Integrating the formula for the sum of the first $n$ natural numbers I was messing around with some math formulas today and came up with a result that I found pretty neat, and I would appreciate it if anyone could explain it to me. 
The formula for an infinite arithmetic sum is $$\sum_{i=1}^{n}a_i=\frac{n(a_1+a_n)}{2},$$ so if you want to find the sum of the natural numbers from $1$ to $n$, this equation becomes $$\frac{n^2+n}{2},$$ and the roots of this quadratic are at $n=-1$ and $0$. What I find really interesting is that $$\int_{-1}^0 \frac{n^2+n}{2}dn=-\frac{1}{12}$$ There are a lot of people who claim that the sum of all natural numbers is $-\frac{1}{12}$, so I was wondering if this result is a complete coincidence or if there's something else to glean from it. 
 A: We have Faulhaber's formula:
$$\sum_{k=1}^n k^p = \frac1{p+1}\sum_{j=0}^p (-1)^j\binom{p+1}jB_jn^{p+1-j},~\mbox{where}~B_1=-\frac12$$
$$\implies f_p(x)=\frac1{p+1}\sum_{j=0}^p(-1)^j\binom{p+1}jB_jx^{p+1-j}$$
We integrate the RHS from $-1$ to $0$ to get
$$I_p=\int_{-1}^0f_p(x)~\mathrm dx=\frac{(-1)^p}{p+1}\sum_{j=0}^p\binom{p+1}j\frac{B_j}{p+2-j}$$
Using the recursive definition of the Bernoulli numbers,
$$I_p=(-1)^p\frac{B_{p+1}}{p+1}=-\frac{B_{p+1}}{p+1}$$
Using the well known relation $B_p=-p\zeta(1-p)$, we get
$$I_p=\zeta(-p)$$
So no coincidence here!
A: Funny that you should use that method to demonstrate that the sum of all natural numbers is $-\frac{1}{12}$. I've seen the same claim demonstrated in a completely different way.
In case you aren't familiar with it, the Riemann Zeta function is defined as
$$\zeta(s)=\sum_{n=1}^\infty n^{-s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+...$$
Your infinite sum of positive integers is given by $\zeta(-1)$.
The Riemann zeta function satisfies the reflection formula
$$\zeta(1-s)=2(2\pi)^{-s}\cos\big(\frac{\pi s}{2}\big)\Gamma(s)\zeta(s)$$
Which suggests the same thing, if you evaluate it at $s=2$:
$$\zeta(1-2)=2(2\pi)^{-2}\cos\big(\frac{2\pi}{2}\big)\Gamma(2)\zeta(2)$$
$$\zeta(-1)=\frac{1}{2\pi^2}\cos(\pi)\Gamma(2)\zeta(2)$$
$$\zeta(-1)=\frac{1}{2\pi^2}\cdot-1\cdot 1! \cdot \frac{\pi^2}{6}$$
$$\zeta(-1)=-\frac{1}{12}$$
However, I am somewhat inclined to think that this is merely a coincidence, as strange as that may sound. This is because the other "strange" values that can be obtained using the $\zeta$ function, such as
$$\zeta(0)=-\frac{1}{2}$$
$$\zeta(-2)=0$$
are not "confirmed" or obtained from your integral in any analogous way. I would advise you to look for some ways to produce these values from your integral so that you (and any more potential answerers) will have more to work with.
Just because the integral of your function between its two zeroes is equal to this magical number does not mean that there is any sort of connection. For example, consider the formula for the sum of the first $n$ square numbers:
$$\frac{n(n+1)(2n+1)}{6}$$
This has zeroes at $0, -\frac{1}{2},$ and $1$. When I take the integral from $-1$ to $0$ of this, I get
$$\int_{-1}^0 \frac{n(n+1)(2n+1)}{6}dn=0$$
Which is equal to our strange value for $\zeta(-2)$, which also is a sum of squares. But wait, there's more! 
$\zeta(0)$ is the sum of $1+1+1+...$. The partial sum for this would be the sum of $1$, $n$ times, and so this sum is given by $n$. And guess what?
$$\int_{-1}^0 n dn=-\frac{1}{2}$$
Which also agrees with our strange value for $\zeta(0)$.
And this just keeps working! $\zeta(-3)$ is, supposedly, $\frac{1}{120}$. The sum of the first $n$ cubes has the formula
$$\frac{n^2(n+1)^2}{4}$$
and guess what?
$$\int_{-1}^0 \frac{n^2(n+1)^2}{4} dn=\frac{1}{120}$$
Now I've given you lots of examples to work with, and perhaps to draft a proof that the "weird" values of $\zeta(-s)$ is equal to the integral, from $-1$ to $0$, of the formula for the sum of the first $n$ $s$th powers. And hopefully this answer was helpful enough to you to be deserving of that hefty $50$-point bounty... ahem...
A: We can generalize this to general summations for convergent and divergent series as follows. For a convergent series we have:
$$\sum_{k=1}^{\infty}f(k) = \int_p^{\infty}f(x) dx + \int_{p-1}^p S(x) dx\tag{1}$$
where $p$ is an arbitrary real or complex number and $S(x)$ is the analytically continued partial sum defined by analytically continuing the partial sum:
$$S(N) = \sum_{k=1}^N f(k)\tag{2}$$
from the integers to the complex plane using Carlson's theorem and imposing the conditions assumed in that theorem. It then follows that $f(x)$ also needs to be defined using Carlson's theorem if it is only specified for integer arguments and it then follows from (2) that $f(x)$ is given in terms of $S(x)$ via:
$$f(x) = S(x) - S(x-1)$$
To prove (1), we write:
$$ \sum_{k=0}^{\infty} f(k) = \lim_{x\to \infty}  S(x)$$
The first integral on the r.h.s. of (6) can be written as:
$$\begin{split}\int_p^{\infty}f(x) dx &= \lim_{R\to\infty} \left(\int_p^{R}S(x)dx - \int_{p-1}^{R-1}S(x) dx\right)\\ &= \lim_{R\to\infty} \int_{R-1}^R S(x) dx - \int_{p-1}^p S(x) dx\end{split}$$
The integral from $R-1$ to $R$ of $S(x)$ obviously tends to the same limit for $R$ to infinity as $S(x)$ for $x$ to infinity. The last term in (1) cancels out the integral from $p-1$ to $p$ in the above expression.
For a divergent series we can use (1) to evaluate the series by appealing to analytical continuation without going about doing this explicitly. The idea is then that one can imagine modifying the function $f(x)$ using a parameter, so that the summation would become convergent for some range of that parameter, and then (1) would hold. For the function at hand, the summation is divergent and the integral is then  also divergent per the integral test. So, if we cut off the integral at $R$ then the large R asymptotic will contain terms that tend to infinity.
Analytic continuation to the domain where the summation converges must then change the diverging terms into converging terms. Analytic continuation of the summation to infinity back to the starting point therefore amounts to dropping the terms that tend to infinity as a function of $R$. This means that the integral in (1) should be cut off to $R$ and we should then take the constant part of that integral. The formula for the sum of a divergent series then becomes:
$$\sum_{k=1}^{\infty}f(k) = \text{Constant part of }\int_{p}^{R}f(x) dx + \int_{p-1}^pS(x) dx$$
where "constant part" means $R$-independent part. In the case at hand the constant term will vanish for $p = 0$, therefore the summation will be given by $\int_{-1}^0S(x)dx$.
