Question

In Problems 13to 16, find the Fourier cosine transform of the function in the indicated problem, and write f(x)as the Fourier integral [ use equation (12.15)]. Verify that the cosine integral forf(x) is the same as the exponential integral found previously.

16. Problem 11

Short answer

The Fourier cosine transform of the function in the problem 11, written in the Fourier integral is:

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

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 11 is :

$f\left(x\right)=\left\{\begin{array}{l}\begin{array}{cc}\cos x,& -\pi /2<x<\pi /2\end{array}\\ \begin{array}{cc}0,& \left|x\right|>\pi /2\end{array}\end{array}\right.$

Find the Fourier integral

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

$\begin{array}{c}g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\left(\cos x\right)\cos \left(\alpha x\right)dx\\ =\frac{1}{\sqrt{2\pi }}{\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}[\cos \left(x(1+\alpha )\right)-\cos \left(x(1-\alpha )\right)]dx\\ =\frac{1}{\sqrt{2\pi }}{[\frac{\sin \left(x(1+\alpha )\right)}{1+\alpha }+\frac{\sin \left(x(1-x)\right)}{1-\alpha }]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx\\ =\frac{1}{\sqrt{2\pi }}\frac{\sin (\pi (1+\alpha )/2)(1-\alpha )+\sin (\alpha (1-\alpha )/2)(1+\alpha )}{1-{\alpha }^2}\end{array}$

$g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\frac{\cos (\pi \alpha /2)}{1-{\alpha }^2}$

Then the function will be:

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

Find the Fourier cosine of the function

The solution of the Problem 11 is:

$f\left(x\right)=\frac{1}{\pi }{\int }_{-\infty }^{\infty }\frac{\cos (\pi \alpha /2)}{1-{\alpha }^2}{e}^{ix\alpha }d\alpha$

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

This implies that,

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

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

So, the Fourier cosine transform is:

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