Question 3.2: Find the function corresponding to the Fourier transform g(k...

The transform g(k) in the figure is equal to g in the interval  k_{0} − \Delta k ≤ k ≤ k_{0} + \Delta k , and it is zero otherwise. Using Eq. (3.90), we get

ψ(x) =\frac{1}{\sqrt{2\pi } }\int_{-\infty }^{\infty }g(k)e^{ikx}  dk.                             (3.90)

ψ(x) =\frac{g}{\sqrt{2\pi } }\int_{k_{0}-\Delta k }^{k_{0}+\Delta k } e^{ikx}dk.
The integral in this equation may be evaluated by defining the change of variable
u = ikx.

Solving this last equation for k in terms of u and taking the differential dk, we obtain
\begin{matrix}k=\frac{u}{i x}, & dk=\frac{du}{i x}\end{matrix}
The expression for ψ(x) can thus be written
ψ(x) = \frac{g}{\sqrt{2\pi } }· \frac{1}{ix} \int_{ix(k_{0}-\Delta k)}^{ix(k_{0}+\Delta k)} e^{u}d u = \frac{g}{\sqrt{2\pi } }· \frac{1}{ix} \left[e^{ix(k_{0}+\Delta k)}-e^{ix(k_{0}-\Delta k)}\right].
The exponential function  e^{i k_{0} x} may be factored from the term within square brackets giving
ψ(x) = \frac{g}{\sqrt{2\pi } }· \frac{1}{ix}e^{ik_{0}x} \left[e^{i\Delta k x}-e^{-i\Delta k x}\right].
Finally, using Eq. (3.10) which expresses the sine function in terms of exponentials, we obtain

\sin (k x)=\frac{e^{i k x }- e^{- i k x}}{2i}                                      (3.10)
ψ(x) = \left[\frac{2g}{\sqrt{2\pi } } \frac{\sin \Delta k x}{x} \right]e ^{ik_{0}x}                           (3.92)