Bernoulli Numbers

We can define a set of numbers, the so called Bernoulli Numbers, via the equation,

$$\frac{x}{e^x - 1} = \sum_{n=0}^{\infty}\frac{B_n}{n!}x^2$$ (2.1)

Complex analysis allows us to extract the $B_n \ni$

$$B_n = \frac{n!}{2\pi i}\oint_C \frac{1}{e^z - 1} \frac{dz}{z^n}$$ (2.2)

We can construct a contour which avoids the central pole (wrapping over the positive real axis, around a clockwise infinitesimal circle centered at the origin, and back along the positive real axis to rejoin the the main contour of clockwise orientation. We then have,

$$B_n = \frac{n!}{2\pi i}(-2\pi i)\sum_{p} Res[\pm p2\pi i]$$ (2.3)

where

$$Res(f;z_0)=\lim_{z\rightarrow z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)]$$ (2.4)

With an application of l'Hopital's rule,

$$\lim_{z\rightarrow p2\pi i}\frac{z-p2\pi i}{e^z - 1} = \lim_{z\rightarrow p2\pi i}\frac{1}{e^z}=1$$ (2.5)

Thus we note that the odd residues beyond one will cancel each other out

$$B_n = \frac{n!}{2\pi i } (-2\pi i) 2 \sum_{p=1}^{\infty}\frac{1}{... ... = -\frac{(-1)^{n/2}2n!}{4\pi^2}\left(\sum^{\infty}_{p=1}p^{-n}=\zeta(n)\right)$$ (2.6)

For example, $B_0=1, B_1=-1/2, B_2=1/6, B_3=0, B_4=-1/30$, and so on. Bernoulli functions are derived in the same way, defined by

$$\frac{x e^{xs}}{e^x - 1} = \sum^{\infty}_{n=0}B_n(s)\frac{x^n}{n!}$$ (2.7)

This is easily solved by expanding the extra $e^{xs}$ factor and relating the results to the previous results for the Bernoulli numbers, i.e.,

$$B_n(s)=\frac{n!}{2\pi i}\oint \frac{e^z}{e^z -1}\frac{dz}{z^n} = ... ...eft(\frac{1}{z^n} + \frac{s}{z^{n-1}} + \frac{s^2}{2!z^{n-2}} + \cdots\right)dz$$ (2.8)

For example,

$$B_3(s) = B_3 + B_2 \frac{3!}{2!} s + B_1 \frac{3!}{1!2!}s^2 + s^3 \frac{3!}{0!3!}B_0 = s^3 -\frac{3}{2}s^2 + \frac{1}{2}s$$ (2.9)