When one knows the series expansion of a function, $$ f(z)=\sum_n f_n z^n,$$ is there a way to express the following function $$ F(z)=\sum_n {(n!)^2\over (2n)!} f_n z^n $$ in some simple way, like in terms of an integral transformation?
One obvious way is to find some function $\mu(\zeta)$ satisfying $$\int_a^b d\zeta\, \mu(\zeta)\, \zeta^n = {(n!)^2\over (2n)!}$$ so that $$F(z)=\int_a^b d\zeta\, \mu(\zeta)\, f(\zeta z)$$ but I have not been able to find such a function.
This can be obtained using an integral representation for the Beta function. Defining: \begin{align} \phi(z)&=\frac{d}{dz}\left[zf\left( z^2 \right)\right]\\ &=f(z^2)+2z^2f'(z^2)\\ &=\sum_n f_n \left( 2n+1 \right)z^{2n} \end{align} one may write \begin{equation} \phi(\sqrt{z})=\sum_n f_n \left( 2n+1 \right)z^n \end{equation} Now, \begin{align} \int_0^1\phi\left( \sqrt{zt\left( 1-t \right)} \right)\,dt&=\sum_n f_n \left( 2n+1 \right)z^n\int_0^1t^n\left( 1-t \right)^n\,dt\\ &=\sum_n f_n \left( 2n+1 \right)z^nB\left( n+1,n+1 \right)\\ &=\sum_n \frac{\left( n! \right)^2}{\left( 2n! \right)}f_n z^n \end{align} Finally, \begin{equation} \int_0^1\left[f\left( zt\left( 1-t \right) \right)+2 zt\left( 1-t \right) f'\left( zt\left( 1-t \right) \right)\right]\,dt=\sum_n \frac{\left( n! \right)^2}{\left( 2n! \right)}f_nz^n \end{equation}