Question

Find the exponential Fourier transform of the given $\mathit{f}\left(x\right)$and write $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$as a Fourier integral.

$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}x,0<x<1\\ 0,otherwise\end{array}$

Short answer

Answer

The exponential Fourier transform of the given function is$g\left(\alpha \right)=\frac{1e^{+i\alpha }\left(1+i\alpha \right)-1}{{\alpha }^2}$and $f\left(x\right)$as a Fourier integral is$f\left(x\right)=\frac{1}{2\mathrm{\pi }}\underset{-\infty }{\overset{\infty }{\int }}\frac{e^{-i\alpha }\left(1+i\alpha \right)-1}{{\alpha }^2}{e}^{i\alpha x}d\alpha$.

Step-by-step solution

Given information

The given function is$f\left(x\right)=\left\{\begin{array}{l}x,0<x<1\\ 0,otherwise\end{array}\right.$ . The exponential Fourier transform of the function is to be found and the function is to be written as a Fourier integral.

The significance  of Fourier transforms

The following are the formulas for the Fourier series transforms.

Here,is called the Fourier transform of.

Find the Fourier transform

Use the formula $g\left(\alpha \right)=\frac{1}{2\mathrm{\pi }}\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)e^{-i\alpha x}dx$to find the Fourier tr$$\begin{gathered} g\left(\alpha \right)=\frac{1}{2\mathrm{\pi }}\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)e^{-i\alpha x}dx \\ =\frac{1}{2\mathrm{\pi }}{\left[\frac{x}{i\alpha }{e}^{-i\alpha x}+\frac{1}{{\alpha }^2}{e}^{-i\alpha x}\right]}_0^1 \\ =\frac{1}{2\mathrm{\pi }}\left[\frac{e^{-i\alpha }}{i\alpha }+\frac{1}{{\alpha }^2}\left(e^{-i\alpha }-1\right)\right] \\ =\frac{1}{2\mathrm{\pi }}\frac{e^{-i\alpha }-1+i\alpha e^{-i\alpha }}{{\alpha }^2} \end{gathered}$$

Solve further to obtain.

Now, writeas a Fourier integral using the formula.