Name of analytic function identity Doing partial integration on $\int 1\times f(x)dx$, one gets
$$\int f(x)dx=xf(x)-\int xf'(x)dx+C$$
$$=xf(x)-\frac{x^2}2 f'(x)+\int \frac{x^2}2 f''(x)+C$$
$$=...=C+x\sum_{n=0}^\infty \frac{(-x)^n}{(n+1)!} f^{(n)}(x)$$
Replacing $f(x)$ by $f'(x)$ and bringing the right hand side to the left, this becomes
$$\sum_{n=0}^\infty \frac{(-x)^n}{n!}f^{(n)}(x)=C$$ 
My questions:
A) Is this correct?
B) This identity was not hard to show and looks pretty nice, so I'm sure it has been found by other people, but I could not find any reference to it. Has it appeared anywhere before? If yes, what is its name?
 A: If $f$ is analytic, we can write
$$f(x)=\sum_{n=0}^\infty \frac{f^{(n)}(0)x^n}{n!}$$
Differentiating $m$  times reveals
$$f^{(m)}(x)=\sum_{n=0}^\infty \frac{f^{(n+m)}(0)x^{n}}{n!}\tag2$$
Using $(2)$, we find that $\int f(x)\,dx$ can be expressed 
$$\begin{align}
\int f(x)\,dx&=\sum_{m=0}^\infty \frac{(-1)^m f^{(m)}(x)x^{m+1}}{(m+1)!}\\\\
&=\sum_{m=0}^\infty \frac{(-1)^m \sum_{n=0}^\infty \frac{f^{(n+m)}(0)x^{n}}{n!}x^{m+1}}{(m+1)!}\\\\
&=\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{(-1)^mf^{(n+m)}(0)}{(m+1)!n!}x^{n+m+1}\\\\
&=\sum_{m=0}^\infty\sum_{p=m}^\infty \frac{(-1)^mf^{(p)}(0)}{(m+1)!(p-m)!}x^{p+1}\\\\
&=\sum_{p=0}^\infty \frac{f^{(p)}(0)x^{p+1}}{(p+1)!}\underbrace{\sum_{m=0}^p (-1)^m\binom{p+1}{m+1}}_{=1}\\\\
&=\sum_{p=0}^\infty \frac{f^{(p)}(0)x^{p+1}}{(p+1)!}\tag3
\end{align}$$
where $(3)$ is the identical to the result obtained upon integrating $(1)$ term by term.  
Hence, the result $\int f(x)\,dx=\sum_{m=0}^\infty \frac{(-1)^m f^{(m)}(x)x^{m+1}}{(m+1)!}$ presented in the OP is merely a different representation of $\int f(x)\,dx $ from the representation given on the right-hand side of $(3)$.  We have shown that those representations are, as expected, equal.
