Solution
Week 56
(10/6/03)
Stirling’s formula
Let’s first prove the result,
N! =
Z ∞
0
xN e−x dx ≡ IN .
(1)
The proof by induction proceeds as follows. Integrating by parts gives
Z ∞
0
N −x
x e
N −x
dx = −x e
¯∞
Z ∞
¯
xN −1 e−x dx.
dx¯¯ + N
0
0
(2)
The first term on the right-hand side is zero, so we have IN = N IN −1 . Therefore,
if IN −1 = (N − 1)!, then IN = N !. Since it is indeed true that I0 = 0! = 1, we see
that IN = N ! for all N .
Let us now write xN e−x as exp(N ln x − x) ≡ exp(f (x)), and then expand f (x)
in a Taylor series about its maximum, which occurs at x = N . Computing the
first two derivatives of f (x), evaluated at N , gives the following Taylor series in the
exponent.
N! =
Z ∞
0
Ã
!
(x − N )2
exp −N + N ln N −
+ · · · dx
2N
N −N
≈ N e
Z ∞
0
Ã
(x − N )2
exp −
2N
!
dx.
(3)
If N is very large, we can let the integral run from −∞ to ∞, with negligible error.
Letting y ≡ x − N , we have
N ! ≈ N N e−N
Z ∞
−∞
e−y
2 /2N
√
= N N e−N 2πN ,
dy
(4)
as desired.
R∞
√
2
Remark: We have used the fact that −∞ e−y dy = π. This can be proved in the
following way, where we make use of a change of variables from cartesian to polar coordinates.
Let I be the desired integral. Then
Z ∞
Z ∞
2
2
I2 =
e−x dx
e−y dy
−∞
−∞
Z ∞Z ∞
2
2
=
e−(x +y ) dx dy
−∞ −∞
2π Z ∞
Z
=
=
=
2
e−r r dr dθ
0
à 0 2 !¯
¯∞
e−r
¯
2π −
¯
2
0
π.
(5)
1
A change of variables then gives
R∞
−∞
e−y
2
/n
dy =
√
nπ.
The calculation of the higher-order corrections is a bit messier, because we have to
keep track of more terms in the Taylor expansion of f (x). Let’s find the order-1/N
correction first. Our strategy will be to write the integrand in eq. (1) as a gaussian
plus small corrections. Computing the first four derivatives of f (x), evaluated at N ,
gives us the following Taylor series in the exponent (letting y ≡ x − N , and letting
the limits of integration run from −∞ to ∞).
N! =
Z ∞
∞
!
Ã
y4
y2
y3
−
+ · · · dy
exp −N + N ln N −
+
2N
3N 2 4N 3
N −N
≈ N e
Z ∞
∞
N −N
≈ N e
Z ∞
∞
Ã
y2
exp −
2N
Ã
y2
exp −
2N
!
Ã
y4
y3
−
exp
2
3N
4N 3
!Ã
!
"
y3
y4
1+
−
3N 2 4N 3
"
dy
#
1 y3
y4
−
+
2 3N 2 4N 3
(6)
#2
!
+ · · · dy.
Since terms with odd powers of y integrate to zero, we obtain (to leading orders in
1/N ),
N −N
N! ≈ N e
Ã
Z ∞
y2
exp −
2N
−∞

!
"
#2
4
3
y
y
1
1 −
+ · · · dy.
+
4N 3 2 3N 2
R
(7)
∞
At this point, we need to know how to calculate integrals of the form −∞
x2n e−ax dx.
R ∞ −ax2
√ −1/2
Using −∞ e
dx = πa
, and successively differentiating with respect to a,
we obtain
Z ∞
√ −1/2
2
πa
,
e−ax dx =
−∞
Z ∞
−∞
Z ∞
−∞
Z ∞
−∞
2
x2 e−ax dx =
2
x4 e−ax dx =
2
x6 e−ax dx =
2
1 √ −3/2
πa
,
2
3 √ −5/2
πa
,
4
15 √ −7/2
πa
.
8
(8)
Letting a ≡ 1/2N here, eq. (7) gives
N −N √
µ
1 3
1 15
N! ≈ N e
π (2N ) −
(2N )5/2 +
(2N )7/2
4N 3 4
18N 4 8
¶
µ
√
1
N −N
.
= N e
2πN 1 +
12N
¶
1/2
(9)
Note that to obtain all the terms of order 1/N , it is necessary to include the
(y 3 /3N 2 )2 term in eq. (7). This is an easy term to forget.
If you like these sorts of calculations, you can go a step further and find the order1/N 2 correction. It turns out that you have to keep terms out to the −y 6 /6N 5 term
2
in the expansion in the first line of eq. (6). Furthermore, you must keep terms out
to the [· · ·]4 /4! term in the expansion in the last line of eq. (6). The relevant extra
terms that take the place of the “· · ·” in eq. (7) then turn out to be
"
Ã
#
y6
1
y4
−
+
−
6N 5
2
4N 3
 Ã
1
y3
+ 3
3!
3N 2
!2 Ã
!2
Ã
y3
+2
3N 2
!
!Ã
Ã
!
y5 
5N 4
y4  1  y3
−
+
4N 3
4!
3N 2
!4 
,
(10)
where we have grouped these terms via square brackets according to which term in
the ez series expansion in the last line of eq. (6) they come from. To do all of the
necessary integrals in the modified eq. (7), we’ll need the next three integrals in the
list in eq. (8). They are
Z ∞
2
−∞
Z ∞
−∞
Z ∞
−∞
x8 e−ax dx =
2
x10 e−ax dx =
2
x12 e−ax dx =
3 · 5 · 7 √ −9/2
πa
,
24
3 · 5 · 7 · 9 √ −11/2
πa
,
25
3 · 5 · 7 · 9 · 11 √ −13/2
πa
.
26
(11)
Putting the terms of√eq. (10) in place of the “· · ·” in eq. (7), we find that the
coefficient of N N e−N 2πN equals 1/N 2 times
13·5 3 1
−
2 +
6 23
2
1
.
=
288
µ
1
2
+
16 15
¶
3·5·7 4
1 3 3·5·7·9 5
1 1 3 · 5 · 7 · 9 · 11 6
2 −
2 +
2
24
3! 36
25
4! 81
26
(12)
Therefore, we may write Stirling’s formula as
µ
√
N ! ≈ N N e−N 2πN 1 +
¶
1
1
+
.
12N
288N 2
(13)
This result of 1/288 is rather fortuitous, because it is the third term in the Taylor
series for e1/12 . This means that we can write N ! as
³
´
√
N ! = N N e−N 2πN e1/12N + O(1/N 3 )
√
≈ N N e−N +1/12N 2πN .
(14)
It turns out that the order-1/N 3 correction is not equal to 1/(3! · 123 ), which is the
next term in the expansion for e1/12 .
3