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I would like to compute the convolution of a function with itself, where the function is $f(x) = \frac{\delta(x)}{x}$. When there is a shift in the delta function it is easy to compute, but this one is giving me trouble

I realized that I do not even know what the limit of the function is as x goes to 0. Wolfram said it goes to 0, but they used L'Hospital's rule, which seemed wrong to me since it is of the form $\frac{\infty}{0}$.

Does anyone know how I would go about computing the convolution, or at least how to determine the limit as x goes to 0?

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  • $\begingroup$ what is $\frac{\delta(x)}{x}$ ? what are your test functions ? is it really a distribution ? $\endgroup$
    – reuns
    Jan 15, 2016 at 18:36
  • $\begingroup$ I don't know if it is a distribution. I would think that $\frac{\delta(x)}{ix}$ corresponds to the time integral of the Fourier Transform of a constant function, which Wolfram also confirmed. I'm not an expert in distribution theory, I'm a signals guy. $\endgroup$
    – soultrane
    Jan 15, 2016 at 18:40
  • $\begingroup$ how do you define that integral of $1$ ? and how do you define its Fourier transform ? and how do you even define the distributions $\delta(x)$ or $1/x$ ? so be careful with distributions and symbolic calculus. $\endgroup$
    – reuns
    Jan 15, 2016 at 18:50
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    $\begingroup$ Not as a distribution. You may try to interpret $1/x$ as the principal value distribution; but then its singular support is precisely the origin, and thus coincide with that of the Dirac delta. This means we cannot expect a meaningful "product". Alternatively, were the product well defined, we would expect that its Fourier transform to equal that of the convolution of the two Fourier transforms. $\hat{\delta} = 1$ while $\hat{p.v. 1/x} = H(x)$. Neither have compact support and so again their "convolution" is not well-defined. $\endgroup$ Jan 15, 2016 at 19:37
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    $\begingroup$ Note however that $\delta(x)/x$ can be interpreted as a well-defined linear functional on smooth functions vanishing at the origin. The pairing $\langle \delta(x)/x,f(x)\rangle$ for such functions can be interpreted to equal $f'(0)$. However in this interpretation, since the test function space is not translation invariant, you cannot define meaningfully the convolution. $\endgroup$ Jan 15, 2016 at 19:42

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