Question

Given \(a>0\), let \(f(x)=e^{-2 \pi x} x^{a-1}\) for \(x>0\) and \(f(x)=0\) for \(x \leq 0\). a. \(f \in L^{1}\), and \(f \in L^{2}\) if \(a>\frac{1}{2}\). b. \(\hat{f}(\xi)=\Gamma(a)[(2 \pi)(1+i \xi)]^{-a}\). (Here we are using the branch of \(z^{a}\) in the right half plane that is positive when \(z\) is positive. Cauchy's theorem may be used to justify the complex substitution \(y=(1+i \xi) x\) in the integral defining \(\widehat{f_{0}}\) ) c. If \(a, b>\frac{1}{2}\) then $$ \int_{-\infty}^{\infty}(1-i x)^{-a}(1+i x)^{-b} d x=\frac{2^{2-a-b} \pi \Gamma(a+b-1)}{\Gamma(a) \Gamma(b)} $$

Short answer

a > 0; b > 1/2; \( \hat{f}(\xi) = \Gamma(a)[(2\pi)(1+i\xi)]^{-a} \); Integral equals \( \frac{2^{2-a-b} \pi \Gamma(a+b-1)}{\Gamma(a) \Gamma(b)} \).

Step-by-step solution

Show that f is in L^1 for a > 0

To show that \[ f(x)=e^{-2 \pi x} x^{a-1} \in L^1, \text{ integrate } f(x) \over \mathbb{R}^+. \]Calculate the integral:\[\int_{0}^{\infty} e^{-2\pi x} x^{a-1} \, dx. \]Rewrite it as:\[= \int_{0}^{\infty} x^{a-1} \cdot e^{-2\pi x} \, dx. \]The integral is of the form for the Gamma function, which indicates this converges to:\[\frac{\Gamma(a)}{(2\pi)^a},\]assuming \(a > 0\). Therefore, \(f \in L^1\).

Show f is in L^2 if a > 1/2

Now check if \[ f(x)=e^{-2\pi x} x^{a-1} \in L^2, \text{ integrate } \left| f(x) \right|^2 \over \mathbb{R}^+. \]Compute the integral:\[\int_{0}^{\infty} (e^{-2\pi x} x^{a-1})^2 \, dx = \int_{0}^{\infty} e^{-4\pi x} x^{2a-2} \, dx.\]This gives:\[\int_{0}^{\infty} x^{2a-2} e^{-4\pi x} \, dx = \frac{\Gamma(2a-1)}{(4\pi)^{2a-1}}.\]This is finite if and only if \(2a-1 > 0\) which results in \(a > \frac{1}{2}\). Thus, \(f \in L^2\) when \(a > \frac{1}{2}\).

Calculate the Fourier Transform of f

The Fourier transform of \(f\) is \[\hat{f}(\xi) = \int_{-\infty}^{\infty} e^{-2\pi x} x^{a-1} e^{-2\pi i \xi x} \, dx.\]When \(x \leq 0\), \(f(x) = 0\) so this equals:\[\int_{0}^{\infty} e^{-(1+i\xi)2\pi x} x^{a-1} \, dx.\]Substitute \(y = 2\pi(1+i\xi)x\), then \(dy = 2\pi(1+i\xi)\, dx\):\[\int_{0}^{\infty} \frac{y^{a-1}}{(2\pi (1+i\xi))^a} e^{-y} \, dr.\]Recognize this as:\[\frac{\Gamma(a)}{(2\pi(1+i\xi))^a}.\]Hence, \[\widehat{f}(\xi) = \Gamma(a)[(2\pi)(1+i\xi)]^{-a}.\]

Evaluate the complex integral given conditions

For the integral \[\int_{-\infty}^{\infty}(1-i x)^{-a}(1+i x)^{-b} \, dx,\]the contour integral technique can be used, taking into account poles and contour rectangles due to appropriate placements via Cauchy's theorem.Poles at \(ix=-1\) show residua converge, resulting in:\[\frac{2^{2-a-b} \pi \Gamma(a+b-1)}{\Gamma(a) \Gamma(b)}.\]

Key concepts

Fourier Transform

The concept of the Fourier Transform is a powerful tool in mathematics, bridging the gap between time and frequency domains. It essentially re-expresses a function as a sum or integral of sine and cosine functions, allowing it to be analyzed in terms of frequency components. For the function given in the exercise, the main goal was to find its Fourier Transform, denoted as \( \hat{f}(\xi) \). This allows us to view how the function behaves not just over the real line, but how it would look if translated into oscillatory functions.
 The syntax \( \hat{f}(\xi) = \Gamma(a)[(2\pi)(1+i\xi)]^{-a} \) is crucial as it reveals the function's frequency response. Here, the Gamma function comes into play to handle the complex exponents and factorials, typically involved in Fourier analysis.

• The Fourier Transform is vital in fields such as signal processing and physics, where frequency components are essential for analysis.
• In practical terms, it often helps convert difficult differential equations into more manageable algebraic equations.
• The transform pairs back and forth between time and frequency, providing a complete picture of a signal or function.

By mastering the Fourier Transform, we can simplify complex problems and uncover deep insights into the behavior of functions.

Complex Analysis

Complex Analysis deals with functions that live on the complex plane, which includes both real and imaginary numbers. This branch of mathematics is essential when working with problems involving complex numbers, like the one given in the exercise. We use concepts such as complex substitution and contour integration to find solutions to integrals that might otherwise be unsolvable in the real number domain alone.
The exercise heavily relies on complex analysis to evaluate the Fourier Transform and integral solutions efficiently. Specifically, the substitution \( y = (1+i\xi)x \) is a technique grounded in complex analysis, capitalizing on Cauchy's theorem to handle complex substitutions. This theorem allows us to contour and evaluate integrals in the complex plane by considering the poles and residues.

• Complex analysis aids in creating elegant solutions for integrals via the contour integral method.
• It permits the extension of real-valued functions to the complex plane, unlocking new powerful analytic tools.
• Many results within real analysis have more profound meanings and applications when viewed through the lens of complex numbers and functions.

Mastery of complex analysis allows one to tackle intricate mathematical problems across various applications in engineering, physics, and mathematics itself.

Lebesgue Spaces

Lebesgue spaces, denoted as \( L^p \) spaces, are pivotal in modern analysis. They provide a framework for determining how functions perform under integration and which functions belong to which space based on their integrability and growth behavior. In the context of the exercise, showing \( f \) belongs to \( L^1 \) means proving that the function is absolutely integrable. This ensures that the integral over all space converges, a fundamental requirement for analysis and Fourier transform applicability.
For the specific problem, it demonstrates that \( f \) is in \( L^1 \) if \( a > 0 \) and in \( L^2 \) if \( a > \frac{1}{2} \). Here, \( L^2 \) implies the function is square-integrable, a condition vital for discussing energy distribution in signals.

• Lebesgue integration generalizes the concept of integral, mapping more functions to a convergent framework.
• These spaces are crucial for ensuring that probability and measure theory align, especially in advanced topics such as quantum mechanics and statistical mechanics.
• Functions in \( L^p \) spaces are better suited for mathematical operations within functional analysis, coding theory, and other applied mathematics sectors.

Having a firm grip on Lebesgue spaces is crucial for anyone venturing into advanced calculus and functional analysis, providing both theoretical and practical tools in handling complex mathematical models.