Is there any handwavy argument that shows that $\int_{-\infty}^{\infty} e^{-ikx} dk = 2\pi \delta(x)$? It should not be a good argument but rather a short one and one that convinces a physicist ( so no need for mathematical rigor ) that shows that $\int_{-\infty}^{\infty} e^{-ikx} dk = 2\pi \delta(x)$ holds? It should only refer to basic calculus (especially no fourier transform ) since I am supposed to give a proof of a related relationship about fourier transforms on a physics homework sheet. 

Consider 
$$
\begin{equation}
\begin{split}
f(x) & = \int\limits_{-\infty}^\infty e^{-ikx} \mathrm{d}k\\
\int\limits_{-\epsilon}^\epsilon f(x) \mathrm{d}x = \int\limits_{-\epsilon}^{\epsilon}\int\limits_{-\infty}^\infty e^{-ikx} \mathrm{d}k \mathrm{d}x & = \int\limits_{-\infty}^{\infty}\int\limits_{-\epsilon}^\epsilon e^{-ikx} \mathrm{d}x \mathrm{d}k \\
& = \int\limits_{-\infty}^{\infty}\int\limits_{-\epsilon}^\epsilon \cos(kx) \, \mathrm{d}x \mathrm{d}k \\
& = \int\limits_{-\infty}^{\infty} \frac{2\sin(k\epsilon)}{k} \, \mathrm{d}k \\
& = 2\pi
\end{split}
\end{equation}
$$
Hope that was short enough.
