11.16. EXERCISES 353

17. This problem outlines an approach to Stirling’s formula following [49] and [7]. Fromthe above problems, Γ(n+1) = n! for n ≥ 0. Consider more generally Γ(x+1)for x > 0. Actually, we will always assume x > 1 since it is the limit as x → ∞

which is of interest. Γ(x+1) =∫

0 e−ttxdt. Change variables letting t = x(1+u) toobtain Γ(x+1) = xx+1e−x ∫ ∞

−1 ((1+u)e−u)x du. Next let h(u) be such that h(0) =

1 and (1+u)e−u = exp(− u2

2 h(u)). Show that the thing which works is h(u) =

2u2 (u− ln(1+u)). Use L’Hospital’s rule to verify that the limit of h(u) as u→ 0 is1. The graph of h is illustrated in the following picture. Verify that its graph is likethis, with an asymptote at u =−1 decreasing and equal to 1 at 0 and converging to 0as u→ ∞.

−1

1

Next change the variables again letting u = s√

2x . This yields, from the original

description of h

Γ(x+1) = xxe−x√

2x∫

−√

x/2exp

(−s2h

(s

√2x

))ds

For s < 1,h(

s√

2x

)> 2−2ln2 = 0.61371 so the above integrand is dominated by

e−(2−2ln2)s2. Consider the integrand in the above for s > 1. Show that the exponent

part is

(√

2√

xs− x ln

(1+ s

√2x

))

The expression(√

2√

xs− x ln(

1+ s√

2x

))is increasing in x. You can show this

by fixing s and taking a derivative with respect to x. Therefore, it is larger than√

2√

1s− ln(

1+ s√

21

)and so

exp

(−s2h

(s

√2x

))≤ exp

(−

(√

2√

1s− ln

(1+ s

√21

)))=

(1+ s√

2)

e−√

2s

Thus, there exists a dominating function for X[−√ x

2 ,∞](s)exp

(−s2h

(s√

2x

))and

these functions converge pointwise to exp(−s2

)as x→ ∞ so by the dominated con-

vergence theorem,

limx→∞

∫∞

−√

x/2exp

(−s2h

(s

√2x

))ds =

∫∞

−∞

e−s2ds =

√π

See Problem 16. This yields a general Stirling’s formula, limx→∞Γ(x+1)

xxe−x√

2x=√

π .

11.16. EXERCISES 35317. This problem outlines an approach to Stirling’s formula following [49] and [7]. Fromthe above problems, (n+ 1) =n! for n > 0. Consider more generally ['(x+ 1)for x > 0. Actually, we will always assume x > | since it is the limit as x — oowhich is of interest. P(x+1) = Jp e ‘t*dt. Change variables letting t = x(1+u) toobtain P(x+1) =x°tle™ f* ((1+u)e“)" du. Next let h(w) be such that h(0) =1 and (1+u)e~" = exp (-$ (u)) . Show that the thing which works is h(u) =5 (u—In(1+u)). Use L’Hospital’s rule to verify that the limit of h(u) as u > 0 is1. The graph of / is illustrated in the following picture. Verify that its graph is likethis, with an asymptote at u = —1 decreasing and equal to | at 0 and converging to 0as U —> ©,—1Next change the variables again letting u = sf? . This yields, from the originaldescription of hT(x+1) awe tva | exp (-* (+\/2)) dsFors <1,h (s\/?) > 2—21n2 = 0.61371 so the above integrand is dominated bye7 (2-21n2)s?part is. Consider the integrand in the above for s > 1. Show that the exponent- [vivir (: n2))The expression ( /2,/xs—xIn (: +s?) is increasing in x. You can show thisby fixing s and taking a derivative with respect to x. Therefore, it is larger thanV2V1s—In (1+5y/7) and soexp (-*: (+\2)) exp (- [vv (: vo?)))= (1 +sv2) esThus, there exists a dominating function for A Jel (s) exp (<n (s\/2)) and>these functions converge pointwise to exp (—s*) as x —> ee so by the dominated con-vergence theorem,CO 2 PCO 2lim exp ( —sh V2 d =| ds= Jaes <( Ss (: °)) Ss e s=V/nSee Problem 16. This yields a general Stirling’s formula, lim,—,.. ere = Jt.IA