91Ó°ÊÓ

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.

$\begin{array}{l}g\left(\mathrm{Î\pm }\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}\underset{0}{\overset{1}{∫}}\sin \left(\mathrm{Î\pm }x\right)xdx\\ g\left(\mathrm{Î\pm }\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}[−\frac{x}{\mathrm{Î\pm }}\cos \left(\mathrm{Î\pm }x\right)+\frac{1}{{\mathrm{Î\pm }}^2}\sin \left(\mathrm{Î\pm }x\right)]{|}_0^1\end{array}$

Simplify further

$\begin{array}{l}g\left(\mathrm{Î\pm }\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}[−\frac{\cos \mathrm{Î\pm }}{\mathrm{Î\pm }}+\frac{\sin \mathrm{Î\pm }}{{\mathrm{Î\pm }}^2}]\\ g\left(\mathrm{Î\pm }\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}\left[\frac{\sin \mathrm{Î\pm }−\mathrm{Î\pm }\cos \mathrm{Î\pm }}{{\mathrm{Î\pm }}^2}\right]\end{array}$

Thus the function is$f\left(x\right)=\frac{2}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{\sin \mathrm{Î\pm }−\mathrm{Î\pm }\cos \mathrm{Î\pm }}{{\mathrm{Î\pm }}^2}\sin \left(\mathrm{Î\pm }x\right)dx$

The the solution to the problem (12.6) is$f\left(x\right)=\frac{1}{i\mathrm{Ï€}}\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\frac{\sin \mathrm{Î\pm }−\mathrm{Î\pm }\cos \mathrm{Î\pm }}{{\mathrm{Î\pm }}^2}{e}^{i\mathrm{Î\pm }x}d\mathrm{Î\pm }$

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

$\begin{array}{l}f\left(x\right)=\frac{1}{i\mathrm{Ï€}}\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\frac{\sin \mathrm{Î\pm }−\mathrm{Î\pm }\cos \mathrm{Î\pm }}{{\mathrm{Î\pm }}^2}i\left(\sin \left(\mathrm{Î\pm }x\right)\right)d\mathrm{Î\pm }\\ f\left(x\right)=\frac{2}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{\sin \mathrm{Î\pm }−\mathrm{Î\pm }\cos \mathrm{Î\pm }}{{\mathrm{Î\pm }}^2}\left(\sin \left(\mathrm{Î\pm }x\right)\right)d\mathrm{Î\pm }\end{array}$

Therefore, the fourier sine transform of the function in the problem 6 is$f\left(x\right)=\frac{2}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{\sin \mathrm{Î\pm }−\mathrm{Î\pm }\cos \mathrm{Î\pm }}{{\mathrm{Î\pm }}^2}\sin \left(\mathrm{Î\pm }x\right)dx$