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Here is a graph of the function $y=-1/x$:

enter image description here

If we add infinitely many similar functions with a shift of $\pi/2$ each in both directions, we get $\tan x$. But if we do the same only in one direction, we get "incomplete tangent":

http://storage7.static.itmages.ru/i/14/0910/h_1410326921_7988832_91f3fd7d7d.png

The yellow one is $\operatorname{pg}(x)=\frac 1\pi \psi (\frac x\pi)$, the blue one is $\operatorname{cpg}(x)=-\frac 1\pi \psi (1-\frac x\pi)$. They obey $\operatorname{cpg}(x)+\operatorname{pg}(x)=-\cot(x)$.

Now if we differentiate cpg(x) we get:

$$(\operatorname{cpg}(x))^{(s-1)}=\pi^{-s}\Gamma(s)\zeta(s,1-\frac x\pi)$$

At $x=0$ it would be

$$\operatorname{cpg}^{(s-1)}(0)=\pi^{-s}\Gamma(s)\zeta(s)$$

Compare it with this formula involving Riemann Xi-function:

$$\xi(2s) = \pi^{-s}\Gamma\left(s\right)\zeta(2s)$$

I wonder is the similarity between these two formulas just a coincidence or Riemann Xi function serves as a consecutive derivative (generating function) of some simple or notable function?

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  • $\begingroup$ I don't see what $\tan(x)$ or these other two functions have to do with $1/x$. $\endgroup$ – alex.jordan Nov 19 '14 at 6:42
  • $\begingroup$ @alex.jordan: There is a formula expressing the $($co$)$tangent function in terms of digamma functions; there is another formula, expressing digamma functions in terms of harmonic numbers; and then there's yet another formula, expressing harmonic numbers in terms of $\dfrac1x$. This graphic seems to be a geometric and very visually intuitive embodiment of the afore-mentioned algebraic formulas. $\endgroup$ – Lucian Nov 19 '14 at 7:18
  • $\begingroup$ I prefer considering such identities in complex. We have some meromorphic function. If we can reproduce the poles such that the difference is bounded (or at least it does not increase too rapidly) then we can apply Liouville. For example the function $\cot z=\frac{\sin'}{\sin}(x)$ has simple poles at $k\pi$ with residue $1$. The sum $f(z)=\frac1z+\sum_{k=1}^\infty\left(\frac1{z+k\pi}+\frac1{z-k\pi}\right)$ has the same property. It can be verified that difference is bounded and both functions are odd, so they are equal. You can try the same for $\frac1{\sin z}$ or $\frac{\Gamma'}{\Gamma}$. $\endgroup$ – user141614 Nov 19 '14 at 7:47
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    $\begingroup$ @Lucian I just do not understand what "If we add infinitely many similar functions with a shift of pi/2 each in both directions, we get tanx." means. Do you? $\endgroup$ – alex.jordan Nov 19 '14 at 8:24
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    $\begingroup$ @Lucian I get that; what I don't get is what OP means by "if we do the same only in one direction". $\sum_{n\geq n_0}\frac{1}{x+n}$ is divergent. So is $\sum_{n\leq n_0}\frac{1}{x+n}$. $\endgroup$ – alex.jordan Nov 19 '14 at 13:58

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