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Is this integral transform invertible and what is its inverse $f(t)$ in that case?

$$Fs(\omega)=\int\limits_{-\infty}^\infty f(t)(cs({\omega t}) + i \cdot ss({\omega t}))\mathrm dt$$

where $cs(x)$ and $ss(x)$ are defined below. $cs()$ and $ss()$ are periodical with period $2\pi$.

$$ ss(x) = \begin{cases} \displaystyle 0 & \text{if}\;\; 0 \le x < \frac{\pi}{4} \\ \displaystyle \frac{1}{\sqrt{2}} & \text{if}\;\; x = \frac{\pi}{4} \\ 1 & \text{if}\;\; \frac{\pi}{4} \lt x < \frac{3\pi}{4}\\ \displaystyle \frac{1}{\sqrt{2}} & \text{if}\;\; x = \frac{3\pi}{4} \\ \displaystyle 0 & \text{if}\;\; \frac{3\pi}{4} \lt x < \frac{5\pi}{4} \\ \displaystyle -\frac{1}{\sqrt{2}} & \text{if}\;\; x = \frac{5\pi}{4} \\ -1 & \text{if}\;\; \frac{5\pi}{4} \lt x < \frac{7\pi}{4}\\ \displaystyle -\frac{1}{\sqrt{2}} & \text{if}\;\; x = \frac{7\pi}{4} \\ \displaystyle 0 & \text{if}\;\; \frac{7\pi}{4} \lt x \lt 2\pi \\ \end{cases} $$ and $$ cs(x) = ss\left(x+\frac{\pi}{2}\right) $$

$cs(x)$ and $ss(x)$ can be understood as cosine and sine functions approximated to $-1$, $0$ and $1$, with value $\pm\frac{1}{\sqrt{2}}$ in the discontinuities to preserve the identity $cs(x)^2+ss(x)^2 = 1$.

One way to rewrite ss and cs as series continuous functions is to use their Fourier series. A straightforward way of calculating Fourier series of a piecewise function leads to

$$ cs(x) = \sum _{n=1}^{\infty }\frac {2}{\pi n}(sin(\pi n/4)+sin(3\pi n/4))\cos(nx) $$ $$ ss(x) = \sum _{n=1}^{\infty }\frac {2}{\pi n}(cos(\pi n/4)-cos(3\pi n/4))\sin(nx) $$

which makes the integral transform to take the form

$$ Fs(\omega)=\int\limits_{-\infty}^\infty f(t) \sum _{n=1}^{\infty }\frac{2}{\pi n}\left(\left(sin \left(\frac{\pi n}{4}\right)+sin \left(\frac{3 \pi n}{4}\right)\right) \cos(n \omega t)+ i \cdot \left(cos\left(\frac{\pi n}{4}\right)-cos\left(\frac{3\pi n}{4}\right)\right)\sin(n \omega t)\right)\mathrm dt $$

Maybe this form is easier to work with.

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