Let's start by recalling Euler's formula: $e^{ix}= \cos{x} + i\sin{x}$. Suppose that we wanted to create a complex wave of wavenumber $k$ in position space with amplitude $\phi(k)$. We can write that specific wave as $\phi(k)e^{ikx}$. (We write $e^{ikx}$ because $kx$ gives the angle of oscillation undergone by the wave.) Now, suppose that instead of a single wave form, we have a continuum of waves with a spectrum of wavenumbers. In order to obtain the function in position space, $\psi(x)$, we'd need to integrate $\phi(k)e^{ikx}$ over all real $k$ values. We will also need to multiply the integral by a normalizing constant, $A$, for which we will solve later on.
\begin{equation}
\psi(x)=A \int_{-\infty}^{\infty} \phi(k)e^{ikx} dk
\end{equation}
Using the above equation, any function in momentum space, $\phi(k)$, can be transformed into a function in position space, $\psi(x)$. To go the other way and turn a function in position space, $\psi(x)$, into a function in momentum space, $\phi(k)$, we would have to divide $\psi(x)$ by $e^{ikx}$ to get the relative amplitudes of each wavenumber needed to produce $\psi(x)$. Assuming that $\psi(x)$ is non-zero for more than a single point, we then need to integrate over all real $x$, and finally multiply everything by a normalizing constant, $B$. \psi(x)=A \int_{-\infty}^{\infty} \phi(k)e^{ikx} dk
\end{equation}
\begin{equation}
\phi(k) = B\int_{-\infty}^{\infty} \psi(x)e^{-ikx} dx
\end{equation}
\phi(k) = B\int_{-\infty}^{\infty} \psi(x)e^{-ikx} dx
\end{equation}
Normalizing Constants
One important property of Fourier transformations is that transforming a function between spaces should not alter the size of the functions, hence the normalization constants. In the case of quantum mechanics, this is especially true because all wave functions, regardless of whether in position or momentum space, when multiplied to its complex conjugate and integrated over all its respective space, must be equal to $1$. Therefore, it follows that:\begin{equation}
\int_{-\infty}^{\infty} |\psi(x)|^2 dx = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
(Multiplication of a number by its complex conjugate is denoted by an absolute square.)\int_{-\infty}^{\infty} |\psi(x)|^2 dx = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
\int_{-\infty}^{\infty} \left|A\int_{-\infty}^{\infty} \phi(k)e^{ikx}dk \right|^2 dx = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
|A|^2 \int_{-\infty}^\infty\left[ \int_{-\infty}^{\infty} \phi(k)e^{ikx}dk \right] \left[ \int_{-\infty}^{\infty} \phi^*(k)e^{-ikx}dk \right] dx=\int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
Because $\int_{-\infty}^{\infty} \phi^*(k) e^{-ikx} dk$ is a definite integral, $k$ can be replaced in that expression by any dummy variable of our choosing. For simplicity's sake, let's replace it with $k'$.\int_{-\infty}^{\infty} \left|A\int_{-\infty}^{\infty} \phi(k)e^{ikx}dk \right|^2 dx = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
|A|^2 \int_{-\infty}^\infty\left[ \int_{-\infty}^{\infty} \phi(k)e^{ikx}dk \right] \left[ \int_{-\infty}^{\infty} \phi^*(k)e^{-ikx}dk \right] dx=\int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
|A|^2 \int_{-\infty}^\infty \left[ \int_{-\infty}^{\infty} \phi(k)e^{ikx} dk \right] \left[ \int_{-\infty}^{\infty} \phi^*(k')e^{-ik'x}dk' \right] dx = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
|A|^2 \int_{-\infty}^\infty \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \phi(k)e^{ikx} \phi^*(k')e^{-ik'x}dk'dkdx = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
2\pi|A|^2 \int_{-\infty}^\infty \int_{-\infty}^{\infty} \phi(k)\phi^*(k') \left[\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix(k-k')} dx \right]dk'dk = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
Why in the world did we just multiply by $(2\pi) \left( \frac{1}{2\pi} \right)$ in that last step? Well interestingly enough, what we just created in the square brackets of the above equation is the Dirac delta function, $\delta(k-k')$. To prove this, let's simplify the expression:|A|^2 \int_{-\infty}^\infty \left[ \int_{-\infty}^{\infty} \phi(k)e^{ikx} dk \right] \left[ \int_{-\infty}^{\infty} \phi^*(k')e^{-ik'x}dk' \right] dx = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
|A|^2 \int_{-\infty}^\infty \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \phi(k)e^{ikx} \phi^*(k')e^{-ik'x}dk'dkdx = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
2\pi|A|^2 \int_{-\infty}^\infty \int_{-\infty}^{\infty} \phi(k)\phi^*(k') \left[\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix(k-k')} dx \right]dk'dk = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
$$\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix(k-k')}dx = \left[ \frac{e^{ix(k- k')}}{2\pi i(k-k')} \right]^{x=\infty}_{x=-\infty}$$
$$\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix(k-k')}dx = \left[ \frac{\cos{x(k-k')} + i\sin{x(k-k')}}{2\pi i(k-k')} \right]^{x=\infty}_{x=-\infty}$$
Because cosine is an even function and sine is an odd function, the following can be derived:$$\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix(k-k')}dx = \left[ \frac{\cos{x(k-k')} + i\sin{x(k-k')}}{2\pi i(k-k')} \right]^{x=\infty}_{x=-\infty}$$
$$\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix(k-k')} dx= \left[\frac{\sin{x(k- k')}}{\pi (k- k')}\right]_{x=\infty} $$
Let $\epsilon=x^{-1}$.
$$\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix(k-k')} dx= \left[\frac{\sin{(k-k')/ \epsilon}}{\pi (k-k')} \right]_{\epsilon=0} $$
The right side of the above equation is the nascent sinc representation of the Dirac delta function, $\delta(k-k')$.Let $\epsilon=x^{-1}$.
$$\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix(k-k')} dx= \left[\frac{\sin{(k-k')/ \epsilon}}{\pi (k-k')} \right]_{\epsilon=0} $$
$$\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix(k-k')}dx = \delta(k-k')$$
And so equation $(8)$ can be written as:\begin{equation}
2\pi|A|^2 \int_{-\infty}^\infty \int_{-\infty}^{\infty} \phi(k)\phi^*(k') \delta(k-k') dk'dk = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
Using the symmetrical property of the Dirac delta function, we can also say:2\pi|A|^2 \int_{-\infty}^\infty \int_{-\infty}^{\infty} \phi(k)\phi^*(k') \delta(k-k') dk'dk = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
2\pi|A|^2 \int_{-\infty}^\infty \int_{-\infty}^{\infty} \phi(k)\phi^*(k') \delta(k'-k) dk'dk = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
By the sifting or selector property of the Dirac delta function, if $a < b < c$, then $\int_a^c f(x)\delta(x-b) dx=f(b)$.
2\pi|A|^2 \int_{-\infty}^\infty \int_{-\infty}^{\infty} \phi(k)\phi^*(k') \delta(k'-k) dk'dk = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
2\pi|A|^2 \int_{-\infty}^\infty \phi(k)\phi^*(k) dk = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
A = \frac{1}{\sqrt{2\pi}}
\end{equation}
Thus, the normalizing constant $A$ is solved for. In the exact same manner, $B$ can be derived to also equal $1/\sqrt{2\pi}$. Note that mathematically, these normalizing constants could have been written to be negative or even complex as long as $|B| = |A| = 1/\sqrt{2\pi}$. However, for simplicity and by convention, $A$ and $B$ are chosen to be positive and real numbers (so that they don't invert the functions being transformed). The fully derived equations for Fourier transforms between position space and momentum space are:2\pi|A|^2 \int_{-\infty}^\infty \phi(k)\phi^*(k) dk = \int_{-\infty}^{\infty} |\phi(k)|^2 dk
\end{equation}
\begin{equation}
A = \frac{1}{\sqrt{2\pi}}
\end{equation}
\begin{align}
\psi(x) & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \phi(k)e^{ikx}dk \\
\phi(k) & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \psi(x)e^{-ikx}dx
\end{align}
\psi(x) & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \phi(k)e^{ikx}dk \\
\phi(k) & = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \psi(x)e^{-ikx}dx
\end{align}
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