Using standard asymptotic methods, Gamma(x+1,x) |\oo -t x = | e t dt \| x x -x |\oo -x(s-log(s+1)) = x e x | e ds [t = x(s+1)] \| 0 x -x |\oo -xu^2/2 = x e x | e s' du [u^2/2 = s-log(s+1)] [1] \| 0 Compare this to Gamma(x+1) |\oo -t x = | e t dt \| 0 x -x |\oo -x(s-log(s+1)) = x e x | e ds [t = x(s+1)] \| -1 x -x |\oo -xu^2/2 = x e x | e s' du [u^2/2 = s-log(s+1)] [2] \|-oo Working from ss' = (1+s)u in a fairly straightforward manner, we get oo --- k s = > a u [3] --- k k=0 where a_0 = 0, a_1 = 1, and for k > 1, k-2 1 1 --- a = --- a - - > a a [4] k k+1 k-1 2 --- k-j j+1 j=1 Recursion [4] gives that 1 2 1 3 1 4 1 5 1 6 s = u + - u + -- u - --- u + ---- u + ----- u + ... 3 36 270 4320 17010 therefore, 2 1 2 2 3 1 4 1 5 s' = 1 + - u + -- u - --- u + --- u + ---- u + ... [5] 3 12 135 864 2835 In the asymptotic expansion of [2], the odd order terms of s' get eliminated; however, in the asymptotic expansion of [1], they do not. Thus, to get the terms in the asymptotic expansion of [2], we would throw out the odd order terms in s' and double the even order terms. To aid in using [1] with [5], we compute |\oo -bu^2 k 1 |\oo -bu^2 k-1 | e u du = - | e u d(u^2) \| 0 2 \| 0 1 |\oo -bu (k-1)/2 = - | e u du 2 \| 0 1 -(k+1)/2 k+1 = - b Gamma(---) [6] 2 2 Putting together [1], [5], and [6], we get Gamma(x+1,x) |\oo -t x = | e t dt \| x x -x x 1/2 2 1 ~ x e - [ (2/x) Gamma(1/2) + - (2/x) Gamma(1) 2 3 1 3/2 2 2 + -- (2/x) Gamma(3/2) - --- (2/x) Gamma(2) 12 135 1 5/2 1 3 + --- (2/x) Gamma(5/2) + ---- (2/x) Gamma(3) ] 864 2835 x -x 1 2 1 = x e [ - sqrt(2 pi x) + - + --- sqrt(2 pi x) 2 3 24x 4 1 8 - ---- + ------ sqrt(2 pi x) + ------- ] 135x 576x^2 2835x^2 x -x 1 1 1 = x e sqrt(2 pi x) [ - + --- + ------ + ... ] 2 24x 576x^2 x -x 2 4 8 + x e [ - - ---- + ------- + ... ] [7] 3 135x 2835x^2 We have collected the first half of [7] to contain those terms that come from the even order terms of s' and the second half to contain the terms that come from the odd order terms of s'. By tossing out the terms that come from the odd order terms of s' and doubling, we get, as mentioned earlier, Stirling's Formula: Gamma(x+1) |\oo -t x = | e t dt \| 0 x -x 1 1 ~ x e sqrt(2 pi x) [ 1 + --- + ------ + ... ] [8] 12x 288x^2 Combining [7] and [8] we get that the ratio of Gamma(x+1,x)/Gamma(x+1) 1 1 2 23 23 ~ - + ------------ [ - - ---- + ------- + ... ] [9] 2 sqrt(2 pi x) 3 270x 3024x^2 Rob Johnson