Howto show that function is a representation fot the delta function via complex path integrals? So given is the definition:
$$ f(x):=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}e^{ikx}dk $$
I'm supposed to show that this is a representation of the Dirac delta "function" ($f(x) = \delta(x)$) using complex path integrals. That is I need to show that:
$$f(x) = 0\quad \text{for}\quad x\ne 0$$
$$I = \int_{-\infty}^{+\infty} f(x) dx = 1$$
So my idea for the first equation was to say that in complex space:
$$ 0 = \oint e^{izx}dz = \underbrace{\int_{-\infty}^{+\infty} e^{izx}dz}_{=f(x)} + \int_{γ_\text{arc}}e^{izx}dz$$
Where $\gamma_\text{arc}$ is a half circle around 0 of radius $\infty$ in the complex plane. I suspect the integral $\int_{γ_\text{arc}}e^{izx}dz$ over this path $\gamma_\text{arc}$ is zero for $x\ne 0$ which would make f(x) = 0. But I have no idea how to actually do that.
Also I have no idea for the second identity $\int_{-\infty}^{+\infty} f(x) dx = 1$. 
Please help me. It's not homework but could pop up in a future test. The requirements explicitly state that it has to be solved with complex path integrations.
Edit: Here is an integral similar to my idea.
 A: When $x \neq 0$ you should interpret as Cesaro sum.  Something like
\begin{eqnarray} \lim_{L \to \infty} \left| \frac{1}{L} \int_0^L dl\int_{-l}^l  dk\;  e^{ikx} \right| &=&
\lim_{L \to \infty} \left| \frac{1}{L} \int_0^L dl   \frac{e^{ilx}-e^{-ilx}}{ix} \right| \\
&=& \lim_{L \to \infty} \left| \frac{1}{L} \int_0^{\{ L\}} dl   \frac{e^{ilx}-e^{-ilx}}{ix} \right| \\
&\leq &  \lim_{L \to \infty} \frac{2\cdot 2\pi}{xL}= \fbox{$0$}\end{eqnarray} 
where the finite length integrals move in a circle around 0.  Here $\{ L\} \equiv L \mod 2\pi $.
For the ``concentration of mass" part, you can try integrating once, then you get a pole around $x=0$ (as you should!)
\begin{eqnarray} \lim_{L,M \to \infty}  \int_{-L}^L \int_{-M}^M dx \; dk\;  e^{ikx} &=& 
\lim_{L \to \infty}  \int_{-L}^L  dx \;   \frac{e^{iMx} - e^{-iMx}}{ix} \\
&=& \lim_{\epsilon \to 0} \oint_{\epsilon S^1 \cap \mathbb{H} } dz \;   \frac{e^{iMz} - e^{-iMz}}{iz} \\
&=& 2 \pi i \end{eqnarray}
The integral depends only on the value at a small circle around $x = 0$.  It's like the position have have been complexified or something: $x \mapsto z$ or $x \mapsto x + i\epsilon$ and you get well-defined integrals.
A: Let $\psi(x) = e^{ikx} = \langle k | x\rangle $ be the wavefunction of the quantum mechanical free particle.  Then
$$ \int_{-\infty}^\infty e^{ikx} dk= \int_{-\infty}^\infty  dk |k \rangle \langle k | x \rangle = | x \rangle $$
We get the particle in position basis.  We have used some non-rigorous identities from functional analysis:
\[ \int_{-\infty}^\infty  dk |k \rangle \langle k | = \mathbf{1} \text{ and }\langle x' | x \rangle = \delta(x - x') \]
In quantum mechanics, the Fourier transform corresponds to switching between position basis and momentum basis.
Another interpretation is "the Fourier transform of the uniform measure". 
Let $\mu(k) = dk$.
$$ \hat{\mu}(x) = \int_{-\infty}^\infty e^{ikx} \mu(k) $$
The uniform flat-line signal doesn't oscillate.  So dually the sum over all frequences should average out to a single spike.  You might see this point if view in signal processing.
