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?

  • $\begingroup$ You didn't define your remainder, how do you know it $\to 0$ as $n \to \infty$. See a proof of Taylor sum with remainder, implying that if the remainder $\to 0$ then the function is analytic. $\endgroup$
    – reuns
    Oct 25, 2019 at 18:38

1 Answer 1


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.

  • $\begingroup$ I had a proof similar to this, using the power series directly instead of via Taylor expansion, but this one's a bit better, I think. Anyway, A) definitely answered. $\endgroup$
    – Cecilia
    Nov 8, 2019 at 20:55

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