I tried answering a question that ended up with an expression $$\mathcal F\left\{e^{\left(\frac{2\pi j} {t}\right)}\right\}$$

Now this function we know from famous identity is $$e^{ai} = \cos(a)+i\sin(a)$$ gives $$e^{\left(\frac{2\pi j} {t}\right)} = \cos\left(\frac{2\pi} t\right)+i\sin\left(\frac{2\pi} t\right)$$

having very wobbly behaviour around $t=0$, although still being continuous and differentiable a.e.

Now to question. Would it make sense to create integral transform based on basis functions like this? Which functions could it describe well?

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  • $\begingroup$ What is the variable to have a basis function, in Fourier basis angle or frequency is the variable. $\endgroup$ – Creator Apr 16 at 22:05
  • $\begingroup$ @Creator good question. I wonder what would be most interesting. $\endgroup$ – mathreadler Apr 16 at 22:09
  • $\begingroup$ Probably doesn't answer the question, but this integral exists in the ordinary sense: $$\int_{-\infty}^\infty (e^{2 \pi i/t} - 1) e^{i p t} dt = -\frac {(2 \pi)^{3/2} J_1(2 \sqrt {2 \pi p})} {\sqrt p} H(p).$$ The transform of $1$ is $2 \pi \delta(p)$. $\endgroup$ – Maxim Apr 17 at 11:06
  • $\begingroup$ If your question is about $\int_0^\infty f(t) e^{ia/t}dt$ then it is just the usual Fourier transform $ \int_0^\infty u^{-2}f(1/u) e^{iau}du$ $\endgroup$ – reuns Apr 17 at 11:29

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