For an arbitrary [beta] > 0, the residue theorem
and (2.3) give
106]), we compute the Mellin-Barnes integral using the residue theorem
. The proof of the following lemma has two steps.
We apply the Cauchy residue theorem
as follows: Take a rectangle with vertices at s = c + it, - T < t < T, s = [sigma] + iT, - a < [sigma] < c, s = - a + it, - T < t < T and s = [sigma] - iT, - a < [sigma] < c, where T > 0 is to mean [T.sub.1] > 0 and [T.sub.2] > 0 tending to [infinity] independently but we usually use this convention.
On the other hand, applying Residue theorem
and the residue (4.8),
and the statement follows from the Residue theorem
since the only poles in U of the integrand are the zeros of the denominator and the zeroes of the denominator of [[phi].sup.2](z).
It also discusses elementary aspects of complex analysis such as the Cauchy integral theorem, the residue theorem
, Laurent series, the Riemann mapping theorem, and more advanced material selected from Riemann surface theory.
By virtue of the residue theorem
of elliptic functions, we have
The new algorithm uses directly the residue theorem
in one complex variable, which can be applied more efficiently as a consequence of a rich poset structure on the set of poles of the associated rational generating function for E([alpha])(t) (see Subsection 2.2).
The integral representation can also be obtained by folding the contour to the left, running from -[infinity] back to -[infinity] enclosing the point-[D/2]-1 in an anticlockwise sense and applying the residue theorem
Then the integral can be evaluated by closing the contour in the upper half of the [[xi].sub.3]-plane and by using Cauchy's residue theorem
. This gives
In the same way as in , for 0 [less than or equal to] [alpha] < [pi]/2, using the residue theorem
, we get
Shifting the contour to the segment from [[sigma].sub.0] - iT to [[sigma].sub.0] + iT([3/16] < [[sigma].sub.0] < [1/5]), by the residue theorem
, we have