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We can define a set of numbers, the so called Bernoulli Numbers, via the equation,
 |
(2.1) |
Complex analysis allows us to extract the
 |
(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,
![$\displaystyle B_n = \frac{n!}{2\pi i}(-2\pi i)\sum_{p} Res[\pm p2\pi i]$](img11.png) |
(2.3) |
where
![$\displaystyle Res(f;z_0)=\lim_{z\rightarrow z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mf(z)]$](img12.png) |
(2.4) |
With an application of l'Hopital's rule,
 |
(2.5) |
Thus we note that the odd residues beyond one will cancel each other out
 |
(2.6) |
For example,
, and so on.
Bernoulli functions are derived in the same way, defined by
 |
(2.7) |
This is easily solved by expanding the extra
factor and relating the results to the previous
results for the Bernoulli numbers, i.e.,
 |
(2.8) |
For example,
 |
(2.9) |
Next: Euler-Maclaurin Integration Formula
Up: The legal way I:
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2006-09-15