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 [2], 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