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I had this on an exam yesterday, and I'm not entirely convinced that the statement is true. We were asked to show that the function $\delta (x) = \int_{-∞}^{∞} \frac{1}{t(t-x)} dt$ is a dirac delta function by demonstrating that $I=\int_{-∞}^{∞} f(x)\delta(x) dx$ holds all the necessary properties.

There are three things I believe should be shown: 1) The function should be infinite at a single point. (this function is infinite at $t=x$) 2) It should be zero everywhere else 3) It should satisfy $\int_{-∞}^{∞} f(x)\delta(x) dx=f(0)$.

I showed 2 is true by demonstrating that the Cauchy Principal Value is zero for that integral which means that it's zero everywhere save that one point we avoid.... but I don't see how 3 holds in general. I see that it holds for some functions, like $f(x)=x$, but what about $f(x)=1$, for example. Anyway, is this a delta function or not.... if so, why?

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    No! See this is a tricky point, lol. This is a function of $x$! So we are integrating out the $t$, which means it's NOT a function of $t$ so to say it's infinite at $t=0$ would be a misunderstanding (I believe) in the sense that all one needs to do to compute that integral is go into the complex plane and do a contour integral and get arbitrarily close to $t=0$ and $t=x$.2012-12-20
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    Someone deleted their comment, I should point out that I'm not having a conversation with myself. Previous comment was "isn't it also infinite at t=0?".2012-12-20
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    If $g$ is any function that is $0$ except at one point, then $\int_{-\infty}^\infty f(x)g(x) dx = 0$, no matter what $f(x)$ is and no matter what $g$ does at that one point. So you cannot possibly prove that your function has the three properties, because no function at all has those three properties.2012-12-20
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    The dirac function is *formally* defined by $\int_{-\infty}^\infty f(x)\delta(x)dx=f(0)$, but it's not really a function. It's a measure; the Lebesque integral with respect to $\delta$ returns $f(0)$.2012-12-20
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    Okay, but does my function satisfy the requirements to be a Dirac delta? Function or otherwise? Ie.... Call it a function or whatever, is my thing a Dirac delta.2012-12-20
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    Your third requirement is enough by itself provided you say "for all sufficiently well behaved functions $f$" and specificy that that means test functions, or Schwartz functions, or whichever class of functions is appropriate for the occasion.2012-12-20
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    @dustanalysis I know that is what you are looking for, which is why I posted a comment instead of an answer. I wanted to point that out because I found it helpful to understand exactly what $\delta$ *is* before I started proving things about it.2012-12-20

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