Notes to myself

For others, note ${\displaystyle \tau }$ is one positive full turn in radians and ${\displaystyle \Pi (z)}$ denotes the repeated product ${\displaystyle z}$, the Gauss Pi function. So keep in mind

${\displaystyle e^{ix}=i^{{\frac {4}{\tau }}x}=\left(-1\right)^{{\frac {2}{\tau }}x}=\cos \left(x\right)+i\sin \left(x\right)}$

Reference used (pretty similar, but not the same): https://math.stackexchange.com/a/3751840

We'll start off with extending ${\displaystyle \sum _{k=1}^{x}{\frac {1}{k^{p}}}}$, the generalised harmonic number, into a function. First, we need to recognize two things:

First, by a change of variables or the Laplace transform, we have that:

${\displaystyle {\frac {1}{x^{p+1}}}={\frac {1}{\Pi \left(p\right)}}\int _{0}^{\infty }t^{p}e^{-tx}dt,\quad {\text{for }}x>0{\text{ and }}\Re (p)>-1}$

Second, that ${\displaystyle \sum _{k=1}^{x}e^{-kt}}$ can be extended by using ${\displaystyle \nabla ^{-1}}$. We recognize the geometric series and write:

${\displaystyle \nabla ^{-1}e^{-xt}={\frac {e^{-t}-e^{-\left(x+1\right)t}}{1-e^{-t}}}}$

Now we're ready for the manipulation:

${\displaystyle {\frac {1}{x^{p}}}={\frac {1}{\Pi \left(p-1\right)}}\int _{0}^{\infty }t^{p-1}e^{-tx}dt}$

take the antidifference of both sides:

${\displaystyle \nabla ^{-1}{\frac {1}{x^{p}}}={\frac {1}{\Pi \left(p-1\right)}}\int _{0}^{\infty }t^{p-1}{\frac {e^{-t}-e^{-\left(x+1\right)t}}{1-e^{-t}}}dt}$

Our extension of the generalized harmonic number here is valid for ${\displaystyle \Re (p)>0}$ due to the shift by one unit to the right we made in the Gauss Pi function. From here, you can take the limit as ${\displaystyle x}$ approaches infinity, and obtain the integral for ${\displaystyle \zeta (p)}$. However, we can see that the limit as ${\displaystyle x}$ approaches infinity in the indefinite sum and resulting integral for ${\displaystyle \zeta (p)}$ do not converge for ${\displaystyle p<1}$. There isn't a way to logically directly extend this limit as ${\displaystyle x}$ approaches infinity other than using analytic continuation as Bernhard Riemann did in Ueber die Anzahl der Primzahlen unter einergegebenen Grösse.

We can, however, extend the indefinite sum directly. ${\displaystyle \nabla ^{-1}x^{p}}$ can be extended to a meromorphic function only using the limit of Müller's method and Newton series. In the video, we are given

${\displaystyle \lim _{x\to \infty }\Delta ^{m}f(x)=0}$

${\displaystyle S(x)=\lim _{n\to \infty }\left(\sum _{k=1}^{n-1}(f(k)-f(x+k))+\sum _{k=1}^{m}{\binom {x}{k}}\Delta ^{k-1}f(n)\right)}$

Which are known, valid representations presented by Marichal and Zenaïdi. With that, we plug in ${\displaystyle x^{p}}$:

${\displaystyle S\left(x,p\right)=\sum _{k=1}^{n-1}\left(k^{p}-\left(x+k\right)^{p}\right)+\sum _{j=1}^{p+1}\operatorname {nCr} \left(x,j\right)\sum _{l=0}^{j-1}\left(-1\right)^{j-1-l}\operatorname {nCr} \left(j-1,l\right)\left(n+l\right)^{p}}$ with ${\displaystyle n}$ going to infinity.

Then, we can take the partial derivative of ${\displaystyle \nabla ^{-1}x^{p}}$ with respect to ${\displaystyle x}$ evaluated at ${\displaystyle x=0}$ and obtain a function that extends the Bernoulli numbers and is entire. The location of the trivial zeros changes to odd integers ${\displaystyle 3}$ and greater (given by Faulhauber's formula producing polynomials which have an even multiplicity root at ${\displaystyle x=0}$ for those values of ${\displaystyle p}$ rather than the sine term in the reflexion formula), whilst the location of the nontrivial zeros and critical strip remain in the same location.

Related paper:

https://arxiv.org/abs/2207.12694

Open access book version:

https://link.springer.com/book/10.1007/978-3-030-95088-0

We will now find ${\displaystyle B(s)}$, the extended Bernoulli number function. It is defined as ${\displaystyle {\frac {\partial }{\partial x}}\left(\nabla ^{-1}x^{s}\right){\bigg |}_{x=0}}$

Similarly, we can use the Abel-Plana formula:

${\displaystyle \nabla ^{-1}f(x)=\int _{1}^{x}f\left(t\right)dt+{\frac {f\left(x\right)+f\left(1\right)}{2}}-i\int _{0}^{\infty }{\frac {\left(f\left(x+it\right)-f\left(1+it\right)\right)-\left(f\left(x-it\right)-f\left(1-it\right)\right)}{e^{\tau t}-1}}dt}$

But we run into the problem of the singularity around ${\displaystyle 0}$ for ${\displaystyle {\frac {1}{x^{p}}}}$ for positive ${\displaystyle p}$. So we can use ${\displaystyle F(x)-F(x-1)=f(x)}$:

${\displaystyle \int _{1}^{x+1}f\left(t\right)dt+{\frac {f\left(x+1\right)+f\left(1\right)}{2}}-i\int _{0}^{\infty }{\frac {\left(f\left(x+1+it\right)-f\left(1+it\right)\right)-\left(f\left(x+1-it\right)-f\left(1-it\right)\right)}{e^{\tau t}-1}}dt-f\left(x+1\right)}$

As an aside, notice that this definition of ${\displaystyle F(x)}$ passes through ${\displaystyle (0,0)}$ as this is intentional (for parity rules derived later and for alignment with summation from 1 to x):

${\displaystyle \int _{1}^{1}f\left(t\right)dt+{\frac {f\left(1\right)+f\left(1\right)}{2}}-i\int _{0}^{\infty }{\frac {\left(f\left(1+it\right)-f\left(1+it\right)\right)-\left(f\left(1-it\right)-f\left(1-it\right)\right)}{e^{\tau t}-1}}dt-f\left(1\right)=0}$

Then plug in ${\displaystyle f\left(x\right)=x^{a}}$ and take the partial derivative to get:

${\displaystyle \left(x+1\right)^{a-1}\left(x+1-{\frac {a}{2}}\right)-ia\int _{0}^{\infty }{\frac {\left(x+1+it\right)^{a-1}-\left(x+1-it\right)^{a-1}}{e^{\tau t}-1}}dt}$

Then evaluate at ${\displaystyle x=0}$, changing ${\displaystyle a}$ out for the complex variable ${\displaystyle s}$:

${\displaystyle 1-{\frac {s}{2}}-is\int _{0}^{\infty }{\frac {\left(1+it\right)^{s-1}-\left(1-it\right)^{s-1}}{e^{\tau t}-1}}dt}$

${\displaystyle \Pi \left({\frac {s}{2}}\right)\left({\frac {2}{\tau }}\right)^{\frac {s}{2}}B\left(1-s\right)=\Pi \left({\frac {1-s}{2}}\right)\left({\frac {2}{\tau }}\right)^{\frac {1-s}{2}}B\left(s\right)}$

https://www.desmos.com/calculator/ppmuwcuhaw

Reflexion formulas

${\displaystyle B\left(s\right)={\frac {\Pi \left({\frac {s}{2}}\right)}{\Pi \left({\frac {1-s}{2}}\right)}}\left({\frac {2}{\tau }}\right)^{s-{\frac {1}{2}}}B\left(1-s\right)}$

I prefer

${\displaystyle B\left(s\right)={\frac {\Pi \left({\frac {s}{2}}\right)}{\Pi \left({\frac {1-s}{2}}\right)}}\left({\frac {\tau }{2}}\right)^{{\frac {1}{2}}-s}B\left(1-s\right)}$

${\displaystyle B(s)=2\tau ^{-s}\Pi \left(s\right)\cos \left({\frac {\tau s}{4}}\right){\frac {B\left(1-s\right)}{1-s}}}$

Do consider here that ${\displaystyle \left(-i\right)^{s}+i^{s}=2\cos \left({\frac {\tau s}{4}}\right)}$.

${\displaystyle B\left(s\right)=\tau ^{-s}\Pi \left(s\right)\left(\left(-i\right)^{s}+i^{s}\right){\frac {B\left(1-s\right)}{1-s}}}$

Read

https://www.claymath.org/wp-content/uploads/2023/04/Wilkins-translation.pdf

if you ever forget how you came up with this. Everything lines up quite nicely. Also, remember the exact values of the Riemann Zeta function Euler found with is just an application of this reflexion formula.

Euler Product

${\displaystyle {\frac {\partial }{\partial x}}\left(\nabla ^{-1}x^{s}\right){\bigg |}_{x=0}={\begin{cases}-s\prod _{p}{\frac {1}{1-p^{s-1}}}&{\text{for }}s<0\\-2\tau ^{-s}\Pi \left(s\right)\cos \left({\frac {\tau s}{4}}\right)\prod _{p}^{}{\frac {1}{1-p^{-s}}}&{\text{for }}s>1\end{cases}}}$

https://arxiv.org/pdf/2009.06743 is interesting in that they have similar functional equations, though the ones in these notes were obtained through the calculus of finite differences.

They also provide the Jensen formula valid for ${\displaystyle s\in \mathbb {C} }$ and ${\displaystyle v\in \mathbb {C} }$ with ${\displaystyle \Re (v)\geq 1/2}$

${\displaystyle 2\pi \int _{-\infty }^{\infty }{\frac {\left(v-{\frac {1}{2}}+iz\right)^{s}}{\left(e^{\pi z}+e^{-\pi z}\right)^{2}}}dz}$

Also, let us define the function ${\displaystyle B(s,v)}$ where we replace ${\displaystyle v}$ in the Jensen formula with ${\displaystyle v+1}$:

${\displaystyle B(s,v)={\frac {\partial }{\partial x}}\left(\nabla ^{-1}x^{s}\right){\bigg |}_{x=v}={\frac {\tau }{2}}\int _{0}^{\infty }\left(\left(v+{\frac {1}{2}}\right)^{2}+z^{2}\right)^{\frac {s}{2}}\operatorname {sech} ^{2}\left({\frac {\tau z}{2}}\right)\cos \left(s\arctan \left({\frac {z}{v+{\frac {1}{2}}}}\right)\right)dz}$

So using ${\displaystyle F\left(x\right)={\frac {B\left(a+1,x\right)-B\left(a+1,0\right)}{a+1}}}$, we have:

${\displaystyle F(x)={\frac {\pi }{a+1}}\int _{0}^{\infty }\operatorname {sech} ^{2}\left({\frac {\tau z}{2}}\right)\left(\left(\left(x+r+{\frac {1}{2}}\right)^{2}+z^{2}\right)^{\frac {a+1}{2}}\cos \left((a+1)\arctan \left({\frac {z}{x+r+{\frac {1}{2}}}}\right)\right)-\left({\frac {1}{4}}+z^{2}\right)^{\frac {a+1}{2}}\cos \left((a+1)\arctan(2z)\right)\right)dz-\sum _{m=1}^{r}\left(x+m\right)^{a}\left\{\operatorname {real} \left(x\right)\geq -r-{\frac {1}{2}}\right\}}$

Where this ${\displaystyle F}$ is the inverse backward difference of the basic ${\displaystyle x^{a}}$.

${\displaystyle {\frac {\pi }{a+1}}\int _{0}^{\infty }\operatorname {sech} ^{2}\left(\pi z\right)\operatorname {real} \left(\left(x+0.5+iz\right)^{\left(a+1\right)}-\left(0.5+iz\right)^{\left(a+1\right)}\right)dz}$

${\displaystyle {\frac {1}{h}}\int _{h}^{x+h}f\left(t\right)dt+{\frac {f\left(h\right)-f\left(x+h\right)}{2}}+i\int _{0}^{\infty }{\frac {\left(f\left(h+iht\right)-f\left(h-iht\right)\right)-\left(f\left(x+h+iht\right)-f\left(x+h-iht\right)\right)}{e^{\tau t}-1}}dt}$

Using different step size h in ${\displaystyle F(x)-F(x-h)=f(x)}$ as above doesn't result in anything other than a scaling factor in the resulting equivalent to ${\displaystyle B(s)}$

${\displaystyle h^{s-1}\left(1-{\frac {s}{2}}-is\int _{0}^{\infty }{\frac {\left(1+it\right)^{s-1}-\left(1-it\right)^{s-1}}{e^{\tau t}-1}}dt\right)}$

${\displaystyle {\frac {B'\left(s\right)}{B\left(s\right)}}}$ also has the reflexion

${\displaystyle \psi _{\Pi }\left(s\right)={\frac {\Pi '\left(s\right)}{\Pi \left(s\right)}}}$

${\displaystyle {\frac {B'\left(s\right)}{B\left(s\right)}}+{\frac {B'\left(1-s\right)}{B\left(1-s\right)}}={\frac {1}{2}}\psi _{\Pi }\left({\frac {s}{2}}\right)+{\frac {1}{2}}\psi _{\Pi }\left({\frac {1-s}{2}}\right)-\ln \left({\frac {\tau }{2}}\right)}$

In school I used to write things of the form ${\displaystyle \sin \left(\pi f^{-1}\left(x\right)\right)=0}$ for fun as I knew by the growth in the inverse of the function (for example, I would use ${\displaystyle {\sqrt {x}}}$ for ${\displaystyle x^{2}}$) it would graph roots at the sequence I was interested in. Let us generalize this as ${\displaystyle i^{4f^{-1}\left(s\right)}=e^{\tau if^{-1}\left(s\right)}=1}$. We can use the Lagrange inversion theorem to find ${\displaystyle f^{-1}}$ in most cases. I do not believe in set theory/ZFC, rather, I prefer to write sets in this way (as roots to a known equation) when possible in my non online personal notes, especially if the 'set' has infinite members. I do not believe in the axiom of choice. I believe in functions that have infinite zeros and Hadamard factorization, manipulations/transforms/etc., along with other complex analysis using infinitesimal calculus and the calculus of finite differences upon meromoprhic and holomorphic/analytic functions in general. If you cannot make it conctructively/with HoTT some related dependent type theory basis, I do not accept it as a true construction. I have not changed in this since high school.

The Nørlund–Rice integral is interesting.

As summary, ${\displaystyle \zeta \left(s\right)={\frac {B\left(1-s\right)}{s-1}}}$ and ${\displaystyle B\left(s\right)=-s\zeta \left(1-s\right)}$ regularized such that you remove the point discontinuity: ${\displaystyle B\left(0\right)=1}$.

Here is a somewhat general grapher for the inverse backward difference:

https://www.desmos.com/calculator/anih6sjvmb?backgroundColor=bbb&textColor=235656&invertedColors

The idea is to define a base region ${\displaystyle [h,h+1]}$ then use the Abel-Plana formula on that 'safe' region where it is analytic without branch cuts or branch points and exponential type less than 2pi in the imaginary direction on that interval. We can use interpolation and extend it out from there with the recurrence:

${\displaystyle N=444}$

${\displaystyle h\leq x\leq h+1\left\{-14\leq y\leq 14\right\}}$

${\displaystyle f\left(x\right)=f\left(x\right)=\sin \left(x^{2}\right)}$

${\displaystyle \sum _{n=1}^{x}f\left(n\right)}$

${\displaystyle t=x-h}$

${\displaystyle u\left(x\right)=f\left(x+h\right)}$

${\displaystyle f_{rac}\left(x\right)=x-\operatorname {floor} \left(x\right)}$

${\displaystyle S\left(t\right)=t-\operatorname {floor} \left(t\right)}$

${\displaystyle H(t)=\left\{t\geq 0:1,0\right\}}$

${\displaystyle \operatorname {real} \left(\operatorname {total} \left(u\left(t-\left[0...\operatorname {floor} \left(t\right)\right]\right)\right)\cdot H\left(t\right)+F_{fast}\left(S\left(t\right)-1\right)+u\left(S\left(t\right)\right)H\left(-t\right)-\operatorname {total} \left(u\left(t+\left[1...-\operatorname {floor} \left(t\right)\right]\right)\right)\cdot H\left(-t\right)-C\right)}$

${\displaystyle \operatorname {imag} \left(\operatorname {total} \left(u\left(t-\left[0...\operatorname {floor} \left(t\right)\right]\right)\right)\cdot H\left(t\right)+F_{fast}\left(S\left(t\right)-1\right)+u\left(S\left(t\right)\right)H\left(-t\right)-\operatorname {total} \left(u\left(t+\left[1...-\operatorname {floor} \left(t\right)\right]\right)\right)\cdot H\left(-t\right)-C\right)}$

${\displaystyle C_{orig}=K\left(-0.0000000001\right)}$

${\displaystyle C=\left\{\left|C_{orig}\right|>10^{4}:K\left(-1.0000000001\right),C_{orig}\right\}}$

${\displaystyle K\left(x\right)=\operatorname {real} \left(\operatorname {total} \left(u\left(t-\left[0...\operatorname {floor} \left(t\right)\right]\right)\right)\cdot H\left(t\right)+F_{fast}\left(S\left(t\right)-1\right)+u\left(S\left(t\right)\right)H\left(-t\right)-\operatorname {total} \left(u\left(t+\left[1...-\operatorname {floor} \left(t\right)\right]\right)\right)\cdot H\left(-t\right)\right)}$ and don't show this one.

${\displaystyle F_{base}\left(x\right)=\int _{1}^{x+1}u\left(o\right)do+{\frac {u\left(1\right)-u\left(x+1\right)}{2}}+i\int _{0}^{14}{\frac {\left(u\left(x+1-io\right)-u\left(1-io\right)\right)-\left(u\left(x+1+io\right)-u\left(1+io\right)\right)}{e^{2\pi o}-1}}do}$

${\displaystyle L=\left[0...N-1\right]}$

${\displaystyle x_{1}={\frac {1}{N-1}}L-1}$

${\displaystyle y_{1}=F_{base}\left(x_{1}\right)}$

${\displaystyle g\left(x\right)=\max \left(1,\min \left(N-1,\operatorname {floor} \left(\left(x+1\right)\cdot \left(N-1\right)\right)+1\right)\right)}$

${\displaystyle F_{fast}\left(x\right)=y_{1}\left(g\left(x\right)\right)+{\frac {\left(y_{1}\left(g\left(x\right)+1\right)-y_{1}\left(g\left(x\right)\right)\right)}{\left(x_{1}\left(g\left(x\right)+1\right)-x_{1}\left(g\left(x\right)\right)\right)}}\cdot \left(x-x_{1}\left(g\left(x\right)\right)\right)}$

Here also is a product grapher

https://www.desmos.com/calculator/sgytq6npfv?backgroundColor=bbb&textColor=235656&invertedColors

Here is the even, passes through the origin, minimal exponential type interpolant of the divisor counting function:

${\displaystyle {\frac {\sin \left(\pi x\right)}{\pi }}\sum _{j=1}^{\infty }\left({\frac {\pi }{2j}}\cot \left({\frac {\pi x}{2j}}\right)+{\frac {\pi }{2j-1}}\csc \left({\frac {\pi x}{2j-1}}\right)-{\frac {2}{x}}\right)}$

https://www.desmos.com/calculator/qshorjtlp3

https://arxiv.org/pdf/2509.12297 - Fuchs doesn't use minimal possible exponential type, and it resonates and so fails with the calculus of finite differences.

We now define the classical Bernoulli number function, which we denote by ${\displaystyle C(s)}$, as the derivative of the inverse backward difference of ${\displaystyle x^{s}}$ evaluated at ${\displaystyle x=-1}$:

${\displaystyle C(s)={\frac {\partial }{\partial x}}\left(\nabla ^{-1}x^{s}\right){\bigg |}_{x=-1}={\frac {\partial }{\partial x}}\left(\Delta ^{-1}x^{s}\right){\bigg |}_{x=0}}$

Using the twice left shifted Abel-Plana formula

${\displaystyle {\begin{aligned}\nabla ^{-1}x^{s}&=\int _{1}^{x+2}t^{s}dt-{\frac {\left(x+2\right)^{s}}{2}}+{\frac {1}{2}}\\&\quad -i\int _{0}^{\infty }{\frac {\left(x+2+it\right)^{s}-\left(1+it\right)^{s}-\left(x+2-it\right)^{s}+\left(1-it\right)^{s}}{e^{\tau t}-1}}dt-\left(x+1\right)^{s}\end{aligned}}}$

and differentiating under the integral, we obtain

${\displaystyle C(s)=1-{\frac {s}{2}}-is\int _{0}^{\infty }{\frac {\left(1+it\right)^{s-1}-\left(1-it\right)^{s-1}}{e^{\tau t}-1}}dt\;-\;s\cdot 0^{s-1}}$

where the term ${\displaystyle s\cdot 0^{s-1}}$ is interpreted as the distributional limit from the right, giving

${\displaystyle s\cdot 0^{s-1}={\begin{cases}0,&\Re (s)>1,\\[2pt]1,&s=1,\\[2pt]\infty {\text{ (pole town)}},&\Re (s)\leq 0,\;s\neq 0.\end{cases}}}$

Recognising that the first three terms of ${\displaystyle C(s)}$ are exactly the entire extended Bernoulli function ${\displaystyle B(s)}$ (with ${\displaystyle B(1)={\tfrac {1}{2}}}$), we have the fundamental relation

${\displaystyle C(s)=B(s)-s\cdot 0^{s-1}}$

Therefore ${\displaystyle C(s)=B(s)}$ for ${\displaystyle \Re (s)>1}$, and the classical Bernoulli numbers are recovered:

${\displaystyle C(0)=1,\qquad C(1)=-{\frac {1}{2}},\qquad C(2)={\frac {1}{6}},\qquad C(3)=0,\qquad \ldots }$

TL;DR, the generating function is started in a naive place that obscures the operator...

https://www.wolframalpha.com/input?i2d=true&i=Limit%5BPartial%5BHarmonicNumber%5C(40)x%5C(44)-z%5C(41)%2Cx%5D%2Cx-%3E-1%5D

Creating analytic continuations of arbitrary arithmetic sequences

Prereqs: https://arxiv.org/abs/1208.6079 https://github.com/LiSamoht/Preprint-Paper-on-Indefinite-Summations (for the standard Poisson summation derivation and the general formal integral transform framework; see also Nørlund for the left/right summation and cotangent related kernels). Caution: Beyond these standard introductory elements, the rest of this preprint is highly dubious and should be treated with extreme skepticism.

Uniqueness: Suppose two entire functions ${\displaystyle F_{1},F_{2}}$ of exponential type ${\displaystyle \leq \pi }$, even, interpolating ${\displaystyle d(n)}$ at integers, vanishing at ${\displaystyle 0}$, with sub-polynomial growth. Their difference ${\displaystyle G=F_{1}-F_{2}}$ vanishes at all integers, is even, type ${\displaystyle \leq \pi }$, sub-polynomial growth. By Paley-Wiener, ${\displaystyle {\widehat {G}}}$ is supported in ${\displaystyle [-\pi ,\pi ]}$. Poisson summation gives ${\displaystyle \sum _{k}{\widehat {G}}(\xi +2\pi k)=\sum _{n}G(n)e^{-in\xi }=0}$. Testing with functions supported in ${\displaystyle (-\pi ,\pi )}$ eliminates all shifts except ${\displaystyle k=0}$, so ${\displaystyle {\widehat {G}}=0}$ there. Hence ${\displaystyle \operatorname {supp} {\widehat {G}}\subseteq \{-\pi ,\pi \}}$. A distribution supported on two points is a linear combination of ${\displaystyle \delta }$ and derivatives; evenness forces a symmetric combination. Inverse Fourier transform gives ${\displaystyle G(x)=P(x)\cos(\pi x)+Q(x)\sin(\pi x)}$ with polynomials ${\displaystyle P,Q}$. Sub-polynomial growth on a sequence where the trigonometric factor stays away from zero forces ${\displaystyle P,Q\equiv 0}$, so ${\displaystyle G\equiv 0}$. Thus uniqueness.

Existence: Start from the Whittaker-Shannon cardinal series for an even type-${\displaystyle \pi }$ function: ${\displaystyle F(z)={\frac {\sin(\pi z)}{\pi }}\sum _{n=1}^{\infty }(-1)^{n}d(n)\left({\frac {1}{z-n}}+{\frac {1}{z+n}}\right)}$. Substitute ${\displaystyle d(n)=\sum _{d\mid n}1}$, swap sums, and use the Mittag-Leffler expansions ${\displaystyle \sum _{m=1}^{\infty }\left({\frac {1}{z-dm}}+{\frac {1}{z+dm}}\right)={\frac {\pi }{d}}\cot {\frac {\pi z}{d}}-{\frac {1}{z}}}$ and the alternating version ${\displaystyle \sum _{m=1}^{\infty }(-1)^{m}\left({\frac {1}{z-dm}}+{\frac {1}{z+dm}}\right)={\frac {\pi }{d}}\csc {\frac {\pi z}{d}}-{\frac {1}{z}}}$. Separate ${\displaystyle d}$ into even (${\displaystyle d=2j}$) and odd (${\displaystyle d=2j-1}$). This yields ${\displaystyle F(z)={\frac {\sin(\pi z)}{\pi }}\sum _{j=1}^{\infty }\left({\frac {\pi }{2j}}\cot {\frac {\pi z}{2j}}+{\frac {\pi }{2j-1}}\csc {\frac {\pi z}{2j-1}}-{\frac {2}{z}}\right)}$. The series converges absolutely; multiplication by ${\displaystyle \sin(\pi z)}$ cancels all poles, so ${\displaystyle F}$ is entire, even, interpolates ${\displaystyle d(n)}$ at integers, and has type ${\displaystyle \pi }$. Growth on ${\displaystyle \mathbb {R} }$ is sub-polynomial because the inner summands decay like ${\displaystyle 1/j^{2}}$ and ${\displaystyle \sin(\pi x)}$ is bounded.

Ugly example, the Möbius interpolant:

${\displaystyle {\frac {2x\sin \left(\pi x\right)}{\pi }}\sum _{n=1}^{\infty }{\frac {\mu \left(n\right)\left(-1\right)^{n}}{x^{2}-n^{2}}}}$

We do:

${\displaystyle {\frac {2x\sin \left(\pi x\right)}{\pi }}\sum _{n=1}^{2N+1}{\frac {\mu \left(n\right)\left(-1\right)^{n}}{x^{2}-n^{2}}}={\frac {2x\sin \left(\pi x\right)}{\pi }}\left({\frac {1}{1-x^{2}}}-\sum _{n=1}^{N}\left({\frac {\mu \left(2n\right)}{\left(2n\right)^{2}-x^{2}}}-{\frac {\mu \left(2n+1\right)}{\left(2n+1\right)^{2}-x^{2}}}\right)\right)}$

And it notably doesn't simplify. You could rewrite to terms of a convolution of ${\displaystyle d(n)}$ since we already have that. It's still going to be ugly. Regardless, the rewrite allows you to prove absolute convergence.

${\displaystyle {\frac {2x\sin \left(\pi x\right)}{\pi }}\left({\frac {1}{1-x^{2}}}-\sum _{n=1}^{\infty }\left({\frac {\mu \left(2n\right)}{\left(2n\right)^{2}-x^{2}}}-{\frac {\mu \left(2n+1\right)}{\left(2n+1\right)^{2}-x^{2}}}\right)\right)}$ Then you just use the same proofs as before. Difference is it doesn't simplify nicely to my knowledge.

You can take this a step further just by taking the indefinite sum to get the divisor summatory function and Mertens functions respectively. I have tried this with truncation of the series expansions and both work very well.

I took this to its limit by abusing the prime identity function ${\displaystyle y_{k}=\left[0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,...\right]}$, ${\displaystyle U\left(x\right)=y_{k}\left[x\right]}$:

${\displaystyle f\left(x\right)=-{\frac {2x\sin \left(\pi x\right)}{\pi }}\left(\sum _{n=1}^{50}\left({\frac {U\left(2n\right)}{\left(2n\right)^{2}-x^{2}}}-{\frac {U\left(2n+1\right)}{\left(2n+1\right)^{2}-x^{2}}}\right)\right)}$ in the inverse backward difference grapher:

https://www.desmos.com/calculator/rnu4af7nmc?backgroundColor=bbb&textColor=235656&invertedColors

Because this is the indefinite sum ${\displaystyle F}$ of a bounded function (prime identity, 1 if prime 0 if composite) it too is of exponential type ${\displaystyle \pi }$.

Additionally, ${\displaystyle {\frac {s}{x}}+{\frac {\sin \left(\pi x\right)}{\pi }}\sum _{n=1}^{\infty }\left(-1\right)^{n}\left(\operatorname {floor} \left({\frac {s}{n}}\right)-{\frac {s}{n}}\right)\left({\frac {1}{x-n}}+{\frac {1}{x+n}}\right)}$ Can be used as ${\displaystyle f}$ to find the ${\displaystyle F}$ (right half-plane solution) that interpolates ${\displaystyle \sum _{n=1}^{\operatorname {floor} \left(x\right)}\operatorname {floor} \left({\frac {s}{n}}\right)+f_{rac}\left(x\right)\operatorname {floor} \left({\frac {s}{\operatorname {floor} \left(x\right)+1}}\right)}$. It can be truncated to ${\displaystyle s}$ (an integer, not step size) to obtain a generalized harmonic number type version if desired. The resulting tail ${\displaystyle {\frac {\sin \left(\pi x\right)}{\pi }}\left(\left(\sum _{n=1}^{s}\left(-1\right)^{n}\left(\operatorname {floor} \left({\frac {s}{n}}\right)-{\frac {s}{n}}\right)\left({\frac {1}{x-n}}+{\frac {1}{x+n}}\right)\right)-\left(\sum _{n=1}^{\infty }\left(-1\right)^{n}\left(\operatorname {floor} \left({\frac {s}{n}}\right)-{\frac {s}{n}}\right)\left({\frac {1}{x-n}}+{\frac {1}{x+n}}\right)\right)\right)}$ then simplifies to ${\displaystyle {\frac {s\sin \left(\pi x\right)}{\pi x}}\left(\psi \left({\frac {x}{2}}\right)-\psi \left({\frac {x+1}{2}}\right)+{\frac {\pi }{\sin \left(\pi x\right)}}+{\frac {1}{x}}-2\ln \left(2\right)\right)-{\frac {s\sin \left(\pi x\right)}{\pi }}\sum _{n=1}^{s}{\frac {\left(-1\right)^{n}}{n}}\left({\frac {1}{x-n}}+{\frac {1}{x+n}}\right)}$ – it doesn't contain complicated arithmetic data.

Relation to the Riemann Hypothesis

Let ${\displaystyle \xi (s)={\tfrac {1}{2}}s(s-1)\pi ^{-s/2}\,\Gamma \!{\bigl (}{\tfrac {s}{2}}{\bigr )}\,\zeta (s)}$ be the Riemann xi function. It is entire of order ${\displaystyle 1}$, satisfies ${\displaystyle \xi (s)=\xi (1-s)}$, and its zeros are precisely the non-trivial zeros of ${\displaystyle \zeta (s)}$.

Define ${\displaystyle f(s)=1/\xi (s)}$, a meromorphic function whose poles are exactly the non‑trivial zeros of ${\displaystyle \zeta (s)}$ (and hence of the extended Bernoulli function ${\displaystyle B(s)}$).

Exponential type of ${\displaystyle f}$ in pole‑free strips. Let ${\displaystyle D}$ be a closed vertical strip that is bounded away from all zeros of ${\displaystyle \xi (s)}$. Writing ${\displaystyle \xi (s)={\frac {1}{2}}s(s-1)\pi ^{-s/2}\Gamma (s/2)\zeta (s)}$ gives ${\displaystyle f(s)={\frac {2}{s(s-1)}}\,\pi ^{s/2}\,{\frac {1}{\Gamma (s/2)}}\,{\frac {1}{\zeta (s)}}.}$ From Stirling's formula, for ${\displaystyle s=\sigma +it}$ with ${\displaystyle |t|\to \infty }$ and ${\displaystyle \sigma }$ bounded, ${\displaystyle \left|{\frac {1}{\Gamma (s/2)}}\right|\sim {\frac {1}{\sqrt {2\pi }}}\left({\frac {2}{|t|}}\right)^{{\frac {\sigma }{2}}-{\frac {1}{4}}}e^{{\frac {\pi }{4}}|t|},}$ so the dominant exponential factor is ${\displaystyle e^{{\frac {\pi }{4}}|t|}}$. In a zero‑free strip, ${\displaystyle 1/\zeta (s)=O_{\varepsilon }(e^{\varepsilon |t|})}$ for every ${\displaystyle \varepsilon >0}$ (classical). The algebraic factors are subexponential. Hence, for any ${\displaystyle \varepsilon >0}$ and ${\displaystyle |t|}$ large, ${\displaystyle |f(s)|\leq C_{\varepsilon }e^{({\frac {\pi }{4}}+\varepsilon )|t|}.}$ A matching lower bound is obtained from the fact that ${\displaystyle |\xi (s)|}$ itself has exponential type ${\displaystyle \pi /4}$, so on a suitable vertical line inside the strip, ${\displaystyle |f(s)|\geq c\,e^{({\frac {\pi }{4}}-\varepsilon )|t|}}$ for infinitely many ${\displaystyle t}$. Therefore the exponential type of ${\displaystyle f}$ on any closed pole‑free vertical substrip is exactly ${\displaystyle \pi /4}$.

Because ${\displaystyle \pi /4<2\pi }$, the growth condition required for Nørlund's theory of principal solutions (type ${\displaystyle <2\pi }$ in the imaginary direction) is amply satisfied on every such closed substrip. On maximal open strips whose boundaries contain poles, the function loses the uniform bound and has infinite type, but Nørlund's construction is first carried out on a closed interior strip and then analytically continued – the interior type ${\displaystyle \pi /4}$ is what matters, and, with it being so small, we are quite spoiled for choice in terms of series expansions that are applicable.

The number of principal solutions as a diagnostic

Apply the indefinite sum operator ${\displaystyle \nabla ^{-1}}$ (inverse backward difference) to ${\displaystyle f}$: ${\displaystyle F(s)-F(s-1)=f(s)}$. Nørlund’s theory guarantees a principal solution–unique up to an additive constant–on every maximal open vertical strip that

• contains a segment of the real axis,
• is free of singularities of ${\displaystyle f}$.

The poles of ${\displaystyle f}$ lie on the vertical lines ${\displaystyle \Re s=\Re \rho }$ for each zero ${\displaystyle \rho }$ of ${\displaystyle \xi }$. Because the zeros are symmetric about ${\displaystyle \Re s={\tfrac {1}{2}}}$ and about the real axis, the set ${\displaystyle R=\{\,\Re \rho \mid \xi (\rho )=0\,\}}$ is a non-empty subset of ${\displaystyle (0,1)}$ symmetric about ${\displaystyle {\tfrac {1}{2}}}$. If the Riemann Hypothesis (RH) is true, ${\displaystyle R=\{{\tfrac {1}{2}}\}}$; otherwise it contains at least the three distinct values ${\displaystyle \{1-a,\;{\tfrac {1}{2}},\;a\}}$ with ${\displaystyle a>{\tfrac {1}{2}}}$.

The real axis is partitioned by the elements of ${\displaystyle R}$ into open intervals. Each interval determines a connected component (a maximal vertical strip) that contains a real segment and is pole-free. Nørlund’s theory yields exactly one principal solution on each such component. Hence the total number ${\displaystyle N}$ of distinct principal solutions of ${\displaystyle \nabla ^{-1}(1/\xi )}$ equals the number of those intervals.

${\displaystyle {\begin{array}{c|c}{\text{RH true}}&R=\{{\tfrac {1}{2}}\}\;\Longrightarrow \;N=2\\[4pt]{\text{RH false}}&R\supseteq \{1-a,{\tfrac {1}{2}},a\}\;\Longrightarrow \;N\geq 4\end{array}}}$

Thus one has the strict equivalence

${\displaystyle {{\text{Riemann Hypothesis true}}\;\Longleftrightarrow \;\nabla ^{-1}{\frac {1}{\xi (s)}}{\text{ has exactly two Nørlund principal solutions.}}}}$

And by ${\displaystyle \xi (s)=\xi (1-s)}$ making it so that there are symmetrical zeros, our ${\displaystyle f}$ has symmetrical poles, ${\displaystyle +1}$ for the poles at real part one half for which we already know (G.H. Hardy for the ${\displaystyle 40\%}$ of all zeros, numerical calculation up to large values too), so we can very safely say one half is part of the set ${\displaystyle R}$. So, ignoring conjugates as they don't contribute new poles, we have that the non critical lines poles real parts are even in number. We have an additional real part at one half. So the total number of splitting lines is of the form ${\displaystyle 2N+1}$:

${\displaystyle {{\text{Riemann Hypothesis false}}\;\Longleftrightarrow \;\nabla ^{-1}{\frac {1}{\xi (s)}}{\text{ has exactly four or more Nørlund principal solutions.}}}}$

There is, in total, an even number of principal solutions. This comes from the odd number of real parts splitting the plane (a statement true regardless of the truth or falsehood of RH, as we've shown). The number of solutions, in every case, regardless of the truth or falsehood of RH, is even by simple topology; an important fact to keep in mind.

Outer solutions and their domains

Two principal solutions (under RH, the only two) can be constructed unconditionally. The right outer solution is defined for ${\displaystyle \Re s>1}$ by the telescoping sum ${\displaystyle F_{\mathrm {right} }(s)=\sum _{n=1}^{\infty }{\Bigl (}{\frac {1}{\xi (n)}}-{\frac {1}{\xi (s+n)}}{\Bigr )},\qquad \Re s>1.}$ Because ${\displaystyle f(s)}$ decays rapidly as ${\displaystyle \Re s\to +\infty }$, this series converges absolutely and defines an analytic function satisfying the difference equation. Using the Abel–Plana bounds familiar from Candelpergher’s treatment of ${\displaystyle \eta (s)}$, one shows that the series converges uniformly on every compact subset of the larger half-plane ${\displaystyle \Re s>a_{0}}$, where ${\displaystyle a_{0}=\sup\{\Re \rho :\xi (\rho )=0\}}$. (If RH is true, ${\displaystyle a_{0}={\tfrac {1}{2}}}$; otherwise ${\displaystyle a_{0}}$ is the largest real part of an off-critical zero.) Consequently ${\displaystyle F_{\mathrm {right} }}$ extends analytically to ${\displaystyle \Re s>a_{0}}$ and remains a principal solution there. Importantly, ${\displaystyle F_{\mathrm {right} }}$ is analytic at every point on the vertical line ${\displaystyle \Re s=a_{0}}$ (except possibly at the actual poles ${\displaystyle \rho }$ of ${\displaystyle f}$, which are isolated).

Symmetrically, the left outer solution ${\displaystyle F_{\mathrm {left} }(s)=\sum _{n=1}^{\infty }{\Bigl (}{\frac {1}{\xi (n-s)}}-{\frac {1}{\xi (n)}}{\Bigr )}}$ is a principal solution on the half-plane ${\displaystyle \Re s<1-a_{0}}$ and is analytic for ${\displaystyle \Re s<1-a_{0}}$.

The difference function and numerical evidence

The difference between the two outer solutions is ${\displaystyle D(s)=F_{\mathrm {left} }(s)-F_{\mathrm {right} }(s)}$. Truncated approximations with ${\displaystyle N=244}$ were computed on a rectangle ${\displaystyle -2\leq \Re s\leq 34,\;-2\leq \Im s\leq 34}$ using ${\displaystyle S(u)=\sum _{n=1}^{N}{\Bigl (}{\frac {1}{\xi (n)}}-{\frac {1}{\xi (n+u)}}{\Bigr )}-\sum _{n=1}^{N}{\Bigl (}{\frac {1}{\xi (n-u)}}-{\frac {1}{\xi (n)}}{\Bigr )}.}$ Domain colouring plots show a structure consistent with ${\displaystyle D(s)}$ having poles only along the vertical line ${\displaystyle \Re s={\tfrac {1}{2}}}$ (shifted by integers), as expected if RH is true and the two outer solutions are the only canonical ones.

https://samuelj.li/complex-function-plotter/#2%20-%201%2F(pi%5E(-(1%2Bz)%2F2)*gamma((1%2Bz)%2F2%2B1)*z*zeta(1%2Bz))%20%2B%20sum(1%2F(pi%5E(-n%2F2)*gamma(n%2F2%2B1)*(n-1)*zeta(n))%20-%201%2F(pi%5E(-(n%2Bz)%2F2)*gamma((n%2Bz)%2F2%2B1)*(n%2Bz-1)*zeta(n%2Bz))%2C%20n%2C%202%2C%2014)

https://samuelj.li/complex-function-plotter/#1%2F(pi%5E(-(1-z)%2F2)*gamma((1-z)%2F2%2B1)*(-z)*zeta(1-z))%20-%202%20%2B%20sum(1%2F(pi%5E(-(n-z)%2F2)*gamma((n-z)%2F2%2B1)*(n-z-1)*zeta(n-z))%20-%201%2F(pi%5E(-n%2F2)*gamma(n%2F2%2B1)*(n-1)*zeta(n))%2C%20n%2C%202%2C%2014)

https://samuelj.li/complex-function-plotter/#4%20-%201%2F(pi%5E(-(1%2Bz)%2F2)*gamma((1%2Bz)%2F2%2B1)*z*zeta(1%2Bz))%20-%201%2F(pi%5E(-(1-z)%2F2)*gamma((1-z)%2F2%2B1)*(-z)*zeta(1-z))%20%2B%20sum(1%2F(pi%5E(-n%2F2)*gamma(n%2F2%2B1)*(n-1)*zeta(n))%20-%201%2F(pi%5E(-(n%2Bz)%2F2)*gamma((n%2Bz)%2F2%2B1)*(n%2Bz-1)*zeta(n%2Bz))%2C%20n%2C%202%2C%2014)%20-%20sum(1%2F(pi%5E(-(n-z)%2F2)*gamma((n-z)%2F2%2B1)*(n-z-1)*zeta(n-z))%20-%201%2F(pi%5E(-n%2F2)*gamma(n%2F2%2B1)*(n-1)*zeta(n))%2C%20n%2C%202%2C%2014)

https://github.com/surebeae/rxibdg

Open route: Merge (not speculative, simplest and already outlined by Nørlund's theory and his sources) Prereq: read https://arxiv.org/abs/2512.11405

We already have the rightmost telescoping, and by construction, it would have to match a Hurwitz-Taylor series expansion for the rightmost ${\displaystyle F}$ (out of any point, like real part ${\displaystyle 3/2}$ on the real axis). You then match, term-wise, after using the identity theorem to match the outer rightmost half plane solution to itself, to the expansion about real part ${\displaystyle 1/2}$ on the real axis.

The center of each shifts. ${\displaystyle 1/2}$ to the origin. ${\displaystyle 3/2}$ to real part 1 on the real axis. Why? Because of the Bernoulli polynomial alternating recurrence shifting by ${\displaystyle -1/2}$ with respect to the argument, so ${\displaystyle F}$ doesn't have a "normal" radius, it's shifted. You then use the recurrence ${\displaystyle F(z)-F(z-1)=f(z)}$ to match one to the other. Isolate ${\displaystyle F}$: ${\displaystyle F(z)=F(z-1)+f(z)}$. Use a normal Taylor series expansion on ${\displaystyle f}$. Use the Taylor series expansion for ${\displaystyle 3/2}$ which maps to 1+0i after antidifferencing termwise for ${\displaystyle F(x-1)}$. Use ${\displaystyle 1/2}$ (origin) for ${\displaystyle F(x)}$. If they are two different ${\displaystyle F}$, then this cannot hold. If they're the same ${\displaystyle F}$, it has to. The discs now overlap completely in their convergence because the normal Taylor series expansion of ${\displaystyle 1/\xi (x)}$ is 'clean' inside the radius of convergence. I have not done this yet as it is tedious and difficult. You may and then take all of the credit for it, I just want to be able to move the Indefinite sum article to high priority and communicate the problem of RH in a simple, visual manner to a friend I met on Legends of Equestria. To do this communication, I made https://github.com/surebeae/rxibdg and norlundcalc.

Under the false-RH scenario, Nørlund’s theory guarantees an additional inner solution ${\displaystyle F_{\mathrm {inner} }}$ on the pole-free strip ${\displaystyle S=({\tfrac {1}{2}},a)}$, where ${\displaystyle a}$ is the smallest real part greater than ${\displaystyle {\tfrac {1}{2}}}$ of a zero of ${\displaystyle \xi }$. Because ${\displaystyle f(s)}$ has a pole at ${\displaystyle s=a}$, any solution defined on ${\displaystyle S}$ and extending to the left boundary ${\displaystyle \Re s=a}$ must itself possess a singularity there (the difference equation forces a pole when one steps across the pole of ${\displaystyle f}$).

By contrast, the right outer solution ${\displaystyle F_{\mathrm {right} }}$ is analytic on the whole half-plane ${\displaystyle \Re s>a_{0}}$.

Now, if one could prove that ${\displaystyle F_{\mathrm {inner} }}$ and ${\displaystyle F_{\mathrm {right} }}$ are the same analytic function – for instance, by showing that their series expansions (Laurent series or telescoping sums) coincide on a common domain, such as on ${\displaystyle (1,\infty )}$ after shifting – then we would have a single function that is both analytic at ${\displaystyle \Re s=a}$ (from ${\displaystyle F_{\mathrm {right} }}$) and singular there (from ${\displaystyle F_{\mathrm {inner} }}$). This contradiction would prove that the inner solution cannot exist, hence no off-critical zero can exist; RH would follow.

Thus the whole problem reduces to a concrete analytic task: show that the series expansion of the inner solution (obtained via termwise indefinite summation of a Laurent or Taylor expansion using Hurwitz zetas of ${\displaystyle 1/\xi (s)}$ on ${\displaystyle S}$) can be transformed into the series expansion of ${\displaystyle F_{\mathrm {right} }}$ (same). If the two expansions match, then the functions are identical, leading to the contradiction with the pole set. If they do not match, then ${\displaystyle N\geq 4}$ and RH is false. The equivalence stated earlier therefore remains valid and precise: RH holds if and only if the number of principal solutions is exactly two, and the way to decide that number is to test the equality of the two series representations.

Why/where did it reduce to a cross discipline analytic continuation (of infinitesimal and and finite difference calculus)? Because we defined the open interval ${\displaystyle (1/2,a)}$ carefully, and because implication of G.H. Hardy's 40% of zeros on the line proof implies the zeros are discrete in nature, so they cannot be compact about the real part set close to real part ${\displaystyle 1/2}$ (similar to reductio ad absurdum arguments for 0.9999...=1 or there being a "smallest" positive rational number, which, decent, but could be made stronger). We also rely on proofs/density estimates around real part ${\displaystyle 1/2}$ somewhat, but the theory gives a principal solution regardless of how thin the strip is. As such, it is not so dense with poles upon specific real parts as to not be meromorphic in the normal sense (as, say, the Lacunary function is).

Product route (alternative) (speculative)

The inverse backward product ${\displaystyle \prod _{x}(1/\xi (x))}$ is a natural entire function ${\displaystyle P(s)}$ satisfying ${\displaystyle P(s)\xi (s)=P(s-1)}$. (Note: the logarithmic derivative of ${\displaystyle P(s)}$ is ${\displaystyle L(s)=P'(s)/P(s)}$, which satisfies the difference equation ${\displaystyle \Delta L(s)=-\log \xi (s)}$.)

Nevertheless, the core topological framework—derived from Nørlund's analytic continuation for difference equations—applies equally to the indefinite product. Just as with the principal sums, the number of distinct Nørlund principal products is determined exactly by the number of maximal pole-free vertical strips of ${\displaystyle 1/\xi (s)}$. Therefore, the same equivalence holds: the Riemann Hypothesis is true if and only if the principal indefinite product of ${\displaystyle 1/\xi (s)}$ has exactly two principal solutions.

To progress via this route, one would need to leverage the operator calculus of finite differences to study the series expansions of ${\displaystyle \log \xi (s)}$ across different strips. The core task would be to prove that the principal products generated in each strip merge (or fail to merge), which would settle the number of strips. It is also perhaps more attack-able than the other, as indefinite products do not necessarily inherit poles (e.g. ${\displaystyle 1/x}$).

This method is speculative—it lacks a concrete series expansion to begin the matching process—but it is a natural consequence of the calculus of finite differences.

For the previous merge route, you would simply need to series match ${\displaystyle F_{\mathrm {inner} }}$ (which is ${\displaystyle F_{\mathrm {inner\_right} }}$, there is a ${\displaystyle F_{\mathrm {inner\_left} }}$ but it merges with the left solution under RH, not the right one) with ${\displaystyle F_{\mathrm {right} }}$ (of which I have given an unconditional expansion for already) up to a constant. To get them to overlap, you would simply recur ${\displaystyle F_{\mathrm {inner\_right} }}$ rightward using the recurrence ${\displaystyle F(z)-F(z-1)=f(z)}$ rearranged.

Xi, real part. https://www.desmos.com/calculator/wuvlryaq50

${\displaystyle B\left(x\right)=1-{\frac {x}{2}}-2x\int _{0}^{\infty }{\frac {\sin \left(\left(1-x\right)\tan ^{-1}\left(t\right)\right)}{\left(t^{2}+1\right)^{\frac {1-x}{2}}\left(e^{2\pi t}-1\right)}}dt}$

${\displaystyle \Pi \left(x\right)=x!}$

${\displaystyle K\left(s\right)=\left\{s\leq {\frac {1}{2}}:\pi ^{-{\frac {1-s}{2}}}\Pi \left({\frac {1-s}{2}}\right)B\left(s\right),s>{\frac {1}{2}}:\pi ^{-{\frac {s}{2}}}\Pi \left({\frac {s}{2}}\right)B\left(1-s\right)\right\}}$

Xi. https://www.desmos.com/calculator/rxk65eecvi

${\displaystyle B\left(x\right)=1-{\frac {x}{2}}-ix\int _{0}^{\infty }{\frac {\left(1+it\right)^{x-1}-\left(1-it\right)^{x-1}}{e^{\tau t}-1}}dt}$

${\displaystyle \Pi \left(x\right)=\left\{\operatorname {real} \left(x\right)\leq -{\frac {1}{2}}:{\frac {\tau i}{\left(i^{-2x}-i^{2x}\right)\int _{0}^{\infty }t^{-\left(x+1\right)}e^{-t}dt}},\operatorname {real} \left(x\right)>-{\frac {1}{2}}:\int _{0}^{\infty }t^{x}e^{-t}dt\right\}}$

${\displaystyle K\left(s\right)=\left\{\operatorname {real} \left(s\right)\leq {\frac {1}{2}}:\pi ^{-{\frac {1-s}{2}}}\Pi \left({\frac {1-s}{2}}\right)B\left(s\right),\operatorname {real} \left(s\right)>{\frac {1}{2}}:\pi ^{-{\frac {s}{2}}}\Pi \left({\frac {s}{2}}\right)B\left(1-s\right)\right\}}$

Interactive terminal grapher. Public domain reference implementation. https://codeberg.org/AzulBeae/norlundcalc

Mirror if you can't access codeberg: https://github.com/surebeae/norlundcalc

I should read https://www.homepages.ucl.ac.uk/~ucahrha/conferences/ruijsenaars/ADEs-specialfn.pdf

https://www.researchgate.net/publication/2431655_Bernoulli_Polynomials_Old_and_New_Generalizations_and_Asymptotics

https://www.wolframalpha.com/input?i=plot+Re%28HarmonicNumber%28x%2C+-1%2F2+-+i*imag%28zetazero%281%29%29%29%29+and+Im%28HarmonicNumber%28x%2C+-1%2F2+-+i*imag%28zetazero%281%29%29%29%29+from+x%3D-2+to+24

https://www.wolframalpha.com/input?i=ComplexPlot%5B%28HarmonicNumber%28x%2C+-1%2F2+-+i*imag%28zetazero%281%29%29%29%29%5D