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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-user the second part, at least, makes little sense. $\int_{-\infty}^\infty f(x) = 0$ if $f(x) = 0$ except at $x=0$, since single point don't matter when integrating.2012-10-12
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    @conf-f-use For the first part, try damping the integrand, in a way that doesn't disturb the value on the real line.2012-10-12
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    Is this how physicists do math? None of this seems to make sense to me as a mathematician. The integral defining $f$ does not converge, there is no half circle of radius $\infty$ etc.2012-10-12
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    @fpg it does make sense in the context of Fourier transformations and when integrating over a diverging point in the function. And yes I'm confused too about the whole problem.2012-10-12
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    @con-f-use: No, fgp is right. The problem is that the question you mean to ask isn't about a *function* defined as an integral of *functions*: instead, you are dealing with something like tempered distributions. The facts regarding those are going to have a lot of differences from the facts regarding integrals of functions.2012-10-12
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    I'm beginning to have that feeling too. Thanks to all of you. I'll just tell my professor that his problem was ill-conceived. I think he'll understand. Thanks again.2012-10-12
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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

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