Interpreting $\sum_a^b$ when $a$ and $b$ are complex numbers

Question

I'm doing a problem that involves

$$\sum_{k=ad}^{bd}d^{-1}f(kd^{-1})$$

where $a$ and $b$ are real numbers. You might recognise this as a special case of the numerical integration formula, which is related to the problem. However, the main issue is that I need to define what happens when $a, b \in \Bbb{C}$.

Google won't help me, and I can't find similar questions.

So, how would I compute $\sum_a^b\{a, b \in C\}$?

I don't mean abstractly, as most other people on this forum (who are far smarter than me) would ask, I mean literally for two values $a$ and $b$.

Thank you for your time!

Answer 1

Sometimes $\sum_{k=a}^b f(k)$, where $a\le b$ integers, has a closed form which obviously extends to complex numbers. Use that closed form. But in most cases, there is no such thing.

Example $$ \sum_{k=a}^b k^2 = \frac{2b^3+3b^2+b-2a^3+3a^2-a}{6} $$

Answer 2

For sufficiently nice functions $f$, you can use the Euler-Maclaurin formula:

$$ \sum_{a<k\le b}f(k)=\int_a^bf(x)\,\mathrm{d}x+\sum_{k=0}^\infty \frac{B_{k+1}}{(k+1)!} \Big(f^{(k)}(b)-f^{(k)}(a)\Big). $$

Here, $B_s$ are the Bernoulli numbers with $B_1=+\frac{1}{2}$ and $B_s=0$ for all odd $s>1$. How you want to finagle the $a<k\le b$ condition to plug complex numbers in I'll leave up to you.

"Nice" means not too oscillatory: $\int_a^b|f^{(k)}(x)|\,\mathrm{d}x=o\big((2\pi)^k\big)$ for all $a,b$ should do the trick.