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

Definition

\begin{equation} \tilde{f} (k)= F(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x)\, e^{-ikx}\, dx \end{equation}

The inverse of the Fourier transform is given by:

\begin{equation} f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \tilde{f} (k)\, e^{ikx}\, dk \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{eqnarray*} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \tilde{f} (k)\, e^{ikx}\, dk &=&\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\\ &=&\int_{-\infty}^{\infty} f(x') \left[\frac{1}{2\pi} \int_{-\infty}^{\infty}e^{ik(x-x')}\,dk\right]\, dx'\\ &=&\int_{-\infty}^{\infty} f(x') \delta(x-x')\, dx'\\ &=&f(x) \end{eqnarray*} where in the second to the last line, we have used the integral representation of the delta function.