Section 10.1 Definition of Fourier Transform
¶Consider the (square integrable) function \(f(x)\text{,}\) its Fourier transform is defined by:
Definition
\begin{equation}
\cal{F}(f)
=\tilde{f} (k)= \frac{1}{\sqrt{2\pi}}
\int_{-\infty}^{\infty} f(x)\, e^{-ikx}\, dx\tag{10.1.1}
\end{equation}
The inverse of the Fourier transform is given by:
\begin{equation}
\cal{F}^{-1}(\tilde{f})=f(x)
=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}
\tilde{f} (k)\, e^{ikx}\, dk\tag{10.1.2}
\end{equation}
To show that the inverse Fourier transform is indeed the inverse operation, start with the right-hand-side of the inverse Fourier transform and insert the definition of the Fourier transform
\begin{align}
\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \tilde{f} (k)\, e^{ikx}\, dk
\amp = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}
\left[\frac{1}{\sqrt{2\pi}}
\int_{-\infty}^{\infty} f(x')\, e^{-ikx'}\, dx'\right]\,
e^{ikx}\, dk\notag\\
\amp = \int_{-\infty}^{\infty} f(x')
\left[\frac{1}{2\pi} \int_{-\infty}^{\infty}e^{ik(x-x')}\,dk\right]\,
dx'\notag\\
\amp = \int_{-\infty}^{\infty} f(x') \delta(x-x')\, dx'\notag\\
\amp = f(x)\tag{10.1.3}
\end{align}
where in the second to the last line, we have used the integral representation of the delta function, see Section 6.6, to evaluate the expression in the square brackets.