During my research, I encountered this expression: $$ \sum_{p=0}^{\infty} \frac{1}{(p+1)!} \sum_{k = 0}^{p} k! \, u^k \, B_{p,k}(f_1, \dots, f_{p-k+1}) $$ with the arguments of $B_{p,k}$ being the coefficients of a power series: $$ f(x) = \sum_{i=0}^{\infty} \frac{x^i}{i!}f_i$$ I am aware that this looks similar to the generating function of the Bell polynomials at $x=1$, $$ \exp (u\, (f(x)-f_0) = \sum_{p=0}^{\infty} \frac{x^p}{p!} \sum_{k = 0}^{p} u^k \, B_{p,k}(f_1, \dots, f_{p-k+1})$$ but the additional factor of $k!/(p+1)$ in the second sum throws me off.
I am looking for a simplification of the original expression, probably by use of the generating function. If the solution is actually trivial and I oversaw something, please give me a hint.
I found a way to insert a factor of $k!$. Notice that $u$ can be any number, so I could also scale it by $y$ and write: $$ e^{y\,u\, (f(x)-f_0)} = \sum_{p=0}^{\infty} \frac{x^p}{p!} \sum_{k = 0}^{p} y^k \, u^k \, B_{p,k}(f_1, \dots, f_{p-k+1})$$ After that, multiply both sides by $e^{-y}$ and integrate $y$ with limits $[0,\infty)$. $$ \int_0^{\infty} dy \, e^{y [u\, (f(x)-f_0) - 1]} = \sum_{p=0}^{\infty} \frac{x^p}{p!} \sum_{k = 0}^{p} \int_0^{\infty} dy \,y^k e^{-y} \, u^k \, B_{p,k}(f_1, \dots, f_{p-k+1})$$ On the right-hand side, we find the definition of the gamma function, $\Gamma (k+1) = k!$, on the left-hand side, we can evaluate the integral if $1 > u[f(x) - f_0]$:
$$ \frac{1}{1 - u\, (f(x)-f_0)} = \sum_{p=0}^{\infty} \frac{x^p}{p!} \sum_{k = 0}^{p} k! \, u^k \, B_{p,k}(f_1, \dots, f_{p-k+1})$$
Finally, perform an integration in $x$ from 0 to 1:
$$ \int_0^1 \frac{dx}{1 - u\, (f(x)-f_0)} = \sum_{p=0}^{\infty} \frac{1}{(p+1)!} \sum_{k = 0}^{p} k! \, u^k \, B_{p,k}(f_1, \dots, f_{p-k+1}) \qquad \text{if} \,\, 1 > u[f(x) - f_0]$$