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Usually the Fourier transformation is defined using the imaginary unit $$ F(p) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{ipx} dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) \cos px + i \sin px\ dx $$ Now with split complex numbers one could come up with the idea to replace the imaginary unit $i^2=-1$ with $j^2=1$ and get in analogy a kind of hyperbolic Fourier transformation $$ F_h(p) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) e^{jpx} dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(x) \cosh px + j \sinh px\ dx $$ Does such a transformation make sense? Would it have any use?

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    A potential issue is the growth rate of $\cosh$ and $\sinh$. It seems like $F_h(p)$ will only converge for functions that are at least inverse exponential growth, although I haven't put too much thought into this. – 2017-02-22
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    If you split up your cosh and your sinh into exponentials you'll find that you're essentially performing Laplace transforms. – 2017-02-22
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    For potential reading, [this](https://arxiv.org/pdf/1406.1014.pdf) seems to look at "fourier transforms" in hypercomplex numbers (which the split complex numbers are a simple case of). I haven't read it closely, but on page 8 they seem to go through and look at what properties must be followed for a hypercomplex Algebra (potentially multiple dimensions) to have a fourier transform on some plane spanned by two hypercomplex elements. It seems like the conclusion they reach is for the plane to be isomorphic to $\mathbb C$. – 2017-02-22
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    As the split complex numbers contain no plane isomorphic to $\mathbb C$, their work would suggest the answer is "no". – 2017-02-22

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