Question

Show that the Fourier sine transform of ${\mathit{x}}^{\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}}$ is . Hint: Make the change of variable$\mathit{z}\mathbf{=}\mathit{\alpha }\mathit{x}$ . The integral ${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathit{z}}^{\mathbf{-}\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{z}\mathbf{}\mathit{d}\mathit{z}$can be found by computer or in tables

Short answer

Thus, the required Fourier series is $g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\alpha \frac{1}{2}\frac{\mathrm{\pi }}{2}$

Step-by-step solution

Determine the Fourier transformation for the given function. 

Fourier sine transform of f(x) is,

$$\begin{gathered} g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }f\left(x\right)\sin \alpha xdx \\ =\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }{x}^{\frac{1}{2}}\sin \alpha xdx \end{gathered}$$

Let $z=\alpha x,dz=\alpha dx$

So, we get,

$$\begin{gathered} g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }{x}^{\frac{1}{2}}\sin \alpha xdx \\ =\sqrt{\frac{2}{\pi }}{\int }_0^{\infty }{\left(\frac{z}{\alpha }\right)}^{\frac{1}{2}}\sin z\frac{dz}{\alpha } \\ =\sqrt{\frac{2}{\pi }}{\alpha }^{\frac{1}{2}}{\int }_0^{\infty }\frac{\sin z}{z}dz \\ g\left(\alpha \right)=\sqrt{\frac{2}{\pi }}{\alpha }^{\frac{1}{2}}\frac{\pi }{2} \end{gathered}$$