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            052184472Xc09  CUNY148/Severini  June 2, 2005  12:8





                            278                       Approximation of Integrals

                            so that
                                                                1

                                                     y (0) =          =−L
                                                           −g (ˆ y)y (0)


                            and, hence, that
                                                                 1
                                                           2
                                                          L =        .
                                                               −g (ˆ y)


                              Finally, note that u and g (y(u)) have the same sign so that L ≥ 0. It follows that
                                                                 1
                                                         L =
                                                                     1

                                                              [−g (ˆ y)] 2
                            and
                                                                 h(ˆ y)
                                                        ¯
                                                        h(0) =        1  .

                                                              [−g (ˆ y)] 2
                            Hence, if |h(ˆ y)| > 0, then
                                                               √
                                         u(b)        n           2   1   h(ˆ y)        1
                                           ¯
                                           h(u)exp − u  2  du =  √   2        1  + O   3

                                       u(a)          2             n   [−g (ˆ y)] 2   n 2
                            so that
                                                       √
                                                        (2π)h(ˆ y)      −1
                                         I n = exp{ng(ˆ y)}    1  [1 + O(n )] as n →∞.

                                                       [−ng (ˆ y)] 2
                            If h(ˆ y) = 0, then
                                                   u(b)        n             1
                                                     ¯
                                                    h(u)exp − u  2  du = O   3
                                                 u(a)          2            n 2
                            so that

                                                               1
                                              I n = exp{ng(ˆ y)}O   as n →∞.
                                                                3
                                                              n 2
                            Example 9.8 (Stirlings approximation). Consider approximation of the gamma function
                             (z + 1) for large values of z. Note that, using the change-of-variable y = t/z,wemay
                            write
                                                ∞                    ∞

                                                   z
                                      (z + 1) =   t exp(−t) dt = z z+1  exp{−z[y − log(y)]} dy.
                                               0                    0
                            Hence, we may apply Theorem 9.14 with g(y) = log(y) − y, n = z, and h(y) = 1. It is
                            straightforward to show that the conditions of Theorem 9.14 are satisfied with ˆ y = 1,
                            leading to the result that
                                                 √     z+1/2              −1
                                        (z + 1) =  (2π)z    exp(−z)[1 + O(z )] as z →∞.
                            Since  (z) =  (z + 1)/z, Stirling’s approximation to  (z)isgiven by
                                               √     z−1/2              −1
                                          (z) =  (2π)z    exp(−z)[1 + O(z )] as z →∞.
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