Let $f$ be a smooth function and $g$ integrable. Denote the $n$-th derivade of $f$ by $f^{(n)}$ and the $n$-th integral of $g$ by $g^{(-n)}$.
Integration by parts stands
$$\int fg \ = \ f \int g – \int \left(f^{(1)}\int g\right)= \ \boxed{f^{(0)}g^{(-1)} – \int f^{(1)}g^{(-1)}}.$$
Recently I thought: recursively applying the integration by parts formula to the last sumand at the right hand side of itself yields
$$\int f(x)g(x)dx = \sum\limits_{n=0}^{\infty} (-1)^n f^{(n)}(x) g^{(-(n+1))}(x) +C,$$
which might be a conditionally convergent series and must be computed in that order. Notice the equality "$=$" is almost everywhere.
Is this formula true for all such functions $f$ and $g$? Do you know any source which deals with this idea?
Best Answer
First of all, your $g^{(-n)}$ is defined only up to a polynomial of degree $n-1$, so it's not quite correct to talk about the $n$-th integral. But, to make a definite choice, one may take definite integrals, say $$g^{(0)}(x):=g(x),\quad g^{(-n-1)}(x):=\int_{0}^{x}g^{(-n)}(y)\,dy,$$ and see what happens. We have in this case $\color{gray}{\text{[induction on $0\leqslant k\leqslant n$]}}$ $$g^{(-n-1)}(x)\color{gray}{\left[=\frac{1}{k!}\int_{0}^{x}(x-y)^k g^{(-n-1+k)}(y)\,dy\right]}=\frac{1}{n!}\int_{0}^{x}(x-y)^n g(y)\,dy$$ by the same iterated integration by parts, so your sum becomes $$\int_{0}^{x}g(y)\sum_{n=0}^{\infty}\frac{f^{(n)}(x)}{n!}(y-x)^n\,dy$$ (assuming we can switch the integration and the summation for some reason). Do you recognize it? (At least it surely holds for (real) analytic functions.)