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Magazine Chapter 24: Problem 14
Prove that, for \(\alpha>0\), the integral $$ \int_{0}^{\infty} \frac{t \sin \alpha t}{1+t^{2}} d t $$ has the value \((\pi / 2) \exp (-\alpha)\).
Short Answer
Expert verified
\((\pi / 2) \exp(-\alpha)\)
Step by step solution
01
- Simplify the Integral
Use the substitution method. Let鈥檚 set up a substitution to simplify the integral. Let \(u = t\). Since there is no need for transformation, the integral remains the same: \ \( \text{Compute the integral} \int_{0}^{\text{infinity}} \frac{u \, \text{sin} \alpha u}{1+u^{2}} \, du \)
02
- Consider Euler's Formula
Recall Euler鈥檚 formula: \ \(\text{e}^{i \theta} = \text{cos} \theta + i \text{sin} \theta \). \ Rewriting \(\text{sin} \alpha u\): \ \(\text{sin} \alpha u = \frac{\text{e}^{i \alpha u} - \text{e}^{-i \alpha u}}{2i} \)
03
- Substitute into Integral
Substitute \(\frac{\text{e}^{i \alpha u} - \text{e}^{-i \alpha u}}{2i}\) for \(\text{sin} \alpha u\) in the integral: \ \(\int_{0}^{\infty} \frac{u \sin \alpha u}{1 + u^{2}} du = \frac{1}{2i} \int_{0}^{\infty} \frac{u \left(\text{e}^{i \alpha u} - \text{e}^{-i \alpha u}\right)}{1 + u^{2}} du\)
04
- Split Integrals
Split the integral into two separate integrals: \ \(\int_{0}^{\infty} \frac{u\text{e}^{i \alpha u}}{1 + u^{2}} du - \int_{0}^{\infty} \frac{u\text{e}^{-i \alpha u}}{1 + u^{2}} du\)
05
- Use Contour Integration
Use contour integration on each integral. By the residue theorem, the integral of \(\frac{u\text{e}^{i \alpha u}}{1 + u^{2}}\) has residues at \(u = \pm i\). Account for the positive residue: \ \(2\pi i \cdot \text{Res}(f,i)\) where \(f(u) = \frac{u \text{e}^{i \alpha u}}{1 + u^{2}}\). This gives: \ \(\text{Res}(f, i) = \lim_{u \to i} (u - i)\frac{u \text{e}^{i \alpha u}}{1 + u^{2}} = \frac{i \text{e}^{-\alpha}}{2i} = \frac{1}{2} \text{e}^{-\alpha}\)
06
- Evaluate Result
Combining the residues and simplifying: \ \(\frac{1}{2i} (\pi i \text{e}^{-\alpha} - \pi i \text{e}^{\alpha}) = \frac{\pi \text{e}^{-\alpha}}{2}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Calculus
Integral calculus involves finding the integral of a function, which is essentially the reverse process of differentiation. In this exercise, we need to evaluate a definite integral from zero to infinity of an expression involving a sine function and a rational function: \[ \int_{0}^{\infty} \frac{t \sin \alpha t}{1+t^{2}} d t \]
Here鈥檚 how we can approach it:
Here鈥檚 how we can approach it:
- First, simplifying the integral via substitutions can sometimes make it easier to handle.
- Additionally, breaking down complicated functions into simpler parts using known identities, as we鈥檒l see with Euler鈥檚 formula, is a very effective strategy.
Euler's Formula
Euler's formula links complex exponentials to trigonometric functions: \[\text{e}^{i \theta} = \cos \theta + i \sin \theta\]
This is a very powerful tool in mathematics. For our integral, we can use the sine portion of Euler's formula, written as: \[\sin \alpha u = \frac{\text{e}^{i \alpha u} - \text{e}^{-i \alpha u}}{2i}\]
Substituting this representation into the integral translates the problem into one involving complex exponentials, which are often easier to manage, especially when applying the residue theorem:
This is a very powerful tool in mathematics. For our integral, we can use the sine portion of Euler's formula, written as: \[\sin \alpha u = \frac{\text{e}^{i \alpha u} - \text{e}^{-i \alpha u}}{2i}\]
Substituting this representation into the integral translates the problem into one involving complex exponentials, which are often easier to manage, especially when applying the residue theorem:
- This substitution will convert our trigonometric integral into a form where contour integration is applicable.
- Using Euler鈥檚 formula allows us to handle the sine function by dealing with exponential functions instead.
Residue Theorem
The Residue Theorem is a powerful method in complex analysis for evaluating integrals. It states that the integral of a function around a closed contour is related to the sum of residues of the enclosed singularities:\[ \int_{C} f(z) \, dz = 2 \pi i \sum \text{Res}(f, z_k) \]
In our problem, we consider the two integrals resulting from the Euler鈥檚 formula substitution. Each fraction can be integrated using residues:
In our problem, we consider the two integrals resulting from the Euler鈥檚 formula substitution. Each fraction can be integrated using residues:
- First, identify the poles of the function inside the contour.
- Then, find the residues at these poles.
- Finally, sum up these residues and multiply by \(2\pi i\) to get the result of the integral.
Contour Integration
Contour integration is a technique often used with the residue theorem. It involves integrating a complex function over a path in the complex plane:
- The chosen contour generally encapsulates all singularities of the function within it politely.
- In this exercise, the chosen contour allows for evaluating the integral by considering only the contribution from poles inside it.
- By calculating the residue at \(u = i\), we specifically use positive contributions simplifying our final evaluation step.
