I am looking for a general formula for the following linear operator $T_n[f]$.
$$T_0[f]=1$$
$$T_1[f]=\pi f(x)-\frac12$$
$$T_2[f]=2i\pi f'(x)-\pi f(x) +\frac16$$
etc.
In general, $T_n[f]$ is equal to $B_{n+1}(x)$ (Bernoulli polynomials) where each $x^k$ is replaced with $k i^{k-1}\pi f^{(k-1)}(x)$.
Since $\mathcal{L}^{-1}\left(f^{(k)}(s)\right)=(-1)^k x^k (\mathcal{L}^{-1}f)(x)$ we have
$$ \mathcal{L}^{-1} T_n[f] = \sum k i^{k-1}\pi (-1)^{k-1} x^{k-1}(\mathcal{L}^{-1}f)(x)$$ and $$ T_n[f]=\pi\mathcal{L}\left[B_{n+1}'(-ix)\cdot(\mathcal{L}^{-1} f)(x)\right]. $$