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It seems that the $n$th cumulant of the uniform distribution on the interval $[-1,0]$ is $B_n/n$, where $B_n$ is the $n$th Bernoulli number.

And also $-\zeta(1-n) = B_n/n$, where $\zeta$ is Riemann's $\zeta$ function.

Is there some reason why one should expect these to be the same, as opposed to proofs that convince you that they are?

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Since we're looking at the analytic continuation of $\zeta$, the usual $p$-series concept is out the window - except through formal regularization. – anon Sep 14 '11 at 20:15
Also posted to MathOverflow. – Gerry Myerson Sep 16 '11 at 22:50

The moment generating function of $U([-1,0])$ is $\mathbb{E}(\mathrm{e}^{t u}) = \int_0^1 \mathrm{e}^{-t v} \mathrm{d} v = \frac{1}{t} \left(1 - \mathrm{e}^{-t}\right) $.

The cumulant generating function is the logarithm of the moment generating function: $\mathcal{c}(t) = \log \left(\frac{1}{t} \left(1 - \mathrm{e}^{-t}\right) \right)$.

Now, find $t \frac{\mathrm{d}}{\mathrm{d} t} \left( \mathcal{c}(t) \right)$: $$ t \mathcal{c}^\prime(t) = -1 + \frac{t}{\mathrm{e}^t-1} = \sum_{n \ge 1} \frac{t^n}{n!} B_n $$ And therefore, integrating term-wise we get: $$ \mathcal{c}(t) = \sum_{n \ge 1} \frac{t^n}{n!} \frac{B_n}{n} $$

Of course, this is more of the derivation side...

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This answer doesn't mention the $\zeta$ function. Is there some way to show the values of that function at negative integers are $-1$ times the cumulants, without deriving the two separately and observing the answers are the same? – Michael Hardy Sep 14 '11 at 22:26
Hi @MichaelHardy: Have you found an appropriate answer for your question? If not, please have a look here: Btw, I have a related question:… I guess you know this, don't you? – ltt Nov 21 '13 at 21:52

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