How does a mathematician create a new zeta function?

Imagine a mathematician wants to create this function:

$$F(s)=\sum_{n=0}^{\infty} \frac{1}{(n^2+1)^s}$$

Where would he start?

More precisely, how would he know the exact values of $F(s)$ for $s> 0$? And how would he extend this result for every natural $s$? And for real $s$?

• Maybe you meant $n^\alpha$ instead of $n^2$ in the definition of the "zeta function above. ? – user378947 Jan 1 '17 at 13:40
• @mathbeing Nope. My aim there was to create a new function not studied before, to see the method used to analyse any totally new function – user3141592 Jan 1 '17 at 13:51
• @SimpleArt he changed his function 3 times. And WA is very bad with Dirichlet series – reuns Jan 1 '17 at 14:50
• How does a mathematician create a new zeta function? I believe that also is possible using a property of the Mellin transform, when is applied to an harmonic series. – user243301 Jan 2 '17 at 10:17
• If I may, I have updated my answer... – Simply Beautiful Art Jan 7 '17 at 22:41

Update:

If one may recall the following series:

$$\sum_{n=0}^\infty\frac1{n^2+x}=\frac1{2x}+\frac\pi{2\sqrt x\tanh(\pi\sqrt x)}$$

Then differentiating both sides $k$ times yields

$$\sum_{n=0}^\infty\frac1{(n^2+x)^{k+1}}=\frac1{2x^{k+1}}+\frac{(-1)^k\pi}{2\times k!}\frac{d^k}{dx^k}\frac1{\sqrt x\tanh(\pi\sqrt x)}$$

and evaluation at $x=1$ gives closed forms.

Update:

Using general Leibniz rule, we have

$$\frac{d^k}{dx^k}\frac{\coth(\pi\sqrt x)}{\sqrt x}=\sum_{p=0}^k\binom kp\frac{\Gamma(3/2+n-p)}{-2\sqrt\pi}(-1)^{n-p}x^{-1/2-n+p}(\coth(\pi\sqrt x))^{(p)}$$

We can handle the chain rule with the $n$th derivative of $\coth$ using Faà di Bruno's formula,

$$\small(\coth(\pi\sqrt x))^{(p)}=\sum_{q=1}^n\coth^{(q)}(\pi\sqrt x)B_{p,q}\left(\pi\frac12x^{-1/2},-\pi\frac14x^{-1/2},\dots,\sqrt\pi\frac{\Gamma(3/2+q)}{-4}(-1)^qx^{-1/2-q}\right)$$

where $B_{n,k}$ is the bell polynomial. The $q$th derivative of $\coth$ is then given when $q\ge1$:

$$\coth^{(q)}(x)=2^q(\coth(x)-1)\sum_{r=0}^q\frac{(-1)^rr!S_q^{(r)}}{2^r}(\coth(x)+1)^r$$

Putting all of this together,

$$\tiny F(k+1)=\frac12+\frac{(-1)^k\pi}{2(k!)}\left(\binom kp\frac{\Gamma(3/2+k)}{-2\sqrt\pi}(-1)^k(\coth(\pi)+\sum_{p=1}^k\sum_{q=1}^n\sum_{r=0}^q\binom kp\frac{\Gamma(3/2+k-p)}{-2\sqrt\pi}(-1)^{k+p+r}2^{q-r}(\coth(\pi)-1)(-1)^rr!S_q^{(r)}(\coth(\pi)+1)^r(\pi)B_{p,q}\left(\pi,-\pi\frac14,\dots,\sqrt\pi\frac{\Gamma(3/2+q)}{-4}(-1)^q\right)\right)$$

Old:

$$\frac12[\vartheta_3(0,r)+1]=\sum_{n=0}^\infty r^{n^2}$$

Term by term integration then reveals that

$$\frac1x\int_0^x\frac12[\vartheta_3(0,r)+1]\ dr=\sum_{n=0}^\infty\frac{x^{n^2}}{1+n^2}$$

Repeat this process over and over to get

$$\frac1{x_1}\int_0^{x_1}dx_2\frac1{x_2}\int_0^{x_2}dx_3\dots dx_k\frac1{x_k}\int_0^{x_k}dr\frac12[\vartheta_3(0,r)+1]=\sum_{n=0}^\infty\frac{x_1^{n^2}}{(1+n^2)^k}$$

And of course, evaluate at $x_1=1$ for your function. I do not see a closed form coming out of this, but it may be useful for discerning certain things about your function.

I do note, however, that in the case of $k=1$, the solution is given in this post:

$$\sum_{n=0}^\infty\frac1{1+n^2}=\int_0^1\frac12[\vartheta_3(0,r)+1]\ dr=\frac\pi{2\tanh\pi}+\frac12$$

• Is a better representation of that final nth derivative possible? Perhaps using the Taylor Expansion around $x=1$ for $\Coth(\pi)$ and for $x^{-1/2}$? I'm imagining just plugging $\pi\sqrt{x}$ into the former and then taking the Cauchy Product... Perhaps one of the series would simplify? It just seems that direct calculations of nth derivatives wont be pleasing – Brevan Ellefsen Jan 1 '17 at 16:43
• @BrevanEllefsen Yes, true, but it sure does give a nice way to tackle closed forms. I'll try to think about it and hope that it simplifies. – Simply Beautiful Art Jan 1 '17 at 16:47
• Of course, this only works for integers $k$, whereas the intent of the OP seems to be for real values $s$. – Greg Martin Jan 1 '17 at 22:09
• @GregMartin Ofc, but I see no way of doing that... – Simply Beautiful Art Jan 1 '17 at 22:09
• @BrevanEllefsen If we first apply General Leibniz rule followed by Faà di Bruno's formula and the $n$th derivative of hyperbolic cotangent (working on it), we shall arrive at our destination. – Simply Beautiful Art Jan 7 '17 at 2:21

Accepting, for argument's sake, that it is possible to create a function (the opposing view is that one discovers mathematical objects, rather than creating them), you have created the function simply by writing down that formula. Well, you have created it for real part of $s$ exceeding one-half, which is where the series converges. Knowing "exact values" (again, a term we can argue over, but I'll take it to mean a finite expression in terms of well-known constants, and leave it at that), well, that's generally very difficult. No one knows an exact value for $\zeta(3)$, for example, and for your function I suspect no one knows an exact value for $F(2)$. And as to extending your function to be defined for all complex $s$ (outside of a pole at $s=1/2$), that's what analytic continuation is for, so now you have a keyphrase to search for.

• It appears I know the exact value of $F(1)$ though... – Simply Beautiful Art Jan 1 '17 at 15:16
• Do see my answer below. It should be given by $\frac\pi{2\tanh\pi}+\frac12$ – Simply Beautiful Art Jan 1 '17 at 15:18
• OK, then, make it $F(2)$. – Gerry Myerson Jan 1 '17 at 15:18
• Yes, I'm trying, but that is quite a ways harder... – Simply Beautiful Art Jan 1 '17 at 15:19
• Never mind, $F(2)=\frac12+\frac\pi4\frac{\tanh\pi+2\pi\operatorname{sech}^2\pi}{\tanh^2\pi}$ – Simply Beautiful Art Jan 1 '17 at 16:37

It doesn't have an Euler product.

Expanding $(1+n^{-2})^{-s} =\sum_{k=0}^\infty {-s \choose k} n^{-2k}$ you get $$F(s) =\sum_{n=0}^\infty (n^2+1)^{-s} = 1+2^{-s}+\sum_{k=0}^\infty {-s \choose k} (\zeta(2s+2k)-1)$$ so it is meromorphic on the whole complex plane, with poles at $s = \frac{1}{2}-k, k \in \mathbb{N}$

Using $\Gamma(s) a^{-s} =\int_0^\infty x^{s-1}e^{-ax}dx$ and $\theta(x) = \sum_{n=1}^\infty e^{-x n^2}$ you have $$G(s) = (F(s)-1-2^{-s}) \Gamma(s) = \int_0^\infty x^{s-1}e^{-x} \theta(x)dx, \qquad \Gamma(s) \zeta(2s) = \int_0^\infty x^{s-1}\theta(x) dx$$ where $\Gamma(s) \zeta(2s)-\frac{\sqrt{\pi}}{2(s-1/2)}+\frac{1}{2s}$ is entire.

From these poles location, we can deduce that for arbitrary large $N$ :

$\theta(x) = \frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2}+o(x^N)$ as $x \to 0$ and hence $e^{-x}\theta(x) = \sum_{k\ge 0} \frac{x^k}{k!}(\frac{\sqrt{\pi}}{2}x^{-1/2}-\frac{1}{2})+o(x^N)$ as $x \to 0$

so that $$\lim_{s \to 1/2-k}(s+1/2-k)G(s)= \frac{\sqrt{\pi}}{2k!}, \qquad \lim_{s \to -k}(s-k)G(s)= \frac{-1/2}{(k+1)!}$$

• I wonder what makes this answer more popular. It simply provides an asymptote, but asymptotes are hardly difficult to find. I know the OP has changed the question, but I'm just scratching my head over how useful these asymptotes are. Though, I give +1 for the $s\to-k$ asymptotes, since $F(s)$ isn't naturally defined in that region. – Simply Beautiful Art Jan 1 '17 at 16:53
• @SimpleArt it gives the analytic continuation, the value at $-k$, the poles at $-k+1/2$. Together with the value at positive integers, this is probably all we can tell at this level. And the method is exactly the same with $\zeta(s)$. – reuns Jan 1 '17 at 17:07
• Yeah, probably. And yes, ofc, I can recognize your "$\zeta(s)\Gamma(s)$" combo here. – Simply Beautiful Art Jan 1 '17 at 17:08
• @SimpleArt (Riemann's trick, from this he obtained the functional equation, using the Poisson formula $1+2\theta(x)=\frac{1}{x^{1/2}}(1+2\theta(1/x))$) – reuns Jan 1 '17 at 17:10
• Just picking a nit: I believe it should be $$F(s)=1+2^{-s}+\sum_{k=0}^\infty\binom{-s}{k}(\zeta(2s+2k)-1)$$ – robjohn Jan 1 '17 at 17:19

Perhaps look into the theory of Dirichlet series? Your function is the series corresponding to the sequence $n \mapsto [\exists x\in \mathbb{N}(n=x^2+1)],$ where $[-]$ is the Iverson Bracket (google it.) The first few terms of the sequence are:

$$0,1,1,0,0,1\ldots$$

• @user1952009 Depends if you understand what he is saying. To the right person, this could be a perfectly fine answer. – Simply Beautiful Art Jan 1 '17 at 14:44