# Show that Riemann sum does not depend on rectangle endpoint choice

To calculate a Riemann sum by assuming right endpoints, we have $$S= \lim_{n\to\infty} { \sum_{i=1}^{n} f(x_i^*)\Delta x } = \lim_{n\to\infty} { \sum_{i=1}^{n} f\left( i\Delta x+ a\right)\Delta x .}$$

Then, to calculate $$S$$ using any endpoint, I introduce a parameter $$\alpha, \quad 0\leq\alpha\leq1$$ such that $$S = \lim_{n\to\infty} { \sum_{i=1}^{n} f\left( (i-\alpha)\Delta x+ a\right)\Delta x }.$$ For $$\alpha = 0$$, we have right endpoints, for $$\alpha = 1$$ we use left endpoints, for $$\alpha = \tfrac{1}{2}$$ we use mid point, and so on.

How could I prove that $$S$$ does not depend on $$\alpha$$ for any integrable $$f$$ analytically? I have done so numerically, but am struggling with the analytical method. Since I am not working with a particular function, I can't use the classic textbook formulas for the series for $$\{i^n\}$$. I would be inclined to try to show that partial derivative wrt $$\alpha$$ is 0, but I'm not sure under which conditions I can pass a derivative operator through an infinite sum.

It only supposes that the function $$f$$ is Riemann integrable. Easier proves can be done with further assumptions of $$f$$ like you devise, i.e. if $$f$$ is supposed to be continuously differentiable.