# Invertibility of integral transform of discontinuous function

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.