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Given appropriate constraints, and a continuously differentiable real-valued function $\phi (x)$, the Abel Summation Formula (Wikipedia article here) can be written as

$$\sum _{k=1}^x \phi (k) = \lfloor x\rfloor \phi (x)-\int_1^x \lfloor u\rfloor \phi '(u) \, du$$

Is it possible to generate a similar ASF for the sum $\sum _{i=1}^x \phi (x-i)$? It may be a very simple question (or it may not!), but I keep getting confused by the mix of variables and dummy variables, and defining the sum's limit. As a result, I'm not getting anywhere.

Richard Burke
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1 Answers1

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Note that $$ \sum_{i=1}^x \phi(x-i) = \phi(x-1) + \phi(x-2) + \cdots + \phi(1) + \phi(0) = \sum_{i=0}^{x-1} \phi(i) $$ for which it is easy to derive a suitable formula (or just reference the article to find), $$ \sum_{i=0}^{x-1} \phi(i) = \lfloor x \rfloor \phi(x-1) - \int_0^{x-1} \lfloor u+1 \rfloor \phi'(u) \, \mathrm{d}u $$

PrincessEev
  • 50,606