Howto show that function is a representation fot the delta function via complex path integrals?

Question

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.

Answer 1

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.

Answer 2

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.