Is there any handwavy argument that shows that $\int_{-\infty}^{\infty} e^{-ikx} dk = 2\pi \delta(x)$?

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.

$$\int_{-a}^{a} e^{-ikx} dk = \frac{e^{-ikx}}{-ix} |^{+a}_{-a}= i\frac{e^{-iax}-e^{iax}}{x} $$

$$e^{-iax}-e^{iax}=-2i \sin {(ax)} $$

$$\int_{-a}^{a} e^{-ikx} dk = i\frac{e^{-iax}-e^{iax}}{x} = i\frac{-2i \sin {ax}}{x} =2\frac{\sin {ax}}{x} = f_a(x) $$

You need to find the limit. Lets define $\lim\limits_{ a\to \infty } f_a(x) = 2\pi \delta(x)$

$$\lim\limits_{ a\to \infty } \int_{-a}^{+a} e^{-ikx} dk = \lim\limits_{ a\to \infty } 2\frac{\sin {ax}}{x}= 2\pi \delta(x) $$

we can check some graphs for some $a$ values

$a=100$

$a=10^{100}$

Finally If $ a\to \infty $ then we get the function $2\pi \delta(x)$ below.

The function for $x\neq0 $

$2\pi \delta(x)=0$

but for $x=0 $

$\lim\limits_{ x\to 0 } 2\pi \delta(x)=+\infty$

but the function is very special as Priyatham shew.

For $\epsilon>0$ $$ \int\limits_{-\epsilon}^\epsilon 2\pi \delta(x) \mathrm{d}x = 2\pi$$

Thus Finally we can get the important result for $\delta(x)$ function that

If $\epsilon>0$ then

$$\int\limits_{-\epsilon}^\epsilon \delta(x) \mathrm{d}x = 1$$

Note: The graphs were taken from www.wolframalpha.com