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]]
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.