We can show this because $$\int_{\rho_1}^{\rho_k}\left(\sum_{n=1}^x n\right)dx=-\frac{1}{12}$$ where $\rho_n$ is the $n$th root of the integrand out of $k$ roots.

In fact $$\int_{\rho_1}^{\rho_k}\left(\sum_{n=1}^x n^s\right)dx=\zeta(-s)$$ $s\gt 0$

To find zeta for positive $s$ we can use the functional equation.

I imagined that if we could derive other functional equations then we could use this same method to continue other functions across all real values (as you can see I'm not even talking about complex values because I don't known enough about complex analysis to be sure I'd be accurate).

So we first choose a function $$g(x)=\sum_{n=1}^x f(-s)$$ Integrate $g(x)$, between its smallest and largest roots, then plug that into our (theoretical) functional equation to find the continuation however our functional equation defines it.

I originally wondered if we could do this with $\sum_{n=1}^\infty n^s(n+c)^s$ finding a functional equation that would yield $\sum_{n=1}^\infty\frac{1}{n^s(n+c)^s}$ for some $c$ using the method described. But I bet there are many other interesting functions we could do it with.

Would this work for this function above or for any other specific functions? If so or if not, how come?

  • $\begingroup$ What means "where $\rho_n$ is the $n$th root of the integrand out of $k$ roots" ? And for a fixed $s$, as $N \to \infty$ : $\sum_{n=1}^N n^{-s} = C_0+\sum_{k =0}^K c_k N^{1-s-k}+ O(N^{-s-K})$ and $\zeta(s)$ is the constant term $C_0$ $\endgroup$ – reuns Mar 13 '17 at 11:49
  • $\begingroup$ In the case of $\zeta(s)$, you want to look at $g_N(s) = \sum_{n=1}^N f_n(s)$ where $f_n(s) = n^{-s}$. Note that $f_n(s),g_N(s)$ are analytic on the whole complex plane, while $\zeta(s)= g_\infty(s)$ is not. The functional equation is true only for $\zeta(s)$, for $g_N(s)$ we only have an approximate functional equation $\endgroup$ – reuns Mar 13 '17 at 12:05

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