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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.

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    @con-f-use: That can be risky. As an outsider looking in, my impression of physics is that you're *supposed* to be using tempered distributions (or some other sort of "generalized function") rather than ordinary functions, and corresponding distributional derivatives/limits/whatever to go with them. But, for whatever reason, this fact isn't made clear to many students and they have to pick it up through osmosis. So, the way you were expected to interpret the problem probably makes sense, despite the fact that it is absurd given a more standard interpretation.2012-10-12

2 Answers 2

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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.

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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.