Normalization of eigenfunction to Dirac-delta function In the first chapter of Principles of Quantum Mechanics by R. Shankar, he describes finding the eigenvalues and eigenfunctions of the operator $K=-iD=-i\frac{d}{dx}$. For context, he does this:

What I don't understand is how he arrived at $A=1/\sqrt{2\pi}$. It seems to be because (since this is an infinite-dimensional space) we want to normalize to the Dirac delta function, but I don't understand why
$$\frac1{2\pi}\int_{-\infty}^\infty e^{-i(k-k')x}dx=\delta(k-k').\tag{*}$$
He doesn't really explain this. How does he normalize it?
 A: For a function $f(x)$ the Fourier transform is defined as:
\begin{equation}
\overset{\boldsymbol{\sim}}{f}\left(k\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!f\left(x\right)e^{ikx}\mathrm dx 
\tag{01}\label{01}
\end{equation}
This transformation is invertible, that is:
\begin{equation}
f\left(x\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!\overset{\boldsymbol{\sim}}{f}\left(k\right)e^{\boldsymbol{-}ixk}\mathrm dk 
\tag{02}\label{02}
\end{equation}
With $f\left(x\right)=\delta\left(x\right)$, equation \eqref{01} yields:
\begin{equation}
\overset{\boldsymbol{\sim}}{\delta}\left(k\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!\delta\left(x\right)e^{ikx}\mathrm dx =\dfrac{1}{\sqrt{2\pi}} 
\tag{03}\label{03}
\end{equation}
That is, the Fourier transform of the $\delta$-function is the constant $1/\sqrt{2\pi}$. Equation \eqref{02} gives:
\begin{equation}
\delta\left(x\right)=\dfrac{1}{\sqrt{2\pi}}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!\dfrac{1}{\sqrt{2\pi}}e^{\boldsymbol{-}ixk}\mathrm dk =\dfrac{1}{2\pi}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!e^{\boldsymbol{-}ixk}\mathrm dk
\tag{04}\label{04}
\end{equation}
This is sometimes called the integral definition of the $\delta$-function.
Exchanging the roles of $k$ and $x$ in equation \eqref{04}, the definition becomes:
\begin{equation}
\delta\left(k\right)=\dfrac{1}{2\pi}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!e^{\boldsymbol{-}ikx}\mathrm dx
\tag{05}\label{05}
\end{equation}
Replacing $k$ in equation \eqref{05} with $k-k'$ we arrive at:
\begin{equation}
\delta\left(k-k'\right)=\dfrac{1}{2\pi}\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\!\!e^{\boldsymbol{-}i(k-k')x}\mathrm dx
\tag{06}\label{06}
\end{equation}
which is the equation that Shankar used.
