Solving Contour Integrals Using Cauchy's Integral Formula: A Guide to Evaluating Complex Integrals

In the field of complex analysis, contour integrals are a fundamental concept. This article will delve into how to solve a specific contour integral using the powerful tool known as Cauchy's Integral Formula. We will explore the process through a detailed example, focusing on the identification and handling of singularities within the contour.

Introduction

Contour integrals are integral expressions over a curve (contour) in the complex plane. The integral can be evaluated using various methods, but when the integrand has isolated singularities that lie within the region bounded by the contour, Cauchy's Integral Formula is particularly useful.

Cauchy's Integral Formula

Cauchy's Integral Formula states that for a holomorphic function (f(z)) inside and on a simple closed contour (C) with (z_0) being a point inside the contour, the integral of the function over (C) can be evaluated as:

[ oint_C frac{f(z)}{z - z_0} , dz 2pi i f(z_0) ]

However, for this specific example, we will extend this concept to higher-order derivatives using the Generalized Cauchy Integral Formula, which states:

[ oint_C frac{f(z)}{(z - z_0)^n} , dz frac{2pi i}{(n-1)!} f^{(n-1)}(z_0) ]

An Example Integral and Singularity

Consider the contour integral:

[ I oint_C frac{e^{2z}}{(z 1)^{32}} , dz ]

In this example, the singularity of the integrand occurs at:

[ z -1 ]

This singularity is inside the contour described by (z 2), making the application of the Generalized Cauchy Integral Formula feasible.

Applying the Generalized Cauchy Integral Formula

To evaluate (I), we start by identifying the order of the singularity in the denominator, which is 32. We then apply the Generalized Cauchy Integral Formula:

[ I oint_C frac{e^{2z}}{(z 1)^{32}} , dz ]

[ I frac{2pi i}{(32-1)!} left. frac{d^{31}}{dz^{31}} , e^{2z} right|_{z -1} ]

Simplifying, we find:

[ I frac{2pi i}{31!} cdot frac{d^{31}}{dz^{31}} e^{2z} Bigg|_{z -1} ]

Since the 31st derivative of (e^{2z}) is:

[ frac{d^{31}}{dz^{31}} e^{2z} 2^{31} e^{2z} ]

We substitute this back into the equation:

[ I frac{2pi i}{31!} cdot 2^{31} e^{2(-1)} ]

This simplifies to:

[ I frac{2pi i cdot 2^{31}}{31!} cdot e^{-2} ]

Further simplification yields:

[ I frac{2pi i cdot 2^{31}}{31!} cdot e^{-2} ]

Since (2^{31} 2147483648), we have:

[ I frac{2pi i cdot 2147483648}{31!} cdot e^{-2} ]

Finally, the integral evaluates to:

[ I frac{8pi i}{3e^2} ]

Conclusion

By following the steps of the Generalized Cauchy Integral Formula, we were able to evaluate the given contour integral. This example showcases the utility of Cauchy's Integral Formula in dealing with contour integrals with singularities within the contour. Understanding and applying this theorem is crucial for complex analysis and is a powerful tool in advanced mathematics.