$\def\Si{\operatorname{Si}}$
The converse Ramanujan master theorem finds Taylor series coefficients. The goal of our question is applying it on $h(x)=\cases{f^{-1}(x),x\ge1\\ 0,0\le x<1}$:
where $f(x)=x\csc(x)$. The Mellin transform of $h(x)$ is:
$$\operatorname M_s(h(t))=\int_0^\infty t^{s-1}h(t)dt=\int_1^\infty t^{s-1}h(t)dt=\int_0^\pi t f(t)^{s-1}df(t)$$
Integrating by parts, letting $s=n\in\Bbb N$ using Approximating function of the integral $\dfrac{\sin(x)^k}{x^k}$ gives the $\Si(x)$ function $$\operatorname M_n(h(t))=-\frac1n\int_0^\pi\left(\frac{\sin(x)}x\right)^{-n}dx= \frac{1}{(-n-1)!}\sum\limits_{k=0}^{\left\lfloor \frac{-n-1}2\right\rfloor}\frac{(-1)^{k+1}}n{\binom {-n} k}\left(\frac{-n}{2}-k\right)^{-n-1}\text{Si}((-n-2k)\pi)=\Gamma(n)a_{-n}$$
so possibly:
$$h(t)=\sum_{n=0}^\infty\frac{a_n(-x)^n}{n!},a_n= \frac{1}{\color{red}{\Gamma(-n)}(n-1)!}\sum\limits_{k=0}^ {\left\lfloor \frac{n-1}2\right\rfloor}(-1)^k{\binom n k}\left(\frac{n}{2}-k\right)^{n-1}\text{Si}((n-2k)\pi) \tag 1$$
Unfortunately, $\Gamma(-n)$ complicates convergence. However, as @Tom Copeland mentioned here $\Gamma(-n)$ in $\displaystyle f(x)=\sum_{n=0}^\infty \int_0^\infty t^{-n-1} f(t)dt\frac{(-x)^n}{n!\Gamma(-n)}$ making the sum terms $0$:
Roughly you are neglecting that $1/t$ tends to infinity as $t$ tends to $0$. Review the use of delta functions and the Gelfand-Shilov rep. Other more elementary intros to generalized functions and regularization of integrals are useful (Kanwal maybe)
there is a possibility of fixing $(1)$’s convergence and removing the $\Gamma(-n)$ by regularization/use of generalized functions, but how?
