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Anyone have any ideas of how to progress with this sum? $$ \sum_{j = 1}^{\infty} \dfrac{f^{(j)}(a)}{j!} \left(\zeta(-(j-1)e^{i \pi t}) - \dfrac{1}{(j-1)e^{i \pi t} + 1} \right).$$ $f(x) \in O(x)$ as $x \to \infty$, and it is analytic, $t \in (0,1)$.

If anyone is wondering how I arrived at this series, I used the Euler-Maclaurin expansion for an integral involving f and the Bernoulli numbers gave me the zeta function.

I was thinking if I could express this properly as some sort of Mellin transform, maybe I could use something like Ramujan's master theorem so that I could solve it for any such $f$.

Would appreciate any feedback or ideas.

Gary
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BBadman
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  • The context isn't enough, or at least for me. What exactly led you to this sum? What function did you substitute in the Euler-Maclaurin formula? – Kamal Saleh Dec 15 '23 at 01:42
  • Sorry for the late reply. This is the original form for the Euler-Maclaurin summation. $\displaystyle{\sum_{n=1}^{\infty}} \frac{B^{+}{n}}{n!} \left( \frac{d^{n-1}}{ds^{n-1}} (\frac{f( ({\frac{s}{k})^{e^{i \pi t}}}+ a) - f(a)}{(\frac{s}{k})^{e^{i \pi t}}})\rvert{s = k} - \frac{d^{n-1}}{ds^{n-1}} (\frac{f( ({\frac{s}{k})^{e^{i \pi t}}}+ a) - f(a)}{(\frac{s}{k})^{e^{i \pi t}}})\rvert_{s = 0} \right), $ – BBadman Dec 28 '23 at 01:16
  • This seems like a random combination of functions, but apparently, the function getting differentiated looks reminds me of the definition of the derivative. Maybe that helps with solving the second derivative. – Kamal Saleh Dec 28 '23 at 03:35
  • Well without the derivation I definitely understand how it looks random haha. The reason it's reminding you of the definition of the derivative is because it is the difference quotient of the given analytic function. – BBadman Dec 28 '23 at 03:40

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