Riemann Zeta function's analytic continuation relationship with simple closed formulas Consider the famous output of the analytic continuation of the Riemann zeta function at negative values :
$$
\zeta(-1) = 1+2+3+4... \hspace{3mm} =  \dfrac{-1}{12}
$$
Now consider the closed formula which Gauss discovered for the sum of the first $n$ integers($i,n \in \mathbb{N}$):
$$
\sum_{i=1}^{n}i = 1+2+3+4...+n =  \dfrac{n(n+1)}{2} = \dfrac{n^2}{2} + \dfrac{n}{2}
$$
We Find the roots of the closed form ( using my special analytic Theorem which is based on the principal : " I decide the function is continuous now ") :
$$
n_1 = -1
$$
$$
n_2 = 0
$$
Now we find the right to left definite integral between the two roots (by leaning on the same analytic method I described above):
$$
\int_{0}^{-1} \dfrac{n^2}{2}+\dfrac{n}{2} dn  =  \dfrac{n^3}{6}+\dfrac{n^2}{4}\Biggr|_{0}^{-1} = 0 - \left(\dfrac{-1}{6}+\dfrac{1}{4}\right) = \dfrac{-1}{12}
$$
I can't prove it for all negative values of $\zeta$, I have verified the same relationship is true for $\zeta(-3)$ and the closed form of $n^3$ which integrates into $\dfrac{1}{120}$ if we use my "special analytic method".
Now on a more serious note, can anybody please explain this to me? When I look at the functional equation for the analytic continuation of the zeta function, it seems like an impenetrable fortress, yet it turns out it is possible to produce these results with simple (illogical) assumptions regarding continuity of discrete functions?
 A: The general formula for the partial sums of the $n^{th}$ powers of integers can be written as
$$\displaystyle S(n,x)=\sum_{i=1}^{x} i^n
=\frac{(-1)^n B_{n+1}}{n+1 } \\+ \frac{ \sum_{j=0}^{n+1} \,B_{n+1 - j} \,  \binom {n+1}{j} (x+1)^j    }{n+1} \tag{*}$$
where $B_z$ indicates the $z^{th}$ Bernoulli number. Substituting $n$ with integer values in (*), we get the classical formulas
$$ S(1,x)=\sum_{i=1}^{x} i^1=\frac 12 x(x+1)$$
$$ S(2,x)=\sum_{i=1}^{x} i^2=\frac 16 x(2x^2+3x+1)$$
$$ S(3,x)=\sum_{i=1}^{x} i^3=\frac 14 x^2(x+1)^2\\...$$
and so on. The first term of the RHS in (*) corresponds to the explicit value of the zeta function for negative integers, given by
$${\displaystyle \zeta (-n)={\frac{(-1)^n \,B_{n+1}}{n+1}}}$$
So we have
$$\displaystyle S(n,x)
=\zeta(-n) \\ + \frac{ \sum_{j=0}^{n+1} \,B_{n+1 - j} \,  \binom {n+1}{j} (x+1)^j    }{n+1} \tag{**}$$
Now we can take the definite integral. To avoid confusion, I assume that the definite integral described in the OP corresponds to that calculated in the interval from $-1$ to $0$ (or equivalently, to the opposite value of that calculated in the range from $0$ to $-1$). Accordingly, this is what has been done in the OP to obtain $$-\int_0^{-1} \left(\frac{x^2}{2}- \frac x2 \right) dx \\ =-(-\frac 16 + \frac 14 )=-\frac{1}{12}$$
Taking the definite integral, the equation (**) reduces to
$$\displaystyle \int_{-1}^{0} S(n,x)
=\zeta(-n) \\ + \sum_{j=0}^{n+1} \frac{ B_{n+1-j} \, \binom {n+1}{j}   }{(n+1)(j+1)}$$
So the hypothesis proposed in the OP, that is to say
$$\displaystyle \int_{-1}^{0} S(n,x)
=\zeta(-n) $$
is true for any $n$ satisfying
$$ \sum_{j=0}^{n+1} \frac{ B_{n+1-j}  \binom {n+1}{j}  }{(n+1)(j+1)}=0$$
or equivalently
$$ \sum_{j=0}^{n+1} \frac{ \,B_{n+1-j}  \binom {n+1}{n+1-j}  }{j+1}=0$$
Now we can make the substititions $N=n+1$ and $K=n+1-j$ to get
$$ \sum_{K=0}^{N} \frac{ \,B_{K}  \binom {N}{K}  }{(N-K+1)}=0$$
Multiplying both terms of the fraction to $N+1$, we get
$$ \sum_{K=0}^{N} \frac{ \,B_{K} (N+1)N! }{(N+1)K!(N-K)!(N-K+1)}\\
=\sum_{K=0}^{N} \frac{ \,B_{K} (N+1)! }{(N+1)K!(N-K+1)!}\\
= \frac{1}{N+1} \sum_{K=0}^{N}  \,B_{K}  \binom {N+1}{K} =0$$
which is true for any positive integer $N$ since a well known property of Bernoulli numbers is
$$\sum_{K=0}^{M-1}  \,B_{K}  \binom {M}{K} =0$$
