Contour Integral With Real Poles Misconception

Introduction

In complex analysis, contour integration is a powerful tool for evaluating definite integrals. However, when dealing with integrals that have real poles, there is a common misconception about the choice of contour. In this article, we will discuss the misconception and provide a detailed explanation of the correct approach.

The Misconception

The misconception arises from the fact that many students and even some professionals believe that the choice of contour for a contour integral with a real pole is arbitrary. They think that as long as the contour encloses the pole, the result will be the same. However, this is not the case.

The Contour Integral

Let's consider the integral over the real line:

$$I_1=\int_{-\infty}^{\infty}\mathrm d x\,\frac{e^{ix}}{x}=i\pi$$

This integral can be calculated from any of the two contours about $0$:

• A contour that consists of a large semi-circle in the upper half-plane, centered at the origin, and a line segment along the real axis from $-R$ to $R$, where $R$ is a large positive number.
• A contour that consists of a large semi-circle in the lower half-plane, centered at the origin, and a line segment along the real axis from $-R$ to $R$, where $R$ is a large positive number.

The Correct Approach

To evaluate the integral, we need to use the Cauchy's integral formula, which states that if $f(z)$ is analytic inside and on a simple closed contour $C$, and $a$ is a point inside $C$, then:

$$\oint_C \frac{f(z)}{z-a} \mathrm d z = 2\pi i f(a)$$

In our case, we have:

$$f(z) = e^{iz}$$

and

$$a = 0$$

So, we can write:

$$\oint_C \frac{e^{iz}}{z} \mathrm d z = 2\pi i e^{i0} = 2\pi i$$

However, we need to subtract the contribution from the pole at $z=0$, which is:

$$\mathrm{Res}(f,0) = \lim_{z\to 0} z \cdot \frac{e^{iz}}{z} = 1$$

Therefore, the final result is:

$$\oint_C \frac{e^{iz}}{z} \mathrm d z = 2\pi i - 2\pi i = i\pi$$

The Importance of Contour Choice

As we can see, the choice of contour is crucial in evaluating the integral. If we choose the wrong contour, we may get a different result. In this case, if we choose the contour that consists of a large semi-circle in the lower half-plane, we will get a different result, which is:

$$\oint_C \frac{e^{iz}}{z} \mathrm d z = -i\pi$$

This shows that the choice of contour can affect the result of the integral.

Conclusion

In conclusion, the choice of contour for a contour integral with a real pole is not arbitrary. We need to choose a contour that encloses the pole and is suitable for the problem at hand. The Cauchy's integral formula and the residue theorem are powerful tools for evaluating contour integrals, and we need to use them correctly to get the right result.

Regularization of Contour Integrals

In some cases, we may need to regularize the contour integral by subtracting the contribution from the pole at $z=0$. This is done by using the residue theorem, which states that the value of the integral is equal to $2\pi i$ times the residue of the function at the pole.

Example: Regularization of the Integral

Let's consider the integral:

$$I_2=\int_{-\infty}^{\infty}\mathrm d x\,\frac{e^{ix}}{x^2+1}$$

This integral can be evaluated using the Cauchy's integral formula and the residue theorem. We can write:

$$\oint_C \frac{e^{iz}}{z^2+1} \mathrm d z = 2\pi i \cdot \frac{e^{i0}}{2i} = \pi$$

However, we need to subtract the contribution from the pole at $z=0$, which is:

$$\mathrm{Res}(f,0) = \lim_{z\to 0} z \cdot \frac{e^{iz}}{z^2+1} = 0$$

Therefore, the final result is:

$$\oint_C \frac{e^{iz}}{z^2+1} \mathrm d z = \pi$$

Conclusion

In conclusion, the regularization of contour integrals is an important concept in complex analysis. We need to use the Cauchy's integral formula and the residue theorem to evaluate the integral and subtract the contribution from the pole at $z=0$.

Contour Integration with Multiple Poles

In some cases, we may have multiple poles in the contour integral. In this case, we need to use the residue theorem to evaluate the integral.

Example: Contour Integration with Multiple Poles

Let's consider the integral:

$$I_3=\int_{-\infty}^{\infty}\mathrm d x\,\frac{e^{ix}}{(x^2+1)(x^2+4)}$$

This integral can be evaluated using the Cauchy's integral formula and the residue theorem. We can write:

$$\oint_C \frac{e^{iz}}{(z^2+1)(z^2+4)} \mathrm d z = 2\pi i \cdot \frac{e^{i0}}{(2i)(2i\sqrt{2})} = \frac{\pi}{4\sqrt{2}}$$

However, we need to subtract the contribution from the poles at $z=0$ and $z=2i$, which are:

$$\mathrm{Res}(f,0) = \lim_{z\to 0} z \cdot \frac{e^{iz}}{(z^2+1)(z^2+4)} = \frac{1}{4}$$

$$\mathrm{Res}(f,2i) = \lim_{z\to 2i} (z-2i) \cdot \frac{e^{iz}}{(z^2+1)(z^2+4)} = -\frac{1}{4\sqrt{2}}$$

Therefore, the final result is:

$$\oint_C \frac{e^{iz}}{(z^2+1)(z^2+4)} \mathrm d z = \frac{\pi}{4\sqrt{2}} - \frac{\pi}{4\sqrt{2}} = 0$$

Conclusion

In conclusion, the contour integration with multiple poles is an important concept in complex analysis. We need to use the Cauchy's integral formula and the residue theorem to evaluate the integral and subtract the contribution from the poles at $z=0$ and $z=2i$.

Conclusion

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Q: What is a contour integral with real poles?

A: A contour integral with real poles is a type of integral that involves a function with a real pole, which is a point where the function becomes infinite. The contour integral is a way of evaluating the integral by integrating around a closed curve, known as a contour, that encloses the pole.

Q: Why is the choice of contour important in contour integration?

A: The choice of contour is crucial in contour integration because it determines the value of the integral. If the contour is not chosen correctly, the integral may not be evaluated correctly, leading to incorrect results.

Q: What is the Cauchy's integral formula?

A: The Cauchy's integral formula is a fundamental theorem in complex analysis that states that if $f(z)$ is analytic inside and on a simple closed contour $C$, and $a$ is a point inside $C$, then:

$$\oint_C \frac{f(z)}{z-a} \mathrm d z = 2\pi i f(a)$$

Q: What is the residue theorem?

A: The residue theorem is a theorem in complex analysis that states that the value of a contour integral is equal to $2\pi i$ times the residue of the function at the pole.

Q: What is the residue of a function at a pole?

A: The residue of a function at a pole is the coefficient of the $1/(z-a)$ term in the Laurent series expansion of the function around the pole.

Q: How do I evaluate a contour integral with a real pole?

A: To evaluate a contour integral with a real pole, you need to use the Cauchy's integral formula and the residue theorem. First, you need to choose a contour that encloses the pole. Then, you need to evaluate the integral using the Cauchy's integral formula and subtract the contribution from the pole.

Q: What is the importance of contour integration in complex analysis?

A: Contour integration is a fundamental tool in complex analysis that allows us to evaluate integrals of functions with complex poles. It has numerous applications in physics, engineering, and other fields.

Q: Can you provide an example of contour integration with a real pole?

A: Yes, let's consider the integral:

$$I_4=\int_{-\infty}^{\infty}\mathrm d x\,\frac{e^{ix}}{x^2+1}$$

This integral can be evaluated using the Cauchy's integral formula and the residue theorem. We can write:

$$\oint_C \frac{e^{iz}}{z^2+1} \mathrm d z = 2\pi i \cdot \frac{e^{i0}}{2i} = \pi$$

However, we need to subtract the contribution from the pole at $z=0$, which is:

$$\mathrm{Res}(f,0) = \lim_{z\to 0} z \cdot \frac{e^{iz}}{z^2+1} = 0$$

Therefore, the final result is:

$$\oint_C \frac{e^{iz}}{z^2+1} \mathrm d z = \pi$$

Q: What are some common mistakes to avoid when evaluating contour integrals with real poles?

A: Some common mistakes to avoid when evaluating contour integrals with real poles include:

• Choosing the wrong contour
• Not subtracting the contribution from the pole
• Not using the Cauchy's integral formula and the residue theorem correctly

Q: Can you provide some tips for choosing the correct contour for a contour integral with a real pole?

A: Yes, here are some tips for choosing the correct contour for a contour integral with a real pole:

• Choose a contour that encloses the pole
• Choose a contour that is suitable for the problem at hand
• Avoid choosing a contour that is too complex or too simple

Q: What are some applications of contour integration in physics and engineering?

A: Contour integration has numerous applications in physics and engineering, including:

• Quantum mechanics
• Electromagnetism
• Signal processing
• Control systems

Q: Can you provide some resources for learning more about contour integration?

A: Yes, here are some resources for learning more about contour integration:

• Books: "Complex Analysis" by Serge Lang, "Complex Variables and Applications" by James Ward Brown and Ruel V. Churchill
• Online courses: "Complex Analysis" by MIT OpenCourseWare, "Complex Variables" by Stanford University
• Research papers: Search for papers on arXiv or Google Scholar using keywords like "contour integration", "complex analysis", "residue theorem"