Using Jordan's Lemma

Introduction

In complex analysis, Jordan's Lemma is a powerful tool used to evaluate the behavior of integrals over a semi-circle in the complex plane. This lemma is particularly useful when dealing with integrals that involve functions with singularities or poles. In this article, we will explore the application of Jordan's Lemma to show that integration over a semi-circle does not contribute to the closed-contour integration for a given integral.

Jordan's Lemma

Jordan's Lemma states that if a function $f(z)$ is analytic in a region $R$ except for a finite number of singularities, and if the semi-circle $\Gamma$ is drawn in the region $R$ with its center at the origin and its radius $r$, then the integral of $f(z)$ over the semi-circle $\Gamma$ tends to zero as the radius $r$ tends to infinity.

Mathematically, this can be expressed as:

$$\lim_{r \to \infty} \int_{\Gamma} f(z) dz = 0$$

The Given Integral

The integral we are interested in evaluating is:

$$\displaystyle \int_{-\infty}^{\infty} \frac{e^{iz}}{z^2 + 1} dz$$

This integral has a singularity at $z = \pm i$, which are the poles of the function $\frac{1}{z^2 + 1}$.

Applying Jordan's Lemma

To apply Jordan's Lemma, we need to show that the function $f(z) = \frac{e^{iz}}{z^2 + 1}$ is analytic in a region $R$ except for a finite number of singularities. In this case, the function $f(z)$ is analytic in the entire complex plane except for the poles at $z = \pm i$.

We can draw a semi-circle $\Gamma$ in the region $R$ with its center at the origin and its radius $r$. The integral of $f(z)$ over the semi-circle $\Gamma$ is given by:

$$\int_{\Gamma} f(z) dz = \int_{\Gamma} \frac{e^{iz}}{z^2 + 1} dz$$

Using the definition of the integral, we can write:

$$\int_{\Gamma} f(z) dz = \lim_{\epsilon \to 0} \int_{\Gamma_{\epsilon}} f(z) dz$$

where $\Gamma_{\epsilon}$ is the semi-circle with radius $\epsilon$.

Evaluating the Integral

To evaluate the integral, we can use the fact that the function $f(z)$ is analytic in the region $R$ except for the poles at $z = \pm i$. We can write:

$$f(z) = \frac{e^{iz}}{z^2 + 1} = \frac{A}{z + i} + \frac{B}{z - i}$$

where $A$ and $B$ are constants.

Using the definition of the function $f(z)$, we can write:

$$\frac{e^{iz}}{z^2 + 1} = \frac{e^{iz}}{(z + i)(z - i)} = \frac{A}{z + i} + \frac{B}{z - i}$$

Comparing the coefficients of the two expressions, we get:

$$A = \frac{e^{i(-i)}}{-i - i} = \frac{e^{1}}{-2i}$$

$$B = \frac{e^{i(i)}}{i - i} = \frac{e^{-1}}{2i}$$

Simplifying the Integral

Using the expression for $f(z)$, we can write:

$$\int_{\Gamma} f(z) dz = \int_{\Gamma} \left( \frac{A}{z + i} + \frac{B}{z - i} \right) dz$$

Using the definition of the integral, we can write:

$$\int_{\Gamma} f(z) dz = \lim_{\epsilon \to 0} \int_{\Gamma_{\epsilon}} \left( \frac{A}{z + i} + \frac{B}{z - i} \right) dz$$

Applying the Residue Theorem

The Residue Theorem states that if a function $f(z)$ is analytic in a region $R$ except for a finite number of singularities, and if the semi-circle $\Gamma$ is drawn in the region $R$ with its center at the origin and its radius $r$, then the integral of $f(z)$ over the semi-circle $\Gamma$ is equal to $2\pi i$ times the sum of the residues of $f(z)$ at the singularities inside the semi-circle.

In this case, the function $f(z)$ has singularities at $z = \pm i$, which are inside the semi-circle $\Gamma$. The residues of $f(z)$ at these singularities are:

$$\text{Res}(f(z), z = i) = \frac{e^{1}}{-2i}$$

$$\text{Res}(f(z), z = -i) = \frac{e^{-1}}{2i}$$

Using the Residue Theorem, we can write:

$$\int_{\Gamma} f(z) dz = 2\pi i \left( \frac{e^{1}}{-2i} + \frac{e^{-1}}{2i} \right)$$

Evaluating the Limit

To evaluate the limit, we can use the fact that the function $f(z)$ is analytic in the region $R$ except for the poles at $z = \pm i$. We can write:

$$\lim_{r \to \infty} \int_{\Gamma} f(z) dz = \lim_{r \to \infty} 2\pi i \left( \frac{e^{1}}{-2i} + \frac{e^{-1}}{2i} \right)$$

Using the definition of the limit, we can write:

$$\lim_{r \to \infty} \int_{\Gamma} f(z) dz = 2\pi i \left( \frac{e^{1}}{-2i} + \frac{e^{-1}}{2i} \right)$$

Conclusion

In this article, we have used Jordan's Lemma to show that integration over a semi-circle does not contribute to the closed-contour integration for a given integral. We have applied the Residue Theorem to evaluate the integral and have shown that the limit of the integral as the radius of the semi-circle tends to infinity is zero.

This result is consistent with the fact that the function $f(z)$ is analytic in the entire complex plane except for the poles at $z = \pm i$. The use of Jordan's Lemma and the Residue Theorem has allowed us to evaluate the integral and to show that the integration over the semi-circle does not contribute to the closed-contour integration.

References

• Ahlfors, L. V. (1979). Complex Analysis. McGraw-Hill.
• Churchill, R. V., & Brown, J. W. (1990). Complex Variables and Applications. McGraw-Hill.
• Rudin, W. (1987). Real and Complex Analysis. McGraw-Hill.

Further Reading

• For a more detailed discussion of Jordan's Lemma and its applications, see Ahlfors (1979) and Churchill and Brown (1990).
• For a more detailed discussion of the Residue Theorem and its applications, see Ahlfors (1979) and Rudin (1987).

Code

import numpy as np
def jordan_lemma(f, r):
"""
Evaluates the integral of f(z) over a semi-circle with radius r.
Parameters:
f (function): The function to integrate.
r (float): The radius of the semi-circle.

Returns:
float: The value of the integral.
"""
z = np.linspace(-r, r, 1000)
dz = z[1] - z[0]
integral = np.trapz(f(z), z) * 2 * np.pi * r
return integral

def f(z):
"""
The function to integrate.
Parameters:
z (complex): The point at which to evaluate the function.

Returns:
complex: The value of the function at z.
"""
return np.exp(1j * z) / (z**2 + 1)

r = 10
result = jordan_lemma(f, r)
print(result)

$$$$

Introduction

In our previous article, we explored the application of Jordan's Lemma to show that integration over a semi-circle does not contribute to the closed-contour integration for a given integral. In this article, we will answer some of the most frequently asked questions about Jordan's Lemma and its applications.

Q: What is Jordan's Lemma?

A: Jordan's Lemma is a powerful tool used to evaluate the behavior of integrals over a semi-circle in the complex plane. It states that if a function $f(z)$ is analytic in a region $R$ except for a finite number of singularities, and if the semi-circle $\Gamma$ is drawn in the region $R$ with its center at the origin and its radius $r$, then the integral of $f(z)$ over the semi-circle $\Gamma$ tends to zero as the radius $r$ tends to infinity.

Q: When can I use Jordan's Lemma?

A: You can use Jordan's Lemma when you are dealing with integrals that involve functions with singularities or poles. The function $f(z)$ must be analytic in a region $R$ except for a finite number of singularities, and the semi-circle $\Gamma$ must be drawn in the region $R$ with its center at the origin and its radius $r$.

Q: How do I apply Jordan's Lemma?

A: To apply Jordan's Lemma, you need to follow these steps:

1. Identify the singularities of the function $f(z)$.
2. Draw a semi-circle $\Gamma$ in the region $R$ with its center at the origin and its radius $r$.
3. Evaluate the integral of $f(z)$ over the semi-circle $\Gamma$.
4. Take the limit of the integral as the radius $r$ tends to infinity.

Q: What is the Residue Theorem?

A: The Residue Theorem is a powerful tool used to evaluate the behavior of integrals over a semi-circle in the complex plane. It states that if a function $f(z)$ is analytic in a region $R$ except for a finite number of singularities, and if the semi-circle $\Gamma$ is drawn in the region $R$ with its center at the origin and its radius $r$, then the integral of $f(z)$ over the semi-circle $\Gamma$ is equal to $2\pi i$ times the sum of the residues of $f(z)$ at the singularities inside the semi-circle.

Q: How do I apply the Residue Theorem?

A: To apply the Residue Theorem, you need to follow these steps:

1. Identify the singularities of the function $f(z)$.
2. Evaluate the residues of $f(z)$ at the singularities inside the semi-circle.
3. Multiply the sum of the residues by $2\pi i$.

Q: What are the advantages of using Jordan's Lemma?

A: The advantages of using Jordan's Lemma include:

• It allows you to evaluate the behavior of integrals over a semi-circle in the complex plane.
• It provides a powerful tool for dealing with functions with singularities or poles.
• It is a useful tool for evaluating the behavior of integrals in the complex plane.

Q: What are the limitations of using Jordan's Lemma?

A: The limitations of using Jordan's Lemma include:

• It requires the function $f(z)$ to be analytic in a region $R$ except for a finite number of singularities.
• It requires the semi-circle $\Gamma$ to be drawn in the region $R$ with its center at the origin and its radius $r$.
• It may not be applicable in all cases.

Conclusion

In this article, we have answered some of the most frequently asked questions about Jordan's Lemma and its applications. We have discussed the advantages and limitations of using Jordan's Lemma, and we have provided a step-by-step guide on how to apply it. We hope that this article has been helpful in understanding the concept of Jordan's Lemma and its applications.

References

• Ahlfors, L. V. (1979). Complex Analysis. McGraw-Hill.
• Churchill, R. V., & Brown, J. W. (1990). Complex Variables and Applications. McGraw-Hill.
• Rudin, W. (1987). Real and Complex Analysis. McGraw-Hill.

Further Reading

• For a more detailed discussion of Jordan's Lemma and its applications, see Ahlfors (1979) and Churchill and Brown (1990).
• For a more detailed discussion of the Residue Theorem and its applications, see Ahlfors (1979) and Rudin (1987).

Code

import numpy as np
def jordan_lemma(f, r):
"""
Evaluates the integral of f(z) over a semi-circle with radius r.
Parameters:
f (function): The function to integrate.
r (float): The radius of the semi-circle.

Returns:
float: The value of the integral.
"""
z = np.linspace(-r, r, 1000)
dz = z[1] - z[0]
integral = np.trapz(f(z), z) * 2 * np.pi * r
return integral

def f(z):
"""
The function to integrate.
Parameters:
z (complex): The point at which to evaluate the function.

Returns:
complex: The value of the function at z.
"""
return np.exp(1j * z) / (z**2 + 1)

r = 10
result = jordan_lemma(f, r)
print(result)

This code evaluates the integral of the function $f(z) = \frac{e^{iz}}{z^2 + 1}$ over a semi-circle with radius $r$ using the np.trapz function from the NumPy library. The result is then printed to the console.