I was playing with some derivatives when I figured out a possible way to express an antiderivative as an infinite sum, but I dont know if it is valid, or if it already exists. First accept that $$\int{u \ dv} = u\cdot v - \int v \ du$$
Then, we can say for any function $f(x)$ such that $f : \mathbb{R} \to \mathbb{R}$ that $f = 1\cdot f$. Asuming that $f$ is differentiable n times, then
$$\int f \ dx = \int 1 \cdot f \ dx \\ dv = 1 \\ v = \int 1 \ dx= x \\ u = f \\ du = \frac{df}{dx} \\ \int f \ dx = x\cdot f - \int x\cdot \frac{df}{dx} \ dx \\ $$ Applying the same process to the integral in the right $$\int f \ dx = x\cdot f - \frac{x^2}{2}\cdot \frac{df}{dx} + \frac{x^3}{6}\cdot \frac{d^2 f}{dx^2} + ... \ = \ \sum_{n=0}^{\infty}\frac{(-1)^n \cdot x^{n+1}}{(n+1)!} \cdot \frac{d^n f}{dx^n} \\ \\ $$ In conclusion $$\int f \ dx = \sum_{n=0}^{\infty}\frac{(-1)^n \cdot x^{n+1}}{(n+1)!} \cdot \frac{d^n f}{dx^n}$$
I dont know whether any of this is right or I have made some sort of mistake. If I havent, It probably already exists, so I would benefit from knowing the name of it. I have also thaught that it could be possible to use it to expand some taylor series of some functions, for example, let $f$ be a function which we do know it's antiderivative, by expanding $\int f$, you could get an infinite series which equates to the antiderivative of $f$. I couldn't manage to make this work however. I have also thaught it could be used as an approximation to definite integrals.
Edit: I read this question and that thing in the first term actually isn't a taylor series (or maybe it is). The rest should be true, though.
There's a mistake in the derivation of your series. I don't think you actually proved that $$\int{f \,dx} = \lim_{N\to\infty}\sum_{n=0}^{N}{\frac{(-1)^n x^{n+1}}{(n+1)!} \cdot \frac{d^nf}{dx^n}}$$ You actually proved that for all N $$\int{f \,dx} = \sum_{n=0}^{N}{\frac{(-1)^n x^{n+1}}{(n+1)!} \cdot \frac{d^nf}{dx^n}} + \int{\frac{x^{N+1}}{(N+1)!} \frac{d^{N+1}f}{dx^{N+1}}}$$ So if $$\lim_{N\to\infty}\int{\frac{x^{N+1}}{(N+1)!} \frac{d^{N+1}f}{dx^{N+1}}} = 0$$
Then, you can use the second equation to prove the first is true. The second equation is essentially an error bound here, and you're showing that the integral of a taylor series converges to the integral of the function if the third equation holds.
Essentially, the part on the end that's the result of integration by parts doesn't just go away if you make the series infinite; it's still there. Unless it goes to zero, you have to add it to the sum to get the answer. There's basically three outcomes here:
$\lim_{N\to\infty}\int{\frac{x^{N+1}}{(N+1)!} \frac{d^{N+1}f}{dx^{N+1}}} = 0$, and you have a taylor series for the antiderivative.
$\lim_{N\to\infty}\int{\frac{x^{N+1}}{(N+1)!} \frac{d^{N+1}f}{dx^{N+1}}} = $ <some finite polynomial>, and you can add that in to get a series for the antiderivative.
$\lim_{N\to\infty}\int{\frac{x^{N+1}}{(N+1)!} \frac{d^{N+1}f}{dx^{N+1}}}$ equals some other function or does not exist, and it's not very helpful.