The “inversion theorem” for Fourier transforms states that

$\stackrel{\mathbf{~}}{\mathbf{\varphi }}\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\stackrel{\mathbf{~}}{\mathbf{\Phi }}\mathbf{\left(}\mathbf{k}\mathbf{\right)}{\mathbf{e}}^{\mathbf{i}\mathbf{k}\mathbf{z}}\mathbf{dk}\mathbf{\iff }\stackrel{\mathbf{~}}{\mathbf{\Phi }}\mathbf{\left(}\mathit{k}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\stackrel{\mathbf{~}}{\mathbf{\varphi }}\mathbf{\left(}\mathbf{z}\mathbf{\right)}{\mathbf{e}}^{\mathbf{-}\mathbf{k}\mathbf{z}}\mathit{d}\mathit{z}$

Write the expression for the linear combination of a sinusoidal wave.

$\stackrel{\mathbf{~}}{\mathbf{f}}\mathbf{(}\mathbf{z}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\stackrel{\mathbf{~}}{\mathbf{A}}\mathbf{\left(}\mathbf{k}\mathbf{\right)}{\mathbf{e}}^{\mathbf{i}\mathbf{(}\mathbf{kz}\mathbf{-}\mathbf{\omega }\mathbf{t}\mathbf{)}}\mathbf{dk}$ …… (1)

Here, $k$is the propagation vector and $\omega$is the angular frequency.

Substitute $t=0$in equation (1).

$\tilde{f}(z,0)={\int }_{-\infty }^{\infty }\tilde{A}\left(k\right)e^{i(kz-\omega (0\left)\right)}dk$

$={\int }_{-\infty }^{\infty }\tilde{A}\left(k\right)e^{ikz}dk$

Write the conjugate of the above expression.

$$\begin{gathered} \tilde{f}(z,0{)}^{\infty }={\int }_{-\infty }^{\infty }\tilde{A}(-k{)}^{\infty }{e}^{-ikz}dk \\ \tilde{f}(z,0{)}^{\infty }={\int }_{-\infty }^{\infty }\tilde{A}(k{)}^{\infty }{e}^{-ikz}(-dk) \end{gathered}$$

It is known that,

$f(z,0)=Re\left[\tilde{f}\right(z,0\left)\right]$

$f(z,0)=\frac{1}{2}\left[\tilde{f}(z,0)+\tilde{f}(z,0{)}^{*}\right]$ …… (2)

Substitute $\frac{1}{2}\tilde{A}\left(k\right)e^{ikz}dk\text{for}\tilde{f}(z,0)\text{and}\frac{1}{2}\tilde{A}(-k{)}^{*}{e}^{ikz}dk\text{for}\tilde{f}(z,0{)}^{*}$in equation

(2).

$f(z,0)={\int }_{-\infty }^{\infty }\frac{1}{2}\left[\tilde{A}\left(k\right)+\tilde{A}(-k{)}^{*}\right]e^{ikz}dk$

Substitute $f(z,0)={\int }_{-\infty }^{\infty }\frac{1}{2}\left[\tilde{A}\left(k\right)+\tilde{A}(-k{)}^{*}\right]e^{ikz}dk$in equation (2).

$\frac{1}{2}\left[\tilde{A}\left(k\right)+\tilde{A}(-k{)}^{*}\right]=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }f(z,0)e^{-ikz}dz$…… (3)

Solve the conjugate of $f̃(z,0)$

$$\begin{gathered} \tilde{f}(z,t)={\int }_{-\infty }^{\infty }\tilde{A}\left(k\right)(-i\omega )e^{i(kz-\omega t)}dk \\ \tilde{f}(z,t)={\int }_{-\infty }^{\infty }[-i\omega \tilde{A}(k\left)\right]e^{ikz}dk \\ \tilde{f}(z,0{)}^{*}={\int }_{-\infty }^{\infty }\left[-i\omega \tilde{A}(k{)}^{*}\right]e^{-ikz}dk \\ \tilde{f}(z,0{)}^{*}={\int }_{-\infty }^{\infty }\left[i\omega \tilde{A}(k{)}^{*}\right]e^{-ikz}(-dk) \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} \tilde{f}(z,0{)}^{*}={\int }_{-\infty }^{\infty }\left[i\omega \tilde{A}(-k{)}^{*}\right]e^{-ikz}(dk\left) \\ f\right(z,0)=Re[\tilde{f}(z,0)] \end{gathered}$$

$f(z,0)=\frac{1}{2}\left[\tilde{f}(z,0)+\tilde{f}(z,0{)}^{*}\right]$ …… (4)

Substitute$\frac{1}{2}-i\omega \tilde{A}\left(k\right)e^{ikz}dk\text{for}\tilde{f}(z,0)\text{and}\frac{1}{2}-i\omega \tilde{A}\left(k{)}^{*}{e}^{ikz}dk\text{for}\tilde{f}\right(z,0{)}^{*}$for and for in equation (4).

$$\begin{gathered} f(z,0)={\int }_{-\infty }^{\infty }\frac{1}{2}[-i\omega \tilde{A}(k)+i\omega \tilde{A}(-k\left)\right]e^{ikz}dk \\ f(z,0)=\frac{-i\omega }{2}\left[\tilde{A}\left(k\right)-\tilde{A}(-k{)}^{*}\right] \end{gathered}$$

The inversion theorem for Fourier transformation states that,

$f(z,0)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }f(z,0)e^{-ikz}dz$

Equate both the values of $f(z,0)$

$\frac{1}{2}\left[\tilde{A}\left(k\right)-\tilde{A}(-k{)}^{*}\right]=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\left[\frac{i}{\omega }f(z,0)\right]e^{-ikz}dz$ …… (5)

Add equations (3) and (5).

$\frac{1}{2}\left[\tilde{A}\left(k\right)+\tilde{A}(-k{)}^{*}\right]+\frac{1}{2}\left[\tilde{A}\left(k\right)-\tilde{A}(-k{)}^{*}\right]=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }f(z,0)e^{-ikz}dz+\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\left[\frac{i}{\omega }f(z,0)\right]e^{-ikz}dz$

role="math" localid="1657466135705" $\tilde{A}\left(k\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\left[f(z,0)+\frac{i}{\omega }f(z,0)\right]e^{-ikz}dz$

Therefore, the required expression is$\tilde{A}\left(k\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\left[f(z,0)+\frac{i}{\omega }f(z,0)\right]e^{-ikz}dz$