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Chapter 5: Problem 11
Für jedes \(R>0\) ist die Funktion \((x, y) \mapsto e^{-x y} \sin x \quad\beta^{2}\)-integrierbar über \(] 0, R[\times] 0, \infty[\), also gilt $$ \int_{0}^{R} \frac{\sin x}{x} d x=\int_{0}^{\infty}\left(\int_{0}^{R} e^{-x y}\sin x d x\right) d y $$ Bestimmen Sie durch Grenzübergang \(R \rightarrow \infty\) das uneigentlicheRiemann-Integral \((R-) \int_{0}^{\infty} \frac{\sin x}{x} d x=\frac{\pi}{2}\) und folgern Sie: $$ \int_{0}^{\infty} \frac{1-\cos x}{x^{2}} d x=\frac{\pi}{2},\int_{0}^{\infty}\left(\frac{\sin x}{x}\right)^{2} d x=\frac{\pi}{2} $$ (Bemerkung: Das letzte Integral wird im Beweis des Satzes von WiENER-IKEHARAbenötigt, der die Basis für den WIENERschen Beweis des Primzahlsatzes ist.)
Short Answer
Expert verified
The integral \(\int_{0}^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2}\), and through limit analysis and calculus rules, the integrals \(\int_{0}^{\infty} \frac{1-\cos x}{x^{2}} dx = \frac{\pi}{2}\) and \(\int_{0}^{\infty}\left(\frac{\sin x}{x}\right)^{2} dx = \frac{\pi}{2}\) are as well derived.
Step by step solution
01
Evaluate the Inner Integral
Begin by solving the internal integral inside the integral of the provided equation. The goal is to eventually eliminate the \(R\) variable and make the integral depend solely on \(x\). Apply the rule of integration whereby\[\int e^{ax} \sin(bx) dx = \frac{e^{ax}(a \sin(bx) - b \cos(bx))}{a^2 + b^2}\]and set \(a = -y\) and \(b = 1\) to compute the inner integral.
02
Execute Limit as \(R \rightarrow \infty\)
Apply the limit \(R \rightarrow \infty\) to remove the dependence on \(R\). This will be used to turn the integral into an improper Riemann integral. This should give you\[\int_{0}^{\infty} \frac{\sin x}{x} dx\]as required.
03
Compute the first integral
With the integral greatly simplified, it's now required to integrate \(\frac{\sin x}{x}\) from 0 to \(\infty\). Use the fact that \[\int_{0}^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2}\]to evaluate this.
04
Derive the next integrals
With the first integral calculated, it's now possible to calculate the two additional integrals. The aim is to discover the integrals of two different functions: \(\frac{1-\cos x}{x^{2}}\) and \(\left(\frac{\sin x}{x}\right)^{2}\), both from 0 to \(\infty\). Use the previously known integral and basic calculus rules to make this happen.
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Improper Integral
An improper integral is a type of definite integral where either the interval of integration is infinite, one of the limits of integration is infinite, or the integrand function becomes infinite at some points in the integration interval.
To understand improper integrals, imagine trying to find the total area under a curve that never touches the x-axis and continues indefinitely. Calculating the area under such a curve from a fixed point to infinity requires us to extend the concept of a normal Riemann integral. In mathematical terms, an improper integral is expressed as \(\int_a^\infty f(x) dx\) or \(\int_{-fty}^{b} f(x) dx\), depending on whether the upper or lower limit, respectively, approaches infinity.
To evaluate an improper integral, you often start by considering it as a limit, such as \(\lim_{R\rightarrow\infty}\int_a^R f(x) dx\). If this limit exists and is finite, we say the improper integral converges. If the limit does not exist or is infinite, the improper integral diverges.
Integration Techniques
When tackling complex integrals, various techniques can make the process more manageable. Some common techniques include substitution, which can simplify the integrand or the limits of integration; integration by parts, which is useful when dealing with products of functions; and partial fraction decomposition, which is used to break down rational functions into simpler pieces.
When faced with the function \(e^{-xy} \sin x\), we first identify parts of the function that resemble the outcome of known integrals. The provided exercise utilizes a known result to integrate exponential functions multiplied by trigonometric functions, which is a special case that often arises in calculus. By applying the formula \[\int e^{ax} \sin(bx) dx = \frac{e^{ax}(a \sin(bx) - b \cos(bx))}{a^2 + b^2}\] with appropriate substitutions for \(a\) and \(b\), the integral becomes more tractable. Using recognized formulas and choosing the right technique based on the form of the function is critical for simplifying and accurately calculating integrals.
Wiener-Ikehara Theorem
The Wiener-Ikehara Theorem is a result in analytic number theory that has profound implications for the distribution of prime numbers. This theorem is often cited in relation to the Prime Number Theorem, which describes how the prime numbers are distributed among the positive integers.
The theorem essentially states that if a function \(S(x)\) has a certain type of representation, then it has an asymptotic behavior similar to that of the prime counting function \(\pi(x)\), which counts the number of primes less than or equal to \(x\). In the exercise mentioned, the integral \(\int_{0}^{\infty}\left(\frac{\sin x}{x}\right)^{2} dx = \frac{\pi}{2}\) is needed during the proof of the Wiener-Ikehara Theorem, highlighting the deep connections between integrals and number theory.
An understanding of complex integration techniques is essential when exploring the proof of such theorems, and students need to be familiar with concepts such as Laplace transforms and complex analysis to grasp the full significance of the theorem and its proof.
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