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. 

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