I am working with recursive convolutions of complex functions (fourier transforms), so, I calculate convolution integrals of functions of the form $$g(iy) = \int_{-\infty}^\infty f(ix)f(i(y-x))dx, \ \ \to \ \ h(iy) = \int_{-\infty}^\infty g(ix)f(i(y-x))dx$$ So when I calculate these in mathematica (11.2), and I am a beginner user, I retrieve functions which are complex functions of $y$, while I want to have these as functions of $iy$. I could not find on the forum how to write these functions such that there are only $iy$ terms in the equation instead of $i$'s and $y$'s scattered all over the function...
1 Answer
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I found an answer myself:
if g[y] is the output function, then the following solves it:
Solve[{gs == g[y], s == (I y)}, gs, {y}]
where gs is the function $g(s)$ with $s=iy$ (think of the Laplace variable)
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1$\begingroup$ I don’t understand how exactly this solves the problem. Can you define the functions here in the original question for future users? $\endgroup$ Commented Jul 23, 2020 at 23:21
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y->-I zin the final result, does it solve your problem? $\endgroup$