Question

Show that the Fourier sine transform of \(x^{-1 / 2}\) is \(\alpha^{-1 / 2}\). Hint: Make the change of variable \(z=\alpha x .\) The integral \(\int_{0}^{\infty} z^{-1 / 2} \sin z d z\) can be found by computer or in tables.

Short answer

The Fourier sine transform of \(x^{-1 / 2}\) is \(\alpha^{-1 / 2}\).

Step-by-step solution

Write down the Fourier sine transform formula

The Fourier sine transform of a function f(x) is given by: \[ \mathcal{F}_s[f(x)](\alpha) = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} f(x) \sin(\alpha x) \, dx \]

Substitute f(x)

Substitute the given function f(x) = x^{-1/2} into the Fourier sine transform formula: \[ \mathcal{F}_s[x^{-1/2}](\alpha) = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} x^{-1/2} \sin(\alpha x) \, dx \]

Use the change of variable

Make the change of variable z = \alpha x. Then dx = \frac{dz}{\alpha} and when x = 0, z = 0, and as x \to \infty, z \to \infty. The integral becomes: \[ \int_{0}^{\infty} x^{-1/2} \sin(\alpha x) \, dx = \int_{0}^{\infty} (z/\alpha)^{-1/2} \sin(z) \frac{dz}{\alpha} \] This simplifies to: \[ \frac{1}{\alpha^{1/2}} \int_{0}^{\infty} z^{-1/2} \sin(z) \, dz \]

Use the known integral result

From integral tables or computational tools, we know: \[ \int_{0}^{\infty} z^{-1/2} \sin(z) \, dz = \sqrt{\frac{\pi}{2}} \] Thus, the integral becomes: \[ \frac{1}{\alpha^{1/2}} \sqrt{\frac{\pi}{2}} \]

Simplify the expression

Simplify the expression by canceling out terms: \[ \mathcal{F}_s[x^{-1/2}](\alpha) = \sqrt{\frac{2}{\pi}} \cdot \frac{1}{\alpha^{1/2}} \cdot \sqrt{\frac{\pi}{2}} = \alpha^{-1/2} \]

Key concepts

Fourier sine transform

The Fourier sine transform is an important mathematical tool used to convert a given function into a different domain. Imagine you have a function, and you want to express it in terms of its sine wave components. The Fourier sine transform helps you do this. It is particularly useful for functions that only exist on the positive part of the x-axis (i.e., from 0 to \(\rightarrow\infty\)) and have certain symmetry properties. The formula for the Fourier sine transform of a function f(x) is:
\[ \mathcal{F}_s[f(x)](\alpha) = \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} f(x) \sin(\alpha x) \, dx \]
This formula tells us that the Fourier sine transform of f(x) is an integral involving f(x) and \(\sin(\alpha x)\). By solving this integral, we obtain a new function of \(\alpha\), which represents the original function in the sine frequency domain. This transformation is very valuable in various fields, such as signal processing, physics, and engineering, as it allows us to analyze the frequency content of signals.
With the Fourier sine transform, you can solve complex problems more easily, by converting a difficult-to-handle function into a more manageable form.

Integral tables

Integral tables are a collection of formulas that provide the results of various integrals. They act as a reference for solving integrals quickly without performing the integration process every time. In many mathematics courses and textbooks, integral tables are included as an appendix. For example, in our exercise, we needed the result of the integral: \[ \int_{0}^{\infty} z^{-1/2} \sin(z) \, dz \] Looking it up, we find this integral is equal to \(\sqrt{\frac{\pi}{2}}\). Using integral tables can save a lot of time and effort. When you're working on more complex integrals, these tables can be incredibly helpful. Instead of solving every integral from scratch, you can simply refer to these pre-calculated results. However, it's important to know how to use these tables effectively:

• Identify the form of your integral.
• Look up the closest matching integral in the table.
• Apply any necessary substitutions or changes.

So, integral tables act as a shortcut, making the integration process faster and easier. They are especially useful when dealing with standard integrals in calculus and applied mathematics.

Change of variable

Change of variable is a technique used in integration to simplify the integrand, making it easier to solve. This method involves substituting a new variable in place of the original one to transform the integral into a more familiar or simpler form. In our exercise, we made the change of variable \(z = \alpha x\). Here’s the step-by-step process:

• Identify the new variable and its relation to the original variable. In our case: \(z = \alpha x\).
• Calculate the differential. For our change, dx becomes \(\frac{dz}{\alpha}\).
• Adjust the limits of integration based on the new variable.
• Substitute the new variables and differentials into the integral. Transforming our integral: \[ \int_{0}^{\infty} x^{-1/2} \sin(\alpha x) \, dx \rightarrow \int_{0}^{\infty} (z/\alpha)^{-1/2} \sin(z) \, \frac{dz}{\alpha} \]
• Simplify and solve the integral in the new variable. After substituting, our integral became simpler and allowed us to use the result from the integral tables.

This method is very powerful and commonly used in various applications, such as solving differential equations and evaluating definite integrals with complex boundaries. By making an appropriate change of variable, you can transform a challenging problem into one that is much easier to manage.