Question

In Problems 17to 20, find the Fourier sine transform of the function in the indicated problem, and write f(x)as the Fourier integral [ use equation (12.14)]. Verify that the sine integral for f(x)is the same as the exponential integral found previously.

17.Problem 3

Short answer

The Fourier sine transform of the function in the problem 3, written in the Fourier integral is:

$f\left(x\right)=\frac{2}{\pi }{\int }_0^{\infty }\frac{1-\cos \left(\pi \alpha \right)}{\alpha }\sin \left(\alpha x\right)d\alpha$

Step-by-step solution

Meaning of Fourier Sine and Cosine Transform

The Fourier sine and cosine transforms are versions of the Fourier transform that don't employ complex numbers or require negative frequency in mathematics.

Given parameter

The function of the Problem 3 is:

$f\left(x\right)=\left\{\begin{array}{l}\begin{array}{rr}-1,& -\pi <x<0\end{array}\\ \begin{array}{rr}1,& 0<x<\pi \end{array}\\ \begin{array}{rr}0,& \left|x\right|>\pi \end{array}\end{array}\right.$

Find the Fourier integral

The function of the Fourier sine transform will be calculated as:

$\begin{array}{c}g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}{\int }_0^{\pi }\sin \left(\alpha x\right)dx\\ =-{\sqrt{\frac{2}{\pi }}\frac{\cos \left(\alpha x\right)}{\alpha }|}_0^{\pi }\\ =\sqrt{\frac{2}{\pi }}\frac{1-\cos \left(\alpha x\right)}{\alpha }\end{array}$

Then the function will be:

$f\left(x\right)=\frac{2}{\pi }{\int }_0^{\infty }\frac{1-\cos \left(\alpha \pi \right)}{\alpha }\sin \left(\alpha x\right)d\alpha$

Find the Fourier sine of the function

The solution of the problem 3 is:

$f\left(x\right)=-\frac{i}{\pi }{\int }_{-\infty }^{\infty }\frac{1-\cos \left(\alpha \pi \right)}{\alpha }{e}^{i\alpha x}d\alpha$

As the function in front of the complex exponential is now odd, then the integration will only save the sine part of the complex exponential.

This implies that,

$\begin{array}{c}f\left(x\right)=-\frac{i}{\pi }{\int }_{-\infty }^{\infty }\frac{1-\cos \left(\alpha \pi \right)}{\alpha }\left(i\sin \right(\alpha x\left)\right)d\alpha \\ =\frac{2}{\pi }{\int }_0^{\infty }\frac{1-\cos \left(\pi \alpha \right)}{\alpha }\sin \left(\alpha x\right)d\alpha \end{array}$

It clarifies that the sine integral is same as the exponential integral.

So, the Fourier sine transform is:

$f\left(x\right)=\frac{2}{\pi }{\int }_0^{\infty }\frac{1-\cos \left(\pi \alpha \right)}{\alpha }\sin \left(\alpha x\right)d\alpha$