Question

Given

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\left\{\begin{array}{c}1,\\ -1,\end{array}\begin{array}{ccc}-2& <x& <0\\ 0& <x& <2\end{array}\right\}$

find the exponential Fourier transform $\mathit{g}\left(\alpha \right)$and the sine transform${\mathit{g}}_{\mathbf{s}}\left(\alpha \right)$ .Write f(x) as an integral and use your result to evaluate

${\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\frac{\left(cos2\alpha -1\right)\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{2}\mathbf{\alpha }}{\mathbf{\alpha }}\mathit{d}\mathit{\alpha }$

Short answer

The complex transform is$g\left(\alpha \right)=\frac{i}{\mathrm{\pi \alpha }}\left(1-e^{-i2\alpha }\right)$

and the sine transform is$g_s\left(\alpha \right)=\sqrt{\frac{2}{\mathrm{\pi }}}\frac{\cos \left(2\alpha \right)-1}{\alpha }$

The value of the integral is ${\int }_0^{\infty }\frac{\cos \left(2\alpha \right)-1}{\alpha }\sin \left(2\alpha \right)d\alpha =-\frac{\mathrm{\pi }}{4}$

Step-by-step solution

Given Information.

The given equation is$f\left(x\right)=\left\{\begin{array}{c}1\\ -1\end{array}\begin{array}{ccc}-2& <x& <0\\ 0& <x& <2\end{array}\right\}$

Step 2: Meaning of the Fourier Series and Definition of Dirichlet theorem.

An infinite sum of sines and cosines is used to represent the expansion of a periodic function f(x) into a Fourier series. The orthogonality relationships between the sine and cosine functions are used in the Fourier series.

Dirichlet's theorem states that the sequence contains infinitely many prime numbers.

Find the exponential Fourier transform g(α)  and the sine transform  gs(α)

First the complex Fourier transformation:

$$\begin{gathered} g\left(\alpha \right)=\frac{1}{2\mathrm{\pi }}{\int }_{-2}^0{e}^{-i\alpha x}dx-\frac{1}{2\mathrm{\pi }}{\int }_0^2{e}^{-i\alpha x}dx \\ =\frac{1}{2\mathrm{\pi }}\frac{-1}{i\alpha }{e}^{-i\alpha x}{\left|{}_{-2}{}^0+\frac{1}{2\mathrm{\pi }}\frac{1}{i\alpha }{e}^{-i\alpha x}\right|}_0^2 \\ =\frac{i}{2\mathrm{\pi \alpha }}\left(1-e^{i2\alpha }\right)-\frac{i}{2\mathrm{\pi \alpha }}\left(e^{-i2\alpha }-1\right) \\ =\frac{i}{2\mathrm{\pi \alpha }}\left(1-e^{i2\alpha }-e^{-i2\alpha }+1\right) \\ =\frac{i}{2\mathrm{\pi \alpha }}\left(2-\left(e^{i2\alpha }+e^{-i2\alpha }\right)\right) \\ g\left(\alpha \right)=\frac{i}{\mathrm{\pi \alpha }}\left(1-\cos 2\alpha \right) \end{gathered}$$

Then the sine transform

$$\begin{gathered} g_s\left(\alpha \right)=-\sqrt{\frac{2}{\mathrm{\pi }}}{\int }_0^2\sin \left(\alpha x\right)dx \\ =\sqrt{\frac{2}{\mathrm{\pi }}}{\overline{)\frac{\cos \left(\alpha x\right)}{\alpha }}}_0^2 \\ =\sqrt{\frac{2}{\mathrm{\pi }}}\frac{\cos \left(2\alpha \right)-1}{\alpha } \end{gathered}$$

Find the integral

The function is:

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

At x=2 the function f(x) transform converges to $\frac{-1}{2}$ by Dirichlet theorem, so in the upper integral set x=2

$$\begin{gathered} \frac{-1}{2}=\frac{2}{\mathrm{\pi }}{\int }_0^{\infty }\frac{\cos \left(2\alpha \right)-1}{\alpha }\sin \left(2\alpha \right)d\alpha \\ {\int }_0^{\infty }\frac{\cos \left(2\alpha \right)-1}{\alpha }\sin \left(2\alpha \right)d\alpha =\frac{\mathrm{\pi }}{4} \end{gathered}$$

Therefore, the complex transform is$g\left(\alpha \right)=\frac{i}{\mathrm{\pi \alpha }}\left(i-e^{-i2\alpha }\right)$

and the sine transform is $g_s\left(\alpha \right)=\sqrt{\frac{2}{\mathrm{\pi }}}\frac{\cos \left(2\alpha \right)-1}{\alpha }$

The value of the integral is${\int }_0^{\infty }\frac{\cos \left(2\alpha \right)-1}{\alpha }\sin \left(2\alpha \right)d\alpha =\frac{\mathrm{\pi }}{4}$