Proof That $\sum_{k=1}^{n-1}\int_{k}^{k+1}\left\{x\right\}f´(x)dx=\int_{1}^{n}\left\{x\right\}f´(x)dx$

Question

I am Reading the following notes: Ramanujan summation of divergent series by B Candelpergher (https://hal.univ-cotedazur.fr/hal-01150208v2/document). There, the author derives the Euler-MacLaurin summation formula in the beginning of the first chapter. In one of the steps (on page 18 of the file) he states that $$\sum_{k=1}^{n-1}\int_{k}^{k+1}[x]f´(x)dx=\int_{1}^{n}xf´(x)dx+\int_{1}^{n}\left\{x\right\}f´(x)dx$$ $\left\{x\right\}$ is the fractional part of $x$.

The first integral on the right side I understand, but the second one involving the fractional part of $x$, I can´t really see how it´s true. In other words, he is saying that

$$\sum_{k=1}^{n-1}\int_{k}^{k+1}\left\{x\right\}f´(x)dx=\int_{1}^{n}\left\{x\right\}f´(x)dx$$ Can someone kindly show the proof or the intuition behind it? Thank you in advance.

Answer 1

Ok, I got it!

I took the reverse way to prove it. $$\int_{1}^{n}\left\{x\right\}f´(x)dx= \int_{1}^{n}(x-\left[x\right])f´(x)dx$$ $$ =\int_{1}^{n}x f´(x)dx -\int_{1}^{n} \left[x\right]f´(x)dx$$ $$ =\int_{1}^{n}x f´(x)dx -\sum_{k=1}^{n-1}\int_{k}^{k+1} \left[x\right]f´(x)dx$$ $$=\sum_{k=1}^{n-1} \int_{k}^{k+1}x f´(x)dx -\sum_{k=1}^{n-1}\int_{k}^{k+1} \left[x\right]f´(x)dx$$ $$=\sum_{k=1}^{n-1} \int_{k}^{k+1}(x - \left[x\right])f´(x)dx$$ $$=\sum_{k=1}^{n-1} \int_{k}^{k+1}\left\{x\right\}f´(x)dx$$

Answer 2

Just write out the sum: \begin{align*} \sum_{k=1}^{n-1} \int_k^{k+1} \text{whatever}\, dx &= \int_1^2 \text{whatever}\, dx + \int_2^3 \text{whatever}\, dx + \cdots + \int_{n-1}^n \text{whatever}\, dx \\ &= \int_1^n \text{whatever}\, dx. \end{align*}