# express series exactly as sum of integrals (geometric insight?)

Suppose for a series there exists a function such that $a_n = f(n)$, then even for a non-monotonically decreasing function:

$$A. \sum_{i=n_1+1}^{n_2}f(n)=\int_{n_1}^{n_2}f(x)dx\ +\int_{n_1}^{n_2}(x-\lfloor x \rfloor )f'(x)dx$$

I found this in Arfken's Mathematical Methods for Physicists. The proof given is rather simple and follows from evaluating the right most integral by distributing $\ f'(x)$ and then using the product rule on each resultant expression.

However, it seems to me that the quantity $(x-\lfloor x \rfloor) f'(x)$ has a simple geometric interpretation: multiply the distance from the last integer by the slope at x. In fact it seems one could express $$\int_{n_1}^{n_2}f(\lfloor x \rfloor)dx = \int_{n_1}^{n_2}f(x)dx\ + \int_{n_1}^{n_2}f(\lfloor x \rfloor)-f(x)dx=\int_{n_1}^{n_2}f(x)dx\ +\int_{n_1}^{n_2}(\lfloor x \rfloor - x)f'(c)dx$$

By using the mean value theorem for derivatives. But of course the problem is that $\lfloor x \rfloor \leq c \leq x$. The image below may be useful.

(a) comparison of sum-blocks leading the integral. (b) sum-blocks lagging integral

1) Is there any geometric reasoning that would force $c=x$? Or, is there a simple geometric insight that allows $(x-\lfloor x \rfloor) f'(x)$ to account for the difference between $f(x)$ and $f(\lfloor x \rfloor)$

2) Is there a "name" for equation $A$?

I apologize if this is an overly OCD question.

• Note that you have equality after integrating, not pointwise. That is, it's not true that $$f(x) - f(\Floor{x}) = (x - \Floor{x}) f'(x),$$ only that $$\int_{n}^{n+1} (x - \Floor{x}) f'(x)\, dx = f(n+1) - \int_{n}^{n+1} f(x)\, dx,$$ and $f(x) - f(\Floor{x}) = (x - \Floor{x}) f'(c)$ for some $c$ with $n < c < x$. – Andrew D. Hwang Jul 22 '17 at 12:38
• Precisely, it is interesting that even though $f(\lfloor x \rfloor) - f(x) \neq (\lfloor x \rfloor - x)f'(x)$, that somehow the area difference b/w $f(\lfloor x \rfloor) - f(x)$ is recovered by integrating $(\lfloor x \rfloor - x)f'(x)$. Because if you look at each value of $x$, trace out the tangent line at $x$ a distance $(\lfloor x \rfloor - x)$ then the resultant $\Delta f$ doesn't seem account for $f(\lfloor x \rfloor) - f(x)$ in any obvious way (to me at least), yet somehow summing all these $\Delta f$s does recover the area correctly. – Darren McAffee Jul 22 '17 at 15:15

While this is not a complete answer, for me the easiest way to visualize any geometric aspect of $(\lfloor x \rfloor - x)f'(x)$ is to shift it to the unit interval and subtracting off $f(k+1)$ (for whatever integer of interest) so that $f(1)=0$.
Then with any curve going from $(0, f(k))$ to $(1, 0)$ the expression $(\lfloor x \rfloor - x)f'(x) = xf'(x)$ and $xf'(x)$ is just the distance from $f(x)$ to the y-intercept of the tangent line at $(x, f(x))$.
This then makes it the case that integrating these distances to the y-intercepts of the tangent lines of a function $f(x)$ is equal in amount to the integral of that function. I.e. $\int_0^1xf'(x)=-\int_0^1f(x)dx$ (in this nicely shifted case since f(1)=0).
And this feels very much like a Legendre transformation, that is, somehow all the information of a function can be contained in knowing the y-intercepts as a function of slope (or in this case the distances of $f(x)$ to those y-intercepts).