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

Problem 6.

The given function is $f\left(x\right)=\left\{\begin{array}{l}\begin{array}{cc}x,& \left|x\right|<1,\end{array}\\ \begin{array}{cc}0,& \left|x\right|>1.\end{array}\end{array}\right.$

The Fourier transform is a mathematical technique for expressing a function as the summation of sines and cosines functions.

The specified function is$f\left(x\right)=\left\{\begin{array}{l}\begin{array}{cc}x,& \left|x\right|<1,\end{array}\\ \begin{array}{cc}0,& \left|x\right|>1.\end{array}\end{array}\right.$

The fourier sine transform is given below.

$\begin{array}{l}g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\underset{0}{\overset{1}{\int }}\sin \left(\alpha x\right)xdx\\ g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}[-\frac{x}{\alpha }\cos \left(\alpha x\right)+\frac{1}{{\alpha }^2}\sin \left(\alpha x\right)]{|}_0^1\end{array}$

Simplify further

$\begin{array}{l}g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}[-\frac{\cos \alpha }{\alpha }+\frac{\sin \alpha }{{\alpha }^2}]\\ g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\left[\frac{\sin \alpha -\alpha \cos \alpha }{{\alpha }^2}\right]\end{array}$

Thus the function is$f\left(x\right)=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\sin \alpha -\alpha \cos \alpha }{{\alpha }^2}\sin \left(\alpha x\right)dx$

The the solution to the problem (12.6) is$f\left(x\right)=\frac{1}{i\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{\sin \alpha -\alpha \cos \alpha }{{\alpha }^2}{e}^{i\alpha x}d\alpha$

Only consider the sin part of the complex exponential as the function in front of the complex exponential is odd.

$\begin{array}{l}f\left(x\right)=\frac{1}{i\pi }\underset{-\infty }{\overset{\infty }{\int }}\frac{\sin \alpha -\alpha \cos \alpha }{{\alpha }^2}i\left(\sin \left(\alpha x\right)\right)d\alpha \\ f\left(x\right)=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\sin \alpha -\alpha \cos \alpha }{{\alpha }^2}\left(\sin \left(\alpha x\right)\right)d\alpha \end{array}$

Therefore, the fourier sine transform of the function in the problem 6 is$f\left(x\right)=\frac{2}{\pi }\underset{0}{\overset{\infty }{\int }}\frac{\sin \alpha -\alpha \cos \alpha }{{\alpha }^2}\sin \left(\alpha x\right)dx$