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So, I'm not a big expert in this subject but I know $1+2+3...=-\dfrac{1}{12}$ isn't to do with 'real' maths but it's all to do with the zeta function; however I was watching a maths video and the equation:

$$ \frac{x(x+1)}{2} $$

... is actually a perfect equation for the series $1+2+3...$ etc. where $x$ represents $n$ in a series and $y$ is the sum of the series up to $n$. So, you can conclude that:

$$ \sum^{n}_{i=1}1+2+3...=\frac{x(x+1)}{2} $$

However, this is where it gets weird; as you have probably guessed, the roots of the equation is $x=0,-1$ but if I want to find the integral of the roots from $-1$ to $0$ which is under the $x$ axis, I get the following:

$$ \int_{-1}^{0} \frac{x(x+1)}{2}\:dx=-\frac{1}{12} $$

So, my question is why is this the case; what connection is there between the value of the integral under the $x$ axis that this graph has compared to the summation of the series?

Link to Desmos graph for more clarity

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    I am not the downvoter, but I know that there are really many posts here about this topic, e.g., here and more than $100$ posts linked to it. More or less you should find everything here on this site already. – Dietrich Burde Apr 22 '20 at 11:23
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    This https://math.stackexchange.com/a/2350361/42969 might be what you are looking for. – Martin R Apr 22 '20 at 11:25
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    Please formulate the sum with more care (expression within the sum i, result is n(n+1)/2). And an area is never negative, the corresponding integral can of course be negative. – Peter Apr 22 '20 at 11:29
  • @Peter Yes fair point, edited for more clarity. –  Apr 22 '20 at 11:33
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    For OP: I believe the downvotes are a knee-jerk reaction for people here who are (understandably) fed up with questions regarding misconceptions about the $-1/12$ thing. That said, I think the question did show effort from OP and was not poorly written (writing $x$ instead of $n$ in $x(x+1)/2$ is just a math mistake the post itself should not be penalized for). So even if I have no interest in discussing this topic, I have upvoted this. – Ivo Terek Apr 22 '20 at 12:40

1 Answers1

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Your result for this integral is just a consequence of a general formula wich gives the value of the zeta function at negative integer. Integrating Faulhabert* (or maybe bernoulli's, i'm not sure) formula between -1 and 0 leads to the formula.

*The formula is the one that gives the n-th partial sum of k-th power of integers in term of polynomials of degree k+1.

Guillaume Barbey
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