Mellin transform of theta function $\theta$ to show $\zeta(-1)=-\frac{1}{12}$

Question

Edit: I realize a lot of attention is pointed at the proclamation $\zeta(-1)=-\frac{1}{12}$. Im more interested in the derivation of the Zeta function using the methods describe below, thank you!

Before you read, I'm requesting a hint as to how to proceed or where I've gone wrong, not a complete solution, thanks!

I'm tasked to show $\zeta(-1)=-\frac{1}{12}$ using the Mellin transform ${\displaystyle \left\{{\mathcal {M}}f\right\}(s)=\varphi (s)=\int _{0}^{\infty }x^{s-1}f(x)\,dx}$ on the Theta function $\theta$. I'm also given the hint of Wikipedia's own formulation: ${\displaystyle \Gamma \left({\frac {s}{2}}\right)\pi ^{-{\frac {s}{2}}}\zeta (s)={\frac {1}{2}}\int _{0}^{\infty }{\bigl (}\vartheta (0;it)-1{\bigr )}t^{\frac {s}{2}}{\frac {\mathrm {d} t}{t}}}$. It seems straight forward to apply Mathematica, first completing the sum ${\displaystyle {\vartheta (0;it)=1+2\sum _{n=1}^{\infty }q^{n^{2}}}}$ with $q=e^{\pi iit}=e^{-\pi t}$. Then integrating over ${\displaystyle {\frac {1}{2}}\int _{0}^{\infty }{\bigl (}\sum _{n=1}^{\infty }e^{-n^2\pi t}{\bigr )}t^{\frac {s}{2}}{\frac {\mathrm {d} t}{t}}}$ and bring over the factors from the left hand side to get an expression for $\zeta(s)$.

Mathematica returns an incorrect answer for a given $s$ and complains about divergence:

"$-\frac{1}{2t^{3/2}}+\frac{EllipticTheta[3,0,e^{-\pi t}]}{2 t^(3/2))} $ does not converge on {$0,\infty$}."

How do i get around this problem? Or have I done something wrong in the beginning?

Remove["Global`* "]
summa[s_, t_] := Sum[Exp[-n^2* Pi* t], {n, 1, Infinity}]
integral[s_] := 1/2 (Integrate[summa[s, t]*t^(s/2 - 1), {t, 0, Infinity}])
r[s_] := \[Pi]^(s/2)/(2 Gamma[s/2])
\[Zeta][s_] := integral[s]*r[s]
N[\[Zeta][-1]]

You should include the Mathematica code in your question, not in a comment. — Dave L. Renfro, Oct 12 '22 at 15:56
You can't prove something which is false. $1+2+3+\dots$ does not equal $-\frac{1}{12}$. Now if you want to interpret the summation in some other nonstandard sense that's fine, but you need to say so, because in the normal sense it's just not true . — Lorago, Oct 12 '22 at 15:59
@Lorago I would say it differently, $1+2+3 + ... $ does not CONVERGE to $-\frac{1}{12}$ but it certainly EQUALS it. — Sidharth Ghoshal, Oct 12 '22 at 16:02
@DaveL.Renfro I see what you mean, I guess then I'm more importantly intrested in expressing the Zeta-function as a Mellin-transform of the Theta-function and then to get $-\frac{1}{12}$ from $s=-1$ — Batmannilsson, Oct 12 '22 at 16:04
@Batmannilsson wanting to prove that $\zeta(-1)=-\frac{1}{12}$ makes more sense. Remember that $\zeta(-1)$ is defined through analytic continuation, and not as $1+2+3+\dots$. If you update your question accordingly, then I will change my downvote to an upvote — Lorago, Oct 12 '22 at 16:16
That such assignments are called "summation-methods" is unlucky because we do not actually sum up something. We evaluate a function at a point where the corresponding sum does not converge anymore. Even the assignment (the value is assigned to the sum) does not really make sense. It is like claiming that one can buy a plane and ignore that one does not have the money for it. — Peter, Oct 12 '22 at 16:21
@DaveL.Renfro Ok, thanks for the correction. I'm a physics student if that grants any sympathy — Batmannilsson, Oct 12 '22 at 16:24
I'm a physics student if that grants any sympathy --- You probably know a lot more about this topic than I do, which is almost nothing (although I am slightly familiar with various summability ideas, but mostly matrix summability methods). The only reason I know it's $-\frac{1}{12}$ is that this has been coming up in internet math groups (and even in "popular press" internet articles and in people's blogs) fairly continuously for at least the last 15 years, and probably much longer. — Dave L. Renfro, Oct 12 '22 at 16:40
@Lorago does 1+2+4+8... NOT EQUAL -1 as well? What are you trying to say here by insisting 1+2+3+4... not equal to $-\frac{1}{12}$? It's just a divergent series, and it HAS a normalized value and whenever you encounter this divergent series as part of a finite calculation you can use the normalized value instead. In fact you DONT EVEN NEED the Riemann Zeta function to prove this fact. This is just a natural consequence of the existence of taylor series. See my answer here: https://math.stackexchange.com/questions/39802/to-sum-123-cdots-to-frac112/4507524#4507524 — Sidharth Ghoshal, Oct 12 '22 at 17:33
for the sake of people skimming here $1+2+3+...$ is also just the linear term of the $\log$-series expansion of $\frac{1}{1-x}$ which has nothing to do with riemann zeta function and so forcing the OP to restate prove $1+2+3... = -\frac{1}{12}$ instead as $\zeta(-1) = -\frac{1}{12}$ is quite dishonest. — Sidharth Ghoshal, Oct 12 '22 at 17:40
@SidharthGhoshal No. Summation methods are completely arbitrary. We pick natural ones because we're human and it's nice... but I could just as well say $1+2+3+\cdots:=\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}$, since it is an assignment. It cannot be proven. You can disagree with my "summation", but you can't disprove it! $\zeta(-1)=-\frac{1}{12}$ can be proven, as both the LHS and RHS are well-defined and unambiguous — FShrike, Oct 12 '22 at 17:49
@FShrike If you insisted $1+2+3+ ... = \text{Anything Else}$ then you would rapidly run into contradictions, for example the linear $x$ term taylor series of $\frac{1}{1-e^x}$ would suggest that $-\frac{1}{12} = \text{Anything Else}$ and that simply is NOT true. A world where $1+2+3+... \ne -\frac{1}{12}$ can certainly be made, but in SUCH a world your usual assumptions about addition are all false. So yes: your summations are arbitrary but the choice of method has consequences that have ramifications in divergent series whether you intentionally wanted it to or not. — Sidharth Ghoshal, Oct 12 '22 at 17:52
@SidharthGhoshal in the link you provided, you are very careless with when you're allowed to rearrange infinite series and when you are not. Recall that a series is unconditionally convergent (i.e. you're allowed to freely rearrange it) if and only if it is absolutely convergent. Furthermore, what you are saying is just plain wrong. The way you define $1+2+3+\dots$ in the normal sense is as the limit $\lim_{N\to\infty}\sum_{n=1}^N n$, in which case it is almost trivial to show that $1+2+3+\dots=\infty$ (in the sense of a limit). — Lorago, Oct 12 '22 at 22:19
FShrike and Lorago are right. $1+2+3+\cdots =-1/12$ is nonsense as well as $1+2+4+8+\cdots=-1$. Again , we compute the value where the generating sum does not converge anymore , and at that point the sum cannot be used, so the value is only the value of the function , not of the generating sum. — Peter, Oct 13 '22 at 11:20

score 4 · Accepted Answer · answered Oct 13 '22 at 23:40

I'll try and give you some hints to proceed: note that the identity $${ \Gamma \left({\frac {s}{2}}\right)\pi ^{-{\frac {s}{2}}}\zeta (s)={\frac {1}{2}}\int _{0}^{\infty }{\bigl (}\vartheta (0;it)-1{\bigr )}t^{\frac {s}{2}}{\frac {\mathrm {d} t}{t}}}$$ holds only when $\text{Re}(s) > 1$, since the equality stems from the infinite series definition of $\zeta$: if you are confused, try a change of variables in ${\displaystyle {\frac {1}{2}}\int _{0}^{\infty }{\bigl (}e^{-n^2\pi t}{\bigr )}t^{\frac {s}{2}}{\frac {\mathrm {d} t}{t}}}$, which should come out to be $\Gamma\left(\frac{s}{2}\right)\pi^{-s/2}n^{-s}$.

Try to split the integral into two over the intervals $(0, 1)$ and $(1, \infty)$ and apply Poisson summation to get an expression that is now convergent for $\text{Re}(s) > 0$ and which is symmetric in $s$, $1-s$. Evaluate at $s = 2$ and win by analytic continuation.

Thank you! I'll tackle this problem again shortly and accept your answer as a solution when i solve it. — Batmannilsson, Oct 14 '22 at 14:02
I solved the problem with the change of variables $s\rightarrow1-s$ and then i just integrated the exponential and let Mathematica compute the rest and it gave me the right result. — Batmannilsson, Oct 14 '22 at 15:36

Mellin transform of theta function $\theta$ to show $\zeta(-1)=-\frac{1}{12}$

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