Question

Zeigen Sie für eine ungerade Funktion \(f(t)\) die Beziehung $$ \mathcal{F}[f](\omega)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} f(t) e^{-i \omega t} d t=-\frac{i}{\pi} \int_{0}^{\infty} f(t) \sin (\omega t) d t $$

Short answer

The relationship is shown using the properties of odd functions and Euler's formula for complex exponentials.

Step-by-step solution

Identify the Components

We begin by identifying the components of the problem. We have a Fourier transform denoted as \( \mathcal{F}[f](\omega) \) which is defined for a function \( f(t) \). The function \( f(t) \) is given as an odd function, meaning \( f(-t) = -f(t) \). Our goal is to show that this results in the given relationship involving a sine integral.

Express Fourier Transform for Odd Functions

The Fourier transform \( \mathcal{F}[f](\omega) \) of an arbitrary function \( f(t) \) is given by the integral expression: \[ \mathcal{F}[f](\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(t) e^{-i \omega t} \, dt. \]Due to the odd nature of \( f(t) \), rewrite this using the properties of odd functions.

Utilize Odd Function Property

Split the integral into two symmetric parts around zero:\[ \int_{-\infty}^{\infty} f(t) e^{-i \omega t} \, dt = \int_{0}^{\infty} f(t) e^{-i \omega t} \, dt + \int_{- frac{\infty}{0}} f(t) e^{-i \omega t} \, dt. \]Use the odd function property \( f(-t) = -f(t) \) to rewrite the second integral:\[ \int_{- frac{\infty}{0}} f(t) e^{-i \omega t} \, dt = -\int_{0}^{\infty} f(-t) e^{-i \omega (-t)} \, dt = -\int_{0}^{\infty} f(t) e^{i \omega t} \, dt. \]

Combine the Integrals

Combine the two integrals from the previous step:\[ \mathcal{F}[f](\omega) = \frac{1}{2\pi} \left( \int_{0}^{\infty} f(t) e^{-i \omega t} \, dt - \int_{0}^{\infty} f(t) e^{i \omega t} \, dt \right). \]Factor out the common terms:\[ \mathcal{F}[f](\omega) = \frac{1}{2\pi} \int_{0}^{\infty} f(t) (e^{-i \omega t} - e^{i \omega t}) \, dt. \]

Simplify Using Sine Function

Notice that the expression \( e^{-i \omega t} - e^{i \omega t} \) can be rewritten using Euler's formula as:\[ 2i \sin(\omega t). \]Therefore, the expression becomes:\[ \mathcal{F}[f](\omega) = \frac{i}{\pi} \int_{0}^{\infty} f(t) \sin(\omega t) \, dt. \]

Show Final Relationship

Recognize that this is equivalent to:\[ \mathcal{F}[f](\omega) = -\frac{i}{\pi} \int_{0}^{\infty} f(t) \sin(\omega t) \, dt, \]confirming the given relationship. The negative sign arises because of the factor \( i \) from Euler's formula.

Key concepts

Odd Functions

Odd functions are a unique class of functions characterized by their symmetry about the origin. Specifically, if a function is odd, it satisfies the condition that for every point

• \( f(-t) = -f(t) \)

This property implies that the graph of an odd function is symmetric with respect to the origin. An intuitive example of an odd function is the simple polynomial function \( f(t) = t^3 \), which mirrors itself across the origin at negative and positive inputs.

When working with the Fourier Transform, the property of odd functions plays a crucial role in simplifying calculations. For instance, in the given problem, understanding that \( f(t) \) is odd allows us to split and manipulate the integral into more manageable forms.

By recognizing this property, the integral from negative infinity to infinity can be divided into two symmetric parts: one from zero to infinity and the other from negative infinity to zero. The symmetry helps in simplifying and rewriting these parts, ultimately leading to a more elegant expression involving sine integrals.

Sine Integral

The sine integral appears in various areas of mathematics and engineering, particularly when analyzing signals and solving differential equations. The basic idea of a sine integral involves integrating products of a function with the sine trigonometric function. In the context of the Fourier Transform, we use the sine function to simplify the expressions.

Upon realizing that

• \( e^{-i \omega t} - e^{i \omega t} \)

translates to

• \( 2i \sin(\omega t) \)

using Euler's formula, we can incorporate a sine integral. This transformation is crucial because it leverages the symmetry properties of sine functions when dealing with intervals from zero to infinity.

The sine integral in the problem leads us to a cleaner representation that not only involves an imaginary unit but also reflects the integral's inherent symmetry. Such transformations simplify the evaluation of the Fourier Transform, especially when the original function \( f(t) \) is complex or hard to manage directly.

Euler's Formula

Euler's formula is a fundamental bridge between trigonometry and complex exponential functions. It is expressed as:

• \( e^{i \theta} = \cos(\theta) + i \sin(\theta) \)

This offers a powerful way to rewrite complex exponentials in terms of sines and cosines.

In our context, Euler's formula becomes particularly useful when simplifying expressions within the Fourier Transform. In the specific problem given, it's used to transform the expressions \( e^{-i \omega t} \) and \( e^{i \omega t} \) into a simpler form involving sine.

• Recognizing that:
  • \( e^{-i \omega t} - e^{i \omega t} = 2i \sin(\omega t) \)

allows one to see how the Fourier Transform simplifies into a more tractable form. By understanding Euler's formula, we not only simplify mathematical expressions but also unfold new perspectives to tackle complex problems in mathematics and physics.

This transformation is not just a trick; it's foundational in revealing the underlying symmetries and inherent properties of the functions involved, effectively bridging the gap between differential analysis and integral solutions.