Expressing a transformation of series in terms of integral 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.
 A: 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} 
A: Alternately, you can apply an integral for the Hadamard product of two generating functions: namely, $F(z)= \sum_{n \geq 0} f_n z^n$ and the function
$$C(z) := \sum_{n \geq 0} \binom{2n}{n}^{-1} z^n = \frac{4\left(\sqrt{4-z} + \sqrt{z} \sin^{-1}(\sqrt{z}/2)\right)}{(4-z)^{3/2}}. $$
Then the desired integral formula is of the form 
$$\begin{align}(F \circ C)(z) = \sum_{n \geq 0} \binom{2n}{n}^{-1} f_n z^n& = \frac{1}{2\pi} \int_0^{\pi} F(z e^{\imath t}) C(e^{-\imath t}) dt \\ & = \frac{1}{2\pi} \int_0^{\pi} F(e^{\imath t}) C(z e^{-\imath t}) dt \\ & = \frac{1}{2\pi} \int_0^{\pi} F(\sqrt{z} e^{\imath t}) C(\sqrt{z} e^{-\imath t}) dt.\end{align}$$ 
I suppose it depends on context whether any of these integral formulas are easier to evaluate than the beta function integral in the previous post.
