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300                             Appendix D

                             The function ö n (î) as de®ned by equation (D.15) then becomes
                                                        (ÿi) n    î =2 …  1  ÿ(s =4)‡iîs n
                                                                            2
                                                                   2
                                            ö n (î) ˆ            e       e        s ds         (D:17)
                                                             ð
                                                    2(2 ðn!) 1=2 1=4  ÿ1
                                                       n
                               We now evaluate the summation
                                                           1
                                                          X
                                                             ö n (î)ö n (î9)
                                                          nˆ0
                             by substituting equation (D.17) twice, once with the dummy variable of integration s
                             and once with s replaced by t. Since the functions ö n (î) are real, they equal their
                             complex conjugates. These substitutions give
                               1                           …  1  …  1              1      n
                              X                 1   (î ‡î9 )=2     ÿ[(s ‡t )=4]‡i(îs‡î9t)  X (ÿ1)  n
                                                      2
                                                                        2
                                                                      2
                                                        2
                                  ö n (î)ö n (î9) ˆ  e            e                        (st) ds dt
                                              4ð 3=2                                  2 n!
                                                                                        n
                               nˆ0                          ÿ1 ÿ1                  nˆ0
                                              n
                             since (ÿi) 2n  ˆ (ÿ1) . The summation on the right-hand side is easily evaluated using
                             equation (A.1)
                                                       1      n     n
                                                      X   (ÿ1)   st     ÿst=2
                                                                    ˆ e
                                                           n!    2
                                                       nˆ0
                             Noting that
                                                        2
                                                       s ‡ t 2  st  (s ‡ t) 2
                                                              ‡   ˆ
                                                          4     2      4
                             we have
                                   1                           …  1  … 1
                                  X                 1   (î ‡î9 )=2     ÿ[(s‡t) =4]‡i(îs‡î9t)
                                                                            2
                                                            2
                                                          2
                                      ö n (î)ö n (î9) ˆ  e            e               ds dt    (D:18)
                                                  4ð 3=2
                                   nˆ0                          ÿ1 ÿ1
                             The double integral may be evaluated by introducing the new variables u and v
                                              s ‡ t       s ÿ t
                                          u ˆ     ,   v ˆ       or  s ˆ u ‡ v,  t ˆ u ÿ v
                                                2          2
                                                          ds dt ˆ 2du dv
                             The double integral is thereby factored into
                                    …               …
                                     1    2          1                          2
                                                                          e
                                   2    e ÿu ‡i(î‡î9)u  du  e i(îÿî9)v  dv ˆ 2 3 ð 1=2 ÿ(î‡î9) =4  3 2ðä(î ÿ î9)
                                     ÿ1              ÿ1
                             where the ®rst integral is evaluated by equation (A.8) and the second by (C.6).
                             Equation (D.18) now becomes
                                    1
                                    X                [(î ‡î9 )=2]ÿ[(î‡î9) =4]   (îÿî9) =4
                                                                                    2
                                                       2
                                                         2
                                                                  2
                                       ö n (î)ö n (î9) ˆ e           ä(î ÿ î9) ˆ e    ä(î ÿ î9)
                                    nˆ0
                             Applying equation (C.5e), we obtain the completeness relation for the functions ö n (î)
                                                      1
                                                     X
                                                        ö n (î)ö n (î9) ˆ ä(î ÿ î9)            (D:19)
                                                     nˆ0
                             demonstrating, according to equation (3.31), that the set ö n (î) is a complete set.
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