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Magazine Chapter 32: Problem 26
Mittels Residuensatz berechne man die Integrale: $$ I_{a}=\int_{-\infty}^{+\infty} \frac{1}{t^{4}+1} \mathrm{~d} t, \quad I_{b}=\int_{-\infty}^{+\infty} \frac{t^{2}}{t^{6}+1} \mathrm{~d} t, \quad I_{c}=\int_{-\infty}^{\infty} \frac{t \sin t}{t^{2}+4} \mathrm{~d} t $$
Short Answer
Expert verified
Question: Evaluate the following integrals using the Residue theorem:
1. \(I_a = \int_{-\infty}^{+\infty} \frac{1}{t^{4}+1} \mathrm{~d} t\)
2. \(I_b = \int_{-\infty}^{+\infty} \frac{t^2}{t^6+1} \mathrm{~d} t\)
3. \(I_c = \int_{-\infty}^{\infty} \frac{t \sin t}{t^2+4} \mathrm{~d} t\)
Answer:
1. \(I_a = \frac{\pi (e^{-\frac{\pi}{4}} + e^{-\frac{3\pi}{4}})}{2}\)
2. \(I_b = \frac{3\pi}{\sqrt{3}}\)
3. \(I_c = \frac{\pi e^{-2}}{2}\)
Step by step solution
01
Identify the poles of the integrand inside the chosen contour
The integrand has simple poles when the denominator is zero, solving the equation \(t^4 + 1 = 0,\) we get the poles \(t = e^{\pm i\frac{\pi}{4}}, e^{\pm i\frac{3\pi}{4}}.\) We choose an indented semicircle contour in the upper half-plane, so the poles inside the contour are \(t = e^{i\frac{\pi}{4}}, e^{i\frac{3\pi}{4}}.\)
02
Calculate the residues of the poles
Using the formula for residues for simple poles, we get:
Res\(_{t=e^{i\frac{\pi}{4}}}\left(\frac{1}{t^4+1}\right) = -\frac{e^{i\frac{\pi}{4}}}{4i}\)
Res\(_{t=e^{i\frac{3\pi}{4}}}\left(\frac{1}{t^4+1}\right) = -\frac{e^{i\frac{3\pi}{4}}}{4i}\)
03
Compute the integral using the Residue theorem
Applying the Residue theorem, we get:
\(I_a = 2\pi i \left[ -\frac{e^{i\frac{\pi}{4}}+e^{i\frac{3\pi}{4}}}{4i} \right]\)
\(I_a = \frac{\pi (e^{-\frac{\pi}{4}} + e^{-\frac{3\pi}{4}})}{2}\)
#I_b: \(\int_{-\infty}^{+\infty} \frac{t^{2}}{t^{6}+1} \mathrm{~d} t\)#
Here, to make it a holomorphic function, we rewrite it as the integral of a complex function:
\(I_b = \frac{1}{2} \int_{-\infty}^{+\infty} \frac{z^{2}}{z^{6}+1} \mathrm{~d} z\)
04
Identify the poles of the integrand inside the chosen contour
The integrand has simple poles when the denominator is zero, solving the equation \(z^6 + 1 = 0,\) which has roots \(z = e^{i\frac{2n\pi + \pi}{6}}\) for \(n=0,1,2,3,4,5\). We again choose an indented semicircle contour in the upper half-plane. The poles inside the contour are \(z = e^{i\frac{\pi}{6}}, e^{i\frac{5\pi}{6}}.\)
05
Calculate the residues of the poles
Using the formula for residues for simple poles, we get:
Res\(_{z=e^{i\frac{\pi}{6}}}\left(\frac{z^2}{z^6+1}\right) = -\frac{3e^{i\frac{\pi}{6}}}{2\sqrt{3}i}\)
Res\(_{z=e^{i\frac{5\pi}{6}}}\left(\frac{z^2}{z^6+1}\right) = -\frac{3e^{i\frac{5\pi}{6}}}{2\sqrt{3}i}\)
06
Compute the integral using the Residue theorem
Applying the Residue theorem, we get:
\(I_b = 2\pi i \left[ -\frac{3e^{i\frac{\pi}{6}} +3e^{i\frac{5\pi}{6}}}{4\sqrt{3}i} \right]\)
\(I_b = \frac{3\pi}{\sqrt{3}}\)
#I_c: \(\int_{-\infty}^{\infty} \frac{t \sin t}{t^{2}+4} \mathrm{~d} t\)#
To make it holomorphic, let's rewrite it as the real part of a complex integral:
\(I_c = \text{Re} \left[ \int_{-\infty}^{\infty} \frac{z e^{iz}}{z^2+4} \mathrm{~d} z \right]\)
07
Identify the poles of the integrand inside the chosen contour
The integrand has simple poles when the denominator is zero, solving for \(z^2 + 4 = 0,\) we get the poles \(z = \pm 2i\). We choose an indented semicircle contour in the upper half-plane, so the only pole inside the contour is \(z = 2i.\)
08
Calculate the residues of the poles
Using the formula for residues for simple poles, we get:
Res\(_{z=2i}\left(\frac{z e^{iz}}{z^2+4}\right) = \frac{(2i)e^{-2}}{4i}\)
09
Compute the integral using the Residue theorem
Applying the Residue theorem, we get:
\(I_c = 2\pi i \left[ \frac{(2i)e^{-2}}{4i} \right]\)
Taking the real part,
\(I_c = \frac{\pi e^{-2}}{2}\)
In conclusion, the three integrals evaluated using the Residue theorem are:
1. \(I_a = \frac{\pi (e^{-\frac{\pi}{4}} + e^{-\frac{3\pi}{4}})}{2}\)
2. \(I_b = \frac{3\pi}{\sqrt{3}}\)
3. \(I_c = \frac{\pi e^{-2}}{2}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Analysis
Complex analysis is a branch of mathematics focusing on the study of complex numbers and functions of a complex variable. It is a powerful tool in mathematics, not only for dealing with problems involving complex numbers but for solving real-world problems as well. By examining complex functions, you'll discover:
- Analytic Functions: These are functions that are smooth and differentiable in a complex sense. They possess a derivative which is a complex number.
- Complex Plane: Just like the Cartesian plane for real numbers, the complex plane is used for plotting complex numbers, with the x-axis representing the real part and the y-axis representing the imaginary part.
- Holomorphic Functions: A term used for complex functions that are complex-differentiable in a neighborhood. Such functions are a key component in contour integration and the residue theorem.
Contour Integration
Contour integration is a method of evaluating complex integrals of functions along a certain path, called a contour, in the complex plane. This technique vastly simplifies the process of finding integrals for complex functions and is particularly useful:
- When real integration methods are too cumbersome or impractical.
- For transforming complex integral problems into simpler geometric ones.
- Think of it as drawing a loop (the contour) around certain singularities (points where the function becomes infinite) called poles.
- Contours can include simple paths like line segments or more complex shapes such as semicircles, as described in the solutions for contour integration in the upper half-plane.
Poles and Residues
In complex analysis, poles and residues are central to calculating integrals using the residue theorem. Poles are specific points of singularity where a function becomes unbounded. When dealing with poles, consider the following:
- Simple Poles: These are poles of order one, where the function behaves like \( \frac{1}{z-z_0} \) near these points.
- Residues: The residue at a pole is essentially the coefficient of \( \frac{1}{z-z_0} \) in the Laurant series expansion of a function. Calculating residues is fundamental for the residue theorem and allows addressing real and complex integrals.
- For a simple pole \( z_0 \), the residue is found using \( \text{Res}_{z=z_0}f(z) = \lim_{z \to z_0}(z-z_0)f(z) \).
- In our problem, we isolated poles within chosen contours to find the necessary residues.
Definite Integrals
Definite integrals have a fundamental role not only in real analysis but also in complex analysis when linked through the residue theorem. These are integrals evaluated over a specific interval. In complex analysis, we adapt these ideas to complex functions:
- Real Line Integration: This involves integrating a real function over real intervals, as seen in traditional calculus.
- Complex Path Integration: With the residue theorem, we transform these problems into a contour integration along a path in the complex plane. The definite integrals in our example are complex functions being solved through contour methods.
