Calculus: Analytic Geometry and Calculus, with Vectors

12.6 Euler-Maclaurin summation formulas 645

Remark: We proceed to show how this formula can be used to derive % ery impor-

tant formulas involving factorials. Putting z = 0 in (1) gives

(2) log n! _ (n + V) log n - n + 1

B,

)9 L^11

• 1 +j (m 1)!Bm(x)dx
  ,2 (J - 1 n -^1 x

One of the truly great mathematical discoveries is the fact that (2) can be

improved with the aid of the formula

(3) lim(2n)! / -

n.-.m 22'a(n!)2 -li

see Problems 9.59, Problem 6, equation (13). From (3) we obtain

(4) Jim {log (2n)! + log - 2n log 2 - 2 log n! } = 0.

n-' m

In this formula we substitute the expression for log n! given in (2) and the expres-

sion for log (2n)! obtained by replacing n by 2n in (2). The result should not

overwhelm us, because we can overwhelm it. Many of the terms cancel out, the

remaining ones have limits, and (4) reduces to

(5) log -1+

B, - ( (m-1)!Bm(x)dx=0.

,=2(9-1).1^1 X.

Since

(6) ('n (m - 1)!Bm(x)

dx

1)!Bm(x)

,J 1 xm 11 xm dx

1)!Bm(x)

dx

f 1)1Bm(x)

-fn x'" -JO (x+n)m'

adding the left side of (5) to the right side of (2) gives the remarkable Stirling
formula

(7) log n! = log '/2a + (n + 4) log n - n

m B, 1

I

(m - 1)!Bm(x)

+, 2U - 1)9 n'-1-1^0 (n + x)m

dx.

To produce a Stirling formula applying to factorials of noninteger numbers, we

suppose that z > -1 and use the definition

(S)

z. -t=lim ninz

ri.

(z + 1) (z + 2) ... (z + n)'

which appeared in Section 3.3, Problem 11. This is equivalent to

(9)

n

log z! =

Jim

[log n! + z log n - E log (z + k)].

n--.- ka1