91Ó°ÊÓ

The given function is$f\left(x\right)=\left\{\begin{array}{l}\begin{array}{cc}\sin x,& \left|x\right|<\mathrm{Ï€}/2,\end{array}\\ \begin{array}{cc}0,& \left|x\right|>\mathrm{Ï€}/2.\end{array}\end{array}\right.$

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

The fourier sine transform is given below.

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

Simplify further

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

Simplify further

$g\left(\mathrm{Î\pm }\right)=\sqrt{\frac{2}{\mathrm{Ï€}}}\left[\frac{\mathrm{Î\pm }\cos (\mathrm{Î\pm }\mathrm{Ï€}/2)}{1−{\mathrm{Î\pm }}^2}\right]$

Thus the function is$f\left(x\right)=\frac{2}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{\mathrm{Î\pm }\cos (\mathrm{Î\pm }\mathrm{Ï€}/2)}{1−{\mathrm{Î\pm }}^2}\sin \left(\mathrm{Î\pm }x\right)da$

The the solution to the problem (12.12) is$f\left(x\right)=−\frac{i}{\mathrm{Ï€}}\underset{−\mathrm{∞}}{\overset{\mathrm{∞}}{∫}}\frac{\mathrm{Î\pm }\cos (\mathrm{Î\pm }\mathrm{Ï€}/2)}{1−{\mathrm{Î\pm }}^2}{e}^{i\mathrm{Î\pm }x}d\mathrm{Î\pm }$

Only consider the $i\sin \left(\mathrm{Î\pm }x\right)$part of the complex exponential as the function in front of the complex exponential is odd.

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

Therefore, the fourier sine transform of the function in the problem 12 is $f\left(x\right)=\frac{2}{\mathrm{Ï€}}\underset{0}{\overset{\mathrm{∞}}{∫}}\frac{\mathrm{Î\pm }\cos (\mathrm{Î\pm }\mathrm{Ï€}/2)}{1−{\mathrm{Î\pm }}^2}\sin \left(\mathrm{Î\pm }x\right)d\mathrm{Î\pm }$.