Integral transform with reciprocal complex exponential functions?

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? • What is the variable to have a basis function, in Fourier basis angle or frequency is the variable. – Creator Apr 16 at 22:05
• @Creator good question. I wonder what would be most interesting. – mathreadler Apr 16 at 22:09
• 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 $2 \pi \delta(p)$. – Maxim Apr 17 at 11:06
• If your question is about $\int_0^\infty f(t) e^{ia/t}dt$ then it is just the usual Fourier transform $\int_0^\infty u^{-2}f(1/u) e^{iau}du$ – reuns Apr 17 at 11:29