## Section12.1Definition of Fourier Transform

Consider the (square integrable) function $f(x)\text{,}$ its Fourier transform is defined by:

Definition

$$\cal{F}(f) =\tilde{f} (k)= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x)\, e^{-ikx}\, dx\tag{12.1.1}$$

The inverse of the Fourier transform is given by:

$$\cal{F}^{-1}(\tilde{f})=f(x) =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \tilde{f} (k)\, e^{ikx}\, dk\tag{12.1.2}$$

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{12.1.3} \end{align}

where in the second to the last line, we have used the integral representation of the delta function, see Section 11.6, to evaluate the expression in the square brackets.