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 for f(x)is the same as the exponential integral found previously.

15. Problem 9

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 the graph of a function

The function of the Fourier cosine transform will be calculated as

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

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

Then the Fourier integral will be

$f\left(x\right)=\frac{4}{\pi }{\int }_0^{\infty }\frac{1-\cos \left(a\alpha \right)}{{\alpha }^2}\cos \left(\alpha x\right)d\alpha$

The solution to the problem 9 is:

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

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

$\begin{array}{c}f\left(x\right)=\frac{2}{\pi }{\int }_{-\infty }^{\infty }\frac{1-\cos \left(\alpha a\right)}{{\alpha }^2}\cos \left(\alpha x\right)d\alpha \\ =\frac{4}{\pi }{\int }_0^{\infty }\frac{1-\cos \left(\alpha a\right)}{{\alpha }^2}\cos \left(\alpha x\right)d\alpha \end{array}$

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

So, the Fourier cosine transform is:

$f\left(x\right)=\frac{4}{\pi }{\int }_0^{\infty }\frac{1-\cos \left(\alpha a\right)}{{\alpha }^2}\cos \left(\alpha x\right)d\alpha$