Evaluating the Complex Integral of sin(z)/(z 1)^7 Using the Residue Theorem

The computation of complex integrals involving singularities is an essential aspect of complex analysis. One powerful tool for evaluating such integrals is the Residue Theorem. In this article, we will step through the process of evaluating the following complex integral:

Integral to Evaluate

Let's consider the complex integral (int_C frac{sin{z}}{(z 1)^7} dz), where (C) is the circle of radius 5 centered at (z0).

Steps of Evaluation

Identify Singularities Check if the Singularity is Inside the Contour Compute the Residue Compute the Derivative Evaluate the Limit Apply the Residue Theorem

Step 1: Identify Singularities

The given integrand is (frac{sin{z}}{(z 1)^7}). The singularity occurs at (z-1), which is a pole of order 7.

Step 2: Check if the Singularity is Inside the Contour

The radius of the circle (C) is 5. Since (-1) lies within this circle (distance from 0 to -1 is 1, which is less than 5), we can apply the Residue Theorem.

Step 3: Compute the Residue

To find the residue at the pole (z-1), we use the formula for the residue at a pole of order (n):

Res(f, -1)frac{1}{(n-1)!} lim_{z to -1} frac{d^{n-1}}{dz^{n-1}} left[ (z 1)^n f(z) right])

where (f(z) frac{sin{z}}{(z 1)^7}) and (n 7). We need to compute:

Res(f, -1)frac{1}{6!} lim_{z to -1} frac{d^6}{dz^6} left[ (z 1)^7 cdot frac{sin{z}}{(z 1)^7} right] frac{1}{6!} lim_{z to -1} frac{d^6}{dz^6} left[ sin{z} right])

Step 4: Compute the Derivative

The sixth derivative of (sin{z}) is as follows:

(sin{z} to cos{z}) (cos{z} to -sin{z}) (-sin{z} to -cos{z}) (-cos{z} to -sin{z}) (-sin{z} to -cos{z}) (-cos{z} to -sin{z}) (6^{th}) derivative: (sin{z})

Thus, the sixth derivative (frac{d^6}{dz^6} sin{z}) evaluated at (z -1) is:

(frac{d^6}{dz^6} sin{z}) evaluated at (z -1) is (sin{1}).

Step 5: Evaluate the Limit

Now, substitute (z -1):

(Res(f, -1) frac{1}{720} sin{1})

Step 6: Apply the Residue Theorem

By the Residue Theorem, the integral around the contour (C) is:

(int_C frac{sin{z}}{(z 1)^7} dz 2pi i cdot Res(f, -1) frac{2pi i sin{1}}{720})

Final Result

The value of the integral is thus (int_C frac{sin{z}}{(z 1)^7} dz frac{2pi i sin{1}}{720}).

Alternative Approach Using Cauchy's Integral Theorem

Additionally, using Cauchy's integral formula directly, we have:

(int_C frac{sin{z}}{(z 1)^7} dz frac{2pi i}{6!} frac{d^6}{dz^6} sin{z} Bigg|_{z-1} frac{pi i}{360} sin{1})

Conclusion

This article has demonstrated how to evaluate the complex integral (int_C frac{sin{z}}{(z 1)^7} dz) using both the Residue Theorem and Cauchy's integral formula. The Residue Theorem provides a systematic approach to finding the residue at a singularity, which is particularly useful for higher-order poles.

Key Points:

Identifying the singularity and verifying its location within the contour. Using the formula for the residue at higher-order poles. Evaluating derivatives of the integrand at the singular point. Applying Cauchy's integral formula for alternate computation.

The technique of evaluating complex integrals using residues is a fundamental concept in complex analysis, providing a powerful tool for solving various problems in mathematics and physics.