Equation (4-21) expresses a function $\mathbf{}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$as a sum of plane waves, each with a coefficient $\mathit{A}\left(k\right)$. Equation (4-22) finds the coefficients from the given function$\mathbf{}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$ . The equations aren't independent statements; in fact, one is the inverse of the other. Equation (4-22) gives$\mathit{A}\left(k\right)$when $\mathbf{}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$is known, and (4-21) does the reverse. Example 4.7calculates $\mathit{A}\left(k\right)$from a specific$\mathbf{}\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$. Show that when this $\mathit{A}\left(k\right)$ is inserted into (4-21) , the original $\mathit{\psi }\left(x\right)$is returned. Use the Euler formula and the symmetry properties of odd and even functions to simplify your work.

$$\begin{gathered} \because A\left(k\right)=\frac{c}{\mathrm{\pi }}\frac{\sin \left({\displaystyle \frac{k\omega }{2}}\right)}{k},e^{ikx}=\cos (kx)+i\sin (kx) \\ \therefore \psi \left(x\right)={\int }_{-\infty }^{\infty }\frac{C}{\mathrm{\pi }}\frac{\sin \left({\displaystyle \frac{k\omega }{2}}\right)}{k}\cos (kx)+i\sin (kx)dk \end{gathered}$$

As noted in the problem statement, equations 4-21 and 4-22 are inverses of each other, hence, given the expression of $A\left(k\right)$, example 4.7, we should recover the original expression for $\psi \left(x\right)$using equation 4-21. We are given the direction on how to proceed, so we will start by plugin the expression for $A\left(k\right)$into 4-21 and with the help of a table of integration, we can solve for $\psi \left(x\right)$

$$\begin{gathered} \because \psi \left(x\right)={\int }_{-\infty }^{\infty }A\left(k\right)e^{ikx}dk \\ \because A\left(k\right)=\frac{c\sin \left({\displaystyle \frac{k\omega }{2}}\right)}{k},e^{ikx}=\cos \left(kx\right)+i\sin \left(kx\right) \\ \therefore \psi \left(x\right)={\int }_{-\infty }^{\infty }\frac{c}{\mathrm{\pi }}\frac{\sin \left({\displaystyle \frac{k\omega }{2}}\right)}{k}\left(\cos \left(kx\right)+i\sin \left(kx\right)\right)dk \end{gathered}$$

Noting that $A\left(k\right)$is an even function of$k\left(A\left(k\right)=A\left(-k\right)\right)$

$\because \psi \left(x\right)=\frac{c\sin \left({\displaystyle \frac{k\omega }{2}}\right)}{k}\cos \left(kx\right)dk$

From symmetry, the sine term (odd) cancel out and the cosine term (even) doubles

$\therefore \psi \left(x\right)=\frac{2c}{\mathrm{\pi }}{\int }_0^{\infty }\frac{\sin \left({\displaystyle \frac{k\omega }{2}}\right)\cos \left(kx\right)}{k}dk$

Well, a table of integration will be helpful at this point, performing this integral, will give us $0,\frac{\mathrm{\pi }}{2}$for x more than $\frac{\omega }{2}$and less than $\frac{\omega }{2}$respectively. The final expression for $\psi \left(x\right)$will be as follows,

$\therefore \psi \left(x\right)=\left\{\begin{array}{ll}0& \left|x\right|>\frac{\omega }{2}\\ c& \left|x\right|<\frac{\omega }{2}\end{array}\right.$