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P. 64

Infinitely Many Poles
                .

                   has a simple pole at s = 0. Moreover, 1 + e          as  = 0 gives


                                       e as  = −1 = e  (2n−1)πi , n = 0, ±1, ±2, . . . ,

                   implying that
                                                 2n − 1
                                         s = (           )πi, n = 0, ±1, ±2, . . . ,
                                           n
                                                    a
                   are poles of F(s).
                                        as
                   For G(s) = 1 + e , G (s ) = −a ̸= 0, which means that the poles s are
                                              ′
                                                 n
                                                                                                      n
                   simple. Furthermore,
                                                                              1
                                                                  ts
                                               Res(0) = lim se F(s) = ,
                                                           s→0                2
                                                    e ts n            e ts n        e t(  2n−1 )πi
                                                                                         a
                              Res(s ) =                         =           = −                .
                                      n
                                            [s(1 + e )]             as e as n      (2n − 1)πi
                                                     as ′
                                                                       n
                                                           s=s n
                   Consequently, sum of residues =
                               ∞                                        ∞
                        1     ∑           1         2n−1       1    2  ∑       1        2n − 1
                    =     −                      e t(  a  )πi  =  −                 sin         πt. (4.15)
                        2           (2n − 1)πi                 2    π      2n − 1           a
                             n=−∞                                      n=1
                                                                                  iθ
                   Finally, let C be the semicircle given by s = R e , with R = 2nπ/a.
                                                                                n
                                                                                              n
                                   n
                   To make the subsequent reasoning simpler, let us take a = 1. Then the
                   circles C cross the y-axis at the points s = ±2nπi. (See Figure 4.4.) We
                             n
                   wish to consider what happens to the points s on C under the mapping
                                                                                   n
                                   s
                   H(s) = 1 + e .
                     1. In the region 0 < x ≤ x and s on C , the image points ζ = H(s) =
                                                      0
                                                                      n
                               x iy
                        1 + e e all lie to the right and slightly below the point ζ = 2, for y
                        sufficiently close to 2nπ, that is, for n sufficiently large. (Notice that
                        as n increases, the circles C flatten out so that y = Im(s) approaches
                                                          n
                        2nπ from below.) Hence

                                                                  s
                                                            |1 + e | ≥ 2

                        for 0 ≤ x ≤ x .
                                         0
                     2. For s on C with Re(s) = x < 0, the values of the function H(s) =
                                      n
                               x iy
                        1 + e e lie inside the circle |ζ − 1| = 1. As the value of y = Im(s)
                        goes from 2nπ down to (2n − 1)π, the images spiral half a revolution
                        with modulus

                                              s
                                       |1 + e | ≥ 1 + e   2nπ cos φ  cos(2nπ sin φ) > b > 0,

                        for x = 2nπ cos φ, y = 2nπ sin φ. As y = Im(s) goes from (2n − 1)π
                        down to (2n − 2)π, the images H(s) spiral away from the origin half

               . . . .  a revolution. For y < 0, it is the same story but spiralling outward.


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