264 CHAPTER 10. CHANGE OF VARIABLES

13. Stirling’s formula from elementary calculus says that for n ∈ N,

limn→∞

enn!nn+(1/2) =

√2π.

Show Γ(x)≡∫

∞

0 e−ttx−1dt exists whenever x > 0. Also show that Γ(x+1) = xΓ(x).Then show that for n ∈ N, Γ(n+1) = n!. This function is discussed in the nextchapter.

14. For n ∈N, Stirling’s formula says limn→∞Γ(n+1)en

nn+(1/2) =√

2π . Here Γ(n+1) = n!. Theidea here is to show that you get the same result if you replace n with x ∈ (0,∞). Todo this, show

(a) n→ Γ(n+1)en

nn+(1/2) is decreasing on the positive integers. This follows from the prop-erties of the Gamma function and a little work.

(b) Show that x→ Γ(x+1)ex

xx+(1/2) is decreasing on (m,m+1) for m ∈ N. This is a littleharder.

Hint: For x ∈ (m,m+1) , ln(

Γ(x+1)ex

xx+(1/2)

)= x+ lnΓ(x+1)−

(x+ 1

2

)lnx

= x+ ln(x(x−1)(x−2) · · ·(x−m+1)Γ(x−m))−(

x+12

)lnx

= x+m−1

∑k=0

ln(x− k)+ ln(Γ(x−m))−(

x+12

)lnx

Now differentiate and try to show that the derivative is negative for x ∈ (m,m+1).Thus the desired derivative is(

m−1

∑k=0

1x− k

− lnx

)+

1Γ(x−m)

∫∞

0ln(t) tx−(m+1)e−tdt− 1

2x

The first term is negative from the definition of ln(x) . The derivative being negativewill be shown if it is shown that the integral term is negative. Do an integration byparts and split the integrals to obtain∫

∞

0ln(t) tx−(m+1)e−tdt = −

∫ 1

0tσ e−tdt +

∫ 1

0(t−σ)e−ttσ ln(t)

+∫

∞

1tσ e−t (1− (t−σ) ln(t))dt

where σ = (x−m)− 1 ∈ (−1,0) so −σ > 0. The last integral is negative because(t−σ) = t +(−σ)> 1. The other two are obviously negative.