field of complex analysis and its applications.