Last revised 10:19 a.m. April 5, 2016
Essays in analysis
Bill Casselman
University of British Columbia
cass@math.ubc.ca
The Gamma function
I attempt here a somewhat unorthodox introduction to the Gamma function and related matters of harmonic
analysis. If one asks, What is the problem for which the Gamma function is the solution?, there are a host of
good answers. One of the principal roles it plays in current mathematics is as factors of zeta and L-functions
contributed by real completions of number fields. This topic is most often treated along the lines of Tate’s
thesis, and that is the main point of view I adopt here. But the Gamma function also occurs elsewhere in
number theory, so I include some of the classical theory, too.
The first part is concerned with analysis related to Tate’s functional equations. The main result is the
uniqueness of distributions that transform as characters under multiplication. This includes some simple
analysis not far removed from elementary calculus. I recover Tate’s functional equation, as well as a second
formula related to it that involves the Laplace transform.
The second part is concerned with classical formulas involvng Gamma functions, including a discussion of
how to compute explicit values.
My principal references here are [Schwartz:1965]. and [Tate:1950/1967].
Contents
I. Functional equations
1. Calculus exercises
2. The multiplicative Schwartz spaces
3. Multiplicatively equivariant distributions on R and C
4. Parties finies
5. The Gamma function
6. Tate’s functional equation
7. The Laplace transform
8. Another formula for Tate’s factor
II. Classical formulas
9. The Beta function
10. The limit product formula
11. The reflection formula
12. The Euler-Maclaurin formula
13. Stirling’s formula
14. The volumes and areas of spheres
III. References
15. References
Part I. Functional equations
The Gamma function
2
1. Calculus exercises
The fundamental theorem of calculus tells us that if f is a smooth function in an open interval U around 0
then
Z x
f (x) − f (0) =
f ′ (s) ds .
0
If we set s = tx this equation becomes
f (x) = f (0) + x
Z
1
f ′ (tx) dt = f (0) + xf1 (x) .
0
The integral
f1 (x) =
Z
1
f ′ (tx) dt
0
defines a smooth function of x. Induction and a simple calculation give us:
1.1. Proposition. If f is smooth in an interval U around 0 in R then for every m
f (x) =
X xj
· f (j) (0) + xm fm (x)
j!
j<m
with fm smooth in U .
The Schwartz space S(Rn ) is the space of all smooth functions f on Rn such that
n ∂ f −N
∂xn ≪ 1 + |x|
(n = (ni ))
for all n, N ≥ 0 or, equivalently, for which
N ∂ nf kf kN,n = sup 1 + |x| n < ∞
∂x
R
for all non-negative integers N , n. It is a Fréchet space with these semi-norms.
The space S(Rn ) is taken into itself by polynomial multiplication. It thus becomes a module over C[x] =
C[x1 , . . . , xn ]).
1.2. Proposition. Let m be the ideal of C[x] generated by the xi (i = 1, . . . n). Then f in S(Rn ) lies in
mm S(Rn ) if and only if all derivatives of f of order less than m vanish at the origin.
Proof. One way is immediate. For the other, apply a double induction on n and m. For n = 1, if f (k) (0) = 0
for k < m, then by the previous result f (x)/xm is locally smooth. But it and all its derivatives clearly vanish
rapidly at infinity, so it lies in S(R).
Now assume true for dimension n − 1. Let ϕ(y) be a function of compact support on R identically 1 near
0. As a consequence of the case n = 1 the function f (x) − f (x1 , . . . , xn−1 , 0)ϕ(xn ) is divisible by xn , and
hence
f (x) = f (x1 , . . . , xn−1 , 0)ϕ(xn ) + xn fn (x) ,
with fn (x) in S(Rn ). Suppose m = 1, so we assuming that f (0) = 0. The second term vanishes at 0, and to
the first we may apply induction. Hence f lies in mS(R).
For m > 1 we may apply double induction without trouble.
We shall need a variation of Proposition 1.1:
The Gamma function
3
1.3. Proposition. If f is smooth in the interval [0, x] then
f (x) = f (0) + xf ′ (0) + · · · +
xp−2
xp−1
f (p−2) (0) +
f (p−1) (0)
(p − 2)!
(p − 1)!
Z xp−1
Z x Z x1
f (p) (xp ) dxp . . . dx1 .
...
+
0
0
0
Proof. We start with
f (x) = f (0) +
Z
x
f ′ (x1 ) dx1 .
0
and then we extend this by repeating the same process with f ′ (x1 ) etc. to get
Z
x1
f ′′ (x2 ) dx2
f (x1 ) = f (0) +
0
Z x
f (x) = f (0) +
f ′ (x1 ) dx1
0
Z x Z
′
= f (0) + xf (0) +
′
′
x1
′′
f (x2 ) dx2 dx1
Z x Z x1 Z x2
x ′′
′
f ′′′ (x3 ) dx3 dx2 dx1
= f (0) + xf (0) + f (0) +
2!
0
0
0
...
Z x
Z xp−1
xp−1
′
(p−1)
= f (0) + xf (0) + · · · +
f (p) (xp ) dxp . . . dx1 .
f
(0) +
...
(p − 1)!
0
0
0
2
0
If f is a function on the open interval (0, α), I write
f (x) ∼ c0 + c1 x + c2 x2 + · · ·
if
f (x) − (c0 + c1 x + · · · + cm−1 xm−1 ) = O(xm )
for every m.
I define C ∞ [0, α) to be the space of all f in C ∞ (0, α) such each limε→0 f (n) (ε) exists.
1.4. Corollary. Suppose f in C ∞ (0, α). The following are equivalent:
(a) the function f lies in C ∞ [0, α);
(b) it is the restriction to (0, α) of a smooth function defined in C ∞ (−α, α);
(c) it has an asymptotic expansion
f (x) ∼ c0 + c1 x + c2 x2 + · · ·
Proof. For the implication from (a) to (c), apply the Proposition to ε < α instead of 0 and let it tend to 0.
To go from (c) to (b): according to a well known result
to Émile Borel, there exists a smooth function
P attributed
ϕ on all of (−α, α) whose Taylor series at 0 is
ci xi . But then the function that is ϕ in (−α, 0] and f on
[0, α) is also smooth.
1.5. Corollary. Suppose f (x) to be a function in C r+1 (0, ρ], such that for some κr+1
(r+1) κr+1
f
(x) ≤ r
x
The Gamma function
4
for all 0 < x ≤ ρ. Then
f0 = lim f (x)
x→0
exists, and
|f0 | ≤ C κr+1 +
for some C > 0 independent of f .
X ρk f (k) (ρ)
k!
0≤k≤r
In effect, the function f (x) extends to a continuous (but not necessarily differentiable) function on all of [0, ρ].
As an illustration of this result, let ℓ(x) = x log x − x. We have on the one hand
ℓ(x) = x log x − x
ℓ′ (x) = log x
1
ℓ′′ (x) =
x
1
′′′
ℓ (t) = − 2
x
(p − 2)!
xp−1
(r − 1)!
,
ℓ(r+1) (x) = (−1)r+1
xr
ℓ(p) (x) = (−1)p
and on the other limx→0 ℓ(x) = 0. The function ℓ(x) will play a role in the proof of Corollary 1.5.
Proof. I begin by recalling the elementary criterion of Cauchy: If f (x) is continuous in (0, ρ] then limx→0 f (x)
exists if and only if for every ε > 0 we can find δ > 0 such that |f (z) − f (y)| < ε whenever 0 < y, z < δ .
The cases r = 0, r ≥ 1 are treated differently.
• The case r = 0. By assumption, f is C 1 on (0, ρ] and f ′ is bounded by κ1 on that interval. For any y , z in
(0, ρ].
Z z
f (z) − f (y) =
f ′ (x) dx, |f (z) − f (y)| ≤ κ1 |z − y| .
y
Therefore Cauchy’s criterion is satisfied, and the limit f0 = limx→0 f (x) exists. Furthermore
f0 = f (ρ) −
Z
ρ
f ′ (x) dx
0
|f0 | ≤ |f (ρ)| + κ1 ρ .
• The case r > 0. If we apply Proposition 1.3 twice with a shift of origin, we get
f (z) = f (ρ) + (z − ρ)f ′ (ρ) + · · · +
(z − ρ)r (r)
f (ρ) +
r!
Z
ρ
z
...
Z
ρ
xr
f (r+1) (xr+1 ) dxr+1 . . . dx1
Z xr
Z y
(y − ρ)r (r)
f (r+1) (xr+1 ) dxr . . . dx1
...
f (ρ) +
r!
ρ
ρ
Z z
Z xr
r
X
(z − ρ)k − (y − ρ)k (k)
f (r+1) (xr+1 ) dxr+1 . . . dx1
f (ρ) +
...
f (z) − f (y) =
k!
ρ
y
f (y) = f (ρ) + (y − ρ)f ′ (ρ) + · · · +
k=1
The Gamma function
5
In order to apply Cauchy’s criterion, we must now show how to bound
Z
y
z
...
Z
xr
ρ
Z
f (r+1) (xr+1 ) dxr+1 . . . dx1 ≤
z
...
Z
ρ
y
xr
κr+1
dxr+1 . . . dx1 .
xrr+1
But we have already seen the iterated integral
Kz,y,r =
Z
y
z
...
Z
ρ
xr
1
dxr+1 . . . dx1 .
xrr+1
If we set f (x) = ℓ(x) = x log x − x in the calculation above we get
ℓ(y) = y log y − y
Z xr
Z ρ
(r − 1)!
...
=
dxr+1 . . . dx1
xrr+1
ρ
y
r
+
so that
Ky,ρ,r =
Z
ρ
...
y
r−1
(y − ρ)
(y − ρ) (r)
ℓ (ρ) +
r!
(r − 1)!
Z
ρ
xr
ℓ(r−1) (ρ) + · · · + ℓ(ρ) ,
(r − 1)!
dxr+1 . . . dx1
xrr+1
r
r−1
(y − ρ) (r)
(y − ρ)
= ℓ(y) −
ℓ (ρ) −
r!
(r − 1)!
ℓ(r−1) (ρ) − · · · − ℓ(ρ) .
Kz,y,r = Kz,ρ,r − Ky,ρ,r
Since ℓ(x) is continuous on [0, ρ] we may now apply Cauchy’s criterion in the other direction to see that the
limit f0 exists. Furthermore, the bound on f (r+1) together with the equation for f (z) − f (y) enable us to to
see that
r
X
ρ k (k) f (ρ) .
|f0 | ≤ κr+1 |K0,ρ,r+1 | +
k!
0
2. The multiplicative Schwartz spaces
The space S(R) contains as closed subspace the Schwartz space S(R× ) of functions whose Taylor series at 0
vanish identically. A well known theorem attributed to Émile Borel asserts that the sequence
0 −→ S(R× ) −→ S(R) −→ C[[x]] −→ 0
is exact.
Let D be the multiplicatively invariant derivative xd/dx.
2.1. Proposition. Suppose f to be a function in C ∞ (R). The following are equivalent:
(a) the function f lies in S(R× );
(b) for all n ≥ 0, N ∈ Z, |x|N f (n) (x) is bounded;
(c) for all n ≥ 0, N ∈ Z, |x|N Dn f (x) is bounded.
Proof. This follows from Proposition 1.1.
2.2. Corollary. For any s in C multiplication by |x|s is an isomorphism of S(R× ) with itself.
Proof. This follows from Leibniz’s formula for (xsf )(n) .
The Gamma function
6
The space S(R× ) contains the closed subspace S(0, ∞) of functions vanishing on (−∞, 0], and similarly
S(−∞, 0).
The multiplicative group R×
of positive real numbers acts on both of the spaces S(R) and S(0, ∞), as well
>0
as on their continuous linear duals, by the formulas:
µa f (x) = f (a−1 x),
hµa Φ, f i = hΦ, µa−1 f i .
The scale factor a−1 rather than a has been chosen for compatibility with linear representations of non-abelian
is spanned by the differential operator D = xd/dx, and the representations
groups. The Lie algebra of R×
>0
are smooth in the sense that
lim
h→0
µ1+h f − f
= −Df
h
is again in S(R). The − sign here comes about because of the choice of a−1 rather than a. It will continue to
annoy.
According to Corollary 2.2, for every s in C the integral
hΦs , f i =
Z
∞
xs f (x)
0
dx
x
defines a continuous linear functional on S(0, ∞)—in effect a kind of distribution.
2.3. Proposition. The distribution Φs on S(0, ∞) is an eigendistribution for µa with eigencharacter a−s .
Furthermore DΦs = sΦ.
Proof. We have
hµa Φs , f i = hΦs , µa−1 f i
Z ∞
dx
=
xs f (ax)
x
Z0 ∞
dy
=
(y/a)s f (y)
y
0
Z ∞
dy
= a−s
y s f (y)
y
0
−s
= a hΦs , f i
so that µa Φs = a−s Φs as a distribution (as well as a function).
For the second claim:
hDΦs , f i = −hΦs , Df i
Z ∞
xs f ′ (x) dx
=−
0
Z ∞
=s
xs−1 f (x) dx
0
= shΦs , f i
There is a converse to this claim, and there is also a uniqueness theorem for eigendistributions. If f lies in
S(0, ∞), its Mellin transform is
fb(s) = hΦs , f i .
It is uniformly bounded on any horizontally bounded strip |RE(s)| ≤ C . It is also holomorphic in all of C,
and
d
Df = sfb.
The Gamma function
7
It therefore belongs to the space M(0, ∞), the space of all function F (s) holomorphic on all of C such that
(1 + |IM (s)|)N |F (s)| is bounded on any horizontal strip |RE(s)| ≤ C , for all N ≥ 0.
2.4. Proposition. The map f 7→ fb is an isomorphism of S(0, ∞) with M(0, ∞).
Proof. One way is because DΦs = sΦs . The other way involves a simple shift of contours of the inverse
integral
Z
f (x) =
1
2πi
RE (s)=0
fb(s)x−s ds .
For any fixed a the image in M(0, ∞) of multiplication by s − a is the subspace of F such that F (a) = 0,
which is of codimension one. Hence the quotient S(0, ∞)/(D − a)S(0, ∞) is isomorphic to C. The space of
distributions Φ such that DΦ = aΦ is the dual of S(0, ∞)/(D + a)S(0, ∞), hence:
2.5. Corollary. The space of distributions on (0, ∞) such that DΦ = sΦ is spanned by Φs .
It is therefore also true that the space of distributions on (0, ∞) such that µa Φ = a−s Φ for all a in R×
>0 is
spanned by Φs .
Now suppose χ to be a character of R× , which is to say a continuous homomorphism from R× to C× . It will
be of the form x 7→ sgn(x)|x|s . The space S(R× ) is the direct sum of S(0, ∞) and S(−∞, 0), and χ defines
a χ-equivariant distribution on S(R× ):
hΦχ , f i =
Z
χ(x)f (x)
R×
dx
.
|x|
The group R× is the direct product {±1} × (0, ∞). The space of functions f in S(R× ) such that µc f = cm f
for c = ±1 is isomorphic to S(0, ∞). Therefore Corollary 2.5 implies:
2.6. Corollary. The space of distributions on S(R× ) such that µt Φ = χ−1 (t)Φ is spanned by Φχ .
Something similar happens for C× . Here S(C× ) is the subspace of functions in S(C) whose Taylor series at
0 vanish identically. On C I use the invariant measure |dz ∧ dz|, which is 2 dx dy , and define the norm
||z|| = zz .
The space S(C× ) is isomorphic to S × (0, ∞). The space of f in C× such that µc f = cn f for c in S is
isomorphic to S(0, ∞). Hence:
2.7. Corollary. If χ is a character of C× , then the space of distributions on S(C× ) such that µt Φ = χ−1 (t)Φ
is spanned by the distribution
hΦχ , f i =
Z
χ(z)f (z)
C×
|dz ∧ dz|
.
||z||
The Gamma function
8
3. Multiplicatively equivariant distributions on R and C
In this section, let F be R or C. In both cases, the norm ||c|| is the modulus of multiplication by c, so that
meas(c Ω) = ||c|| meas(Ω). If F = R then ||c|| is the usual absolute value, but for C it is cc.
Suppose χ to be a unitary character of F × , and for s in C let χs (c) = χ(c)|| c|| s . For RE(s) > 0 and f in the
Schwartz space S(F ) the integral
hχs , f i =
Z
F
χ(x)|| x|| s−1 f (x) dx
converges and defines a tempered distribution that varies holomorphically in s. Integration by parts allows
us to see that this continues meromorphically to all of C except for simple poles at a subset of non-positive
half-integers.
As earlier, for c in F × define the operator
[µc f ](x) = f (c−1 x)
on functions, and dually on linear functionals
hµc Φ, f i = hΦ, µ1/c f i .
Then
µc χs = χ−1 (c)|| c|| −s χs .
The space S(F × ) embeds as a closed subspace of S(F ). According to Borel’s theorem, the quotient may be
identified with the space of Taylor series C0 at the origin—C[[x]] if F = R and C[[z, z]] if F = C. We have
therefore an exact sequence
0 −→ S(F × ) −→ S(F ) −→ C0 −→ 0 .
A character χ of F × defines a distribution on S(F × ). It may not extend to an equivariant distribution on
S(F ). For example the integral
Z
h1, f i =
f (x)
R×
dx
||x||
defines a multiplicatively invariant distribution on R× , but it does not extend to one on all of R. Nonetheless,
there is a multiplicatively invariant distribution on R, namely
δ0 : f 7−→ f (1) .
In this section, I shall explain how this illustrates a general fact:
3.1. Theorem. If F is R or C, then for each character χ the space of distributions Φ on S(F ) such that
µt Φ = χ−1 (t)Φ
has dimension one.
The proof will take a while.
In his thesis Tate proved that for generic s the Fourier transform of χs is a scalar multiple of χ−1
1−s . For him,
this was an explicit calculation, but [Weil:1966] pointed out that this can be deduced more conceptually from
this Theorem, together with an observation about how the Fourier transform and the multiplicative groups
interact. This idea has now become a standard tool in representation theory.
The Gamma function
9
THE SCHWARTZ SPACE OF THE NON-NEGATIVE REALS. Define S[0, ∞) to be the space of all functions f
in C ∞ (0, ∞) for which (a) all f (n) decrease more rapidly at infinity than any x−n , and (b) all derivatives
f (n) (x) possess a limit as as x → 0.
It becomes a Fréchet space if assigned the norms
kf kn = sup xk f (ℓ) k,ℓ≤n
kf k = sup |f (x)| .
x≥0
Restriction to (0, ∞) defines a continuous linear map from S(R) to S[0, ∞).
3.2. Lemma. The restriction map from S(R) to S[0, ∞) is surjective.
Proof. Suppose given f in S[0, ∞). It follows from Corollary 1.4 there exists a function ϕ in S(R) with the
same Taylor series. The function that is f on [0, ∞) and ϕ on (−∞, 0] lies in S(R) and restricts to f on [0, ∞).
The sequence
0 → S(−∞, 0) → S(R) → S[0, ∞) → 0 .
is therefore exact. According to the Closed Graph Theorem, the surjection from S(R) identifies S[0, ∞) as a
quotient topological vector space.
The space S(0, ∞) is embedded as a closed subspace of S[0, ∞) and the multiplicative group acts smoothly
on both. The distribution Φs is well defined on S(0, ∞). Does there exist an eigendistribution on S[0, ∞)
extending Φs ? The key to understanding what happens is this result:
3.3. Lemma. For RE(s) > 0, f in S(R)
hΦs , f i =
Z
∞
xs f (s)
0
(−1)n+1
dx
=
x
s(s + 1) . . . (s + n)
Z
∞
xs+n f (n+1) (x) dx .
0
Proof. Integration by parts give us
hΦs , f i =
Z
∞
Z0 ∞
xs f (x)
dx
x
xs−1 f (x) dx
∞
Z
1 ∞ s ′
f (x)xs
−
=
x f (x) dx
s
s 0
Z ∞ 0
dx
1
xs+1 f ′ (x)
=−
s 0
x
1
= − hΦs+1 , f ′ i
s
=
0
and continuing:
1
hΦs+2 , f ′′ i
s(s + 1)
...
(−1)n+1
=
hΦs+(n+1) , f (n+1) i .
s(s + 1) . . . (s + n)
=
The Gamma function
10
As a consequence, Φs may be defined on S[0, ∞) for all s not in −N. Thus for every onw of these values
of s we have an eigendistribution with eigencharacter x−s . Is it unique? What happens for s = −n? Set
s = −n + h in the Lemma. We get
(3.4)
hΦs , f i =
−1
(n − h)(n − 1 − h) . . . (1 − h)h
Z
∞
xh f (n+1) (x) dx .
0
Thus (s + n)hΦs , f i as s → −n (and h → 0) has limit
1
−
n!
Z
∞
f (n+1) (x) dx =
0
f (n) (0)
.
n!
(n)
The distribution δ0 is defined to take f to f (0). Its derivative δ0
(n)
at s = −n is therefore (−1)n δ0 /n!.
(n)
3.5. Lemma. The distribution δ0
takes f to (−1)n f (n) (0). The residue of Φs
is an eigendistribution for the character an .
Proof. Since f (n) (ax) = an f (n) (ax).
In other words, the distribution Φs fails to be defined precisely when another eigencharacter arises.
3.6. Proposition. For every s, the space of distributions Φ such that DΦ = sΦ has dimension one.
It is possible to prove this by applying a version of the Mellin transform, but instead I’ll give a simpler
cohomological argument. Consider the short exact sequence
0 → S(0, ∞) → S[0, ∞) → C[[x]] → 0 .
The last map is takes f to its Taylor series at 0, and is surjective by Borel’s Lemma. If T = D + sI this gives
rise to a long exact sequence
0 → S(0, ∞)(T ) → S[0, ∞)(T ) → C[[x]] (T )
→ S(0, ∞)/T · S(0, ∞) → S[0, ∞)/T · S(0, ∞) → C[[x]]/T · C[[x]] → 0 .
Here V (T ) is the subspace of v in V such that T v = 0. The first two terms are always 0, since any function
annihilated by D + a is a scalar multiple of x−a and cannot decrease rapidly at ∞.
The operator D + s takes
c0 + c1 x + c2 x2 + · · · −→ ac0 + c1 (1 + s)x + c2 (2 + s)cx2 + · · · .
When s does not belong to −N, D + s is an isomorphism of C[[x]] with itself, and the third and sixth terms
vanish. This proves the first assertion.
When s = −n the kernel of D + s is spanned by xn , and the cokernel is the space of power series with cn = 0.
To find the connecting map from third to fourth term, suppose f to be in S[0, ∞) with Taylor series just xn .
The image of xn is the image f −(D−n)f in S(0, ∞) modulo (D−n)S(0, ∞). But if f −(D−n)f = (D−n)g
then f lies in (D − a)S[0, ∞) and hence has Taylor series identically 0.
Remark. There is a simple variant of the last part of the argument worth pointing out. Let’s prove that there
is no distribution on [0, ∞) that is both R× -invariant and which restricts to integration against 1/x on R× .
For this, suppose Φ to be a distribution on R that is invariant under multiplication by c in R×
>0 , and suppose
that Φ restricted to S(0, ∞) is a constant times integration. Choose f to be a function in S(R) whose Taylor
series is just the constant 1. Then µc f − f lies in S(R× ), and by invariance
0 = hΦ, µc f − f i = C
Z
R×
f (c−1 x) − f (x) dx/|x| .
But we can certainly find c and f for which the integral does not vanish, hence C = 0.
————–
The Gamma function
11
(n)
3.7. Proposition. The distribution Φs on S[0, ∞) is meromorphic on all of C with residue (−1)n δ0 /n! at
−n. For each s not in −N it spans the space of eigendistribution on S[0, ∞) for the character a
(n)
−N the distribution δ0 spans that space.
−s
. For s in
The full multiplicative group R× acts on its own Schwartz
space S(R ), the subspace of functions in S(R) whose Taylor series at 0 vanish. We now have distributions
THE SCHWARTZ SPACE OF THE REAL LINE .
×
hΦs,m , f i =
Z
∞
f (x)|x|s sgnm (x)
−∞
for ℜ(s) > 0 and m = 0, 1, which are again eigendistributions:
dx
|x|
µa Φs,m = sgnm (a)|a|−s Φs,m .
We can write
Z ∞
Z 0
dx
dx
dx
= (−1)m
+
xs f (x)
|x|s f (x)
|x|
|x|
x
0
−∞
−∞
Z ∞
dx
=
xs f (x) + (−1)m f (−x)
,
x
0
which means that Φs,m extends equivariantly and meromorphically to S(R) over all of C with residue
Z
∞
f (x)|x|s sgnm (x)
δ0(n)
n!
at −n. In particular, there is no pole if the parity of m is different from the parity of n. In this case, because
of (4.1) we get as value at −n
Z ∞ (n)
1
f (x) − f (n) (−x)
dx
hPf(1/xn+1 ), f i =
n! 0
x
(−1)m + (−1)n
which always makes sense because the integrand is still a smooth function. For reasons we’ll see in a moment
this is called the finite part of 1/xn+1 . This defines an extension to S(R) of the integral
Z
R
on R× .
THE COMPLEX NUMBERS .
−n−1
|x|
n−1
sgn
(x)f (x) dx =
Z
x−(n+1) f (x) dx .
R
For n in N and s in C define the character
χs,n (z) = ||z|| s (z/|z|)−n
of C× . Every character of C× is equal to either some χs,n . For RE(s) > 0 and f in S(C) the integral
hχs,n , f i =
Z
C
||z|| s−1 (z/|z|)−nf (z) |dz ∧ dz|
converges and defines a tempered distribution on S(C). In polar coordinates, the integral becomes
Z
C
||z|| s−1 (z/|z|)−n f (z) |dz ∧ dz| = 2
Z
= 4π
∞
r2s−1
0
Z
0
Z
2π
e−inθ f (re−θ ) dθ
0
∞
r2s−1 f n (r) dr
dr
Z 2π
1
e−inθ f (reiθ ) dθ .
f n (r) =
2π 0
Suppose n ≥ 0; the other case is similar. The projection f n (r) has a Taylor series of the form
z n (c0 + c2 ||z|| + c4 ||z|| 2 + · · ·
so Lemma 3.3 tells us that this distribution can be meromorphically continued with poles in −n − N.
Uniqueness remains true, by the same cohomological reasoning as in earlier sections. The residues of the
distribution are those distributions taking f to some coefficient in the Taylor series of f at 0.
The Gamma function
12
4. Parties finies
I want now to look at (3.4) again. It can be rewritten and expanded in powers of h:
Z ∞
1
1 1
·
eh log x f (n+1) (x) dx
hΦs , f i = − · ·
h n! (1 − h/n)(1 − h/n − 1) . . . (1 − h) 0
Z ∞
Z ∞
1 1
= − · · 1 + h Λn + O(h2 ) ·
f (n+1) (x) dx + h
(log x)f (n+1) (x) dx + O(h2 )
h n!
0
0
with
Λn = 1 + 1/2 + 1/3 + · · · + 1/n .
We have already seen that the leading term is f (n) (0)/n!, and now we see that the second term in the
expansion is
Z ∞
−
1
n!
(log x)f (n+1) (x) dx
Λn f (n) (0) +
.
0
The integral can be expressed also as the limit as ε → 0 of
Z
∞
h
i∞ Z
f (n+1) (x) log x dx = f (n) (x) log x
−
ε
ε
= −f (n) (ε) log ε −
= −f (n) (0) log ε −
∞
ε
∞ (n)
Z
f
ε
Z
f (n) (x)
dx
x
∞
ε
(x)
dx
x
f (n) (x)
dx ,
x
since f (ε) − f (0) = O(ε) and limε→0 ε log ε = 0. The second term in the Laurent expansion of hΦs , f i at
s = −n is therefore also
Z ∞ (n)
1
f (x)
(n)
(n)
(4.1)
· lim f (0) log ε − Λn f (0) +
dx .
n! ε→0
x
ε
In order to understand the nature of certain eigenfunctions of D on R, I now recall the notion of ‘parties finies’,
introduced in [Hadamard:1923] in order to understand classical techniques for solving the wave equation in
high dimensions.
The first important observation is that the Dirac distributions are eigendistributions. For n ≥ 0
(n)
Dδ0
(n)
= −n δ0 .
There is, however, another distribution Φ such that DΦ = −nΦ.
4.2. Proposition. We have
µa Pf(1/xn+1 ) = an sgn(a)Pf(1/xn+1 )
D Pf(1/xn+1 ) = −n Pf(1/xn+1 ) .
Proof. Left as exercise.
(n)
The two distributions δ0 and Pf(1/xn+1 ) span the space of eigendistributions Φ on R such that DΦ = −nΦ,
or (equivalently) µa Φ = am Φ, but they are distinguished by what µ−1 does to them:
(n)
µ−1 δ0
(n)
= (−1)n δ0 ,
µ−1 Pf(1/xn+1 ) = −(−1)n Pf(1/xn+1 ) .
The Gamma function
13
This has to be, of course, since a cohomological argument like the one we saw earlier shows that there at most
one R× -equivariant extension to all of S(R) of the distribution which on S(R× ) is given by the formula
Z
∞
−∞
f (x)
dx .
xn+1
I’ll say more here about the construction of parties finies distributions. Suppose f in S(R), and let
f (x) = f0 + f1 x + f2 x2 + · · ·
be its Taylor series at 0, so fm = f (m) (0)/m!. Then
ϕn (x) =
f − (f0 + xf1 + · · · fn xn )
xn+1
is still smooth throughout R, although no longer in in S(R). Then
Z
∞
ε
f (x)
dx =
xn+1
=
Z
f (x)
dx +
xn+1
1
f0 + f1 x + · · · + fn xn
dx +
xn+1
ε
Z
Z
1
ε
∞
1
f (x)
dx
xn+1
Z
1
ϕn (x) dx +
ε
Z
∞
1
f (x)
dx .
xn+1
The last integral is independent of ε. As ε → 0, the second integral has a finite limit. The first integral is
h
−
i1
f0
f1
−
−
·
·
·
−
f
log
x
n
nxn
(n − 1)xn−1
ε
f1
f1
f0
f0
+ · · · + fn log ε
− · · · − fn−1 + n + +
=− −
n
(n − 1)
nε
(n − 1)εn−1
Therefore the limit
lim
ε→0
Z
∞
ε
f
f1
f (x)
0
dx
−
+
+
·
·
·
+
f
log
ε
n
xn+1
nεn
(n − 1)εn−1
exists, and agrees with Pf(1/xn+1 ).
The distribution Pf(1/xn+1 ) on [0, ∞) does not behave equivariantly with respect to scalar multiplication,
because of the log ε term. But on R the finite part is
lim
ε→0
Z
ε
−∞
f (x)
dx +
xn+1
Z
ε
∞
X
n
f (x)
2fn−k
,
dx −
xn+1
kεk
k=1
k odd
and it does behave well, because on (−∞, 0] log ε is replaced by log |ε|.
The Gamma function
14
5. The Gamma function
One function in S[0, ∞) is the restriction of f (x) = e−x to [0, ∞). The Gamma function is defined to be the
integral
Z
∞
Γ(s) =
dx
= hΦs , e−x i .
x
xs e−x
0
for ℜ(s) > 0. The argument extending Φs in the last section is classical in this case. Since here f ′ (x) = −f (x),
we have the functional equation
Γ(s + 1) = s Γ(s)
and since Γ(1) = 1, we see by induction that if s is a positive integer n
Γ(n) = (n − 1)!
The extension formula can be rewritten as
Γ(s + 1)
s
Γ(s) =
so that we can extend the definition of Γ(s) to the region ℜ(s) > −1, except for s = 0. And so on. More
explicitly we have
Γ(s) =
Γ(s + n + 1)
(s + n)(s + n − 1) . . . (s + 1)s
which allows Γ(s) to be defined for ℜ(s) > −n − 1, except at the negative integers, where it will have simple
poles (of order one).
Proposition. For n ≥ 0 the residue of Γ(s) at −n is (−1)n /n!
Another formula for Γ(s) can be obtained by a change of variables t = πx2 :
Γ(s) = 2π
s
Z
∞
0
which can also be written as
s
=π
π −s/2 Γ
s
Γ
or
2
s/2
Z
∞
−∞
2
=
Z
2
e−πx x2s
∞
−∞
dx
x
|x|s e−πx
2
|x|s e−πx
2
dx
|x|
dx
.
|x|
This function of s is often expressed as ζR (s) because of its role in functional equations of ζ functions.
The Gamma function
15
6. Tate’s functional equation
For the moment, let F be either R or C. Here I shall introduce the Fourier transform and its interaction with
the multiplicative group. In fact I shall eventually introduce two varieties of Four transform. The one I work
with in this section takes functions on F into other functions on F , while the one I use in the next, for F = R,
does something slightly different.
If F is R let ψ be the additive character e2πix , while on C take it to be e2πi(z+z) . Recall that on R the measure
dω we are using is dx, while on C it is |dz ∧ dz|.
For f in S(F ) its Fourier transform is
fb(λ) =
Z
f (x)ψ(−λx) dω
F
and this defines an isomorphism of S(F ) with itself. The inverse is
f (x) = ψ̌(x) =
Z
F
b
Another way to express this is that fb = µ−1 f .
fb(x)ψ(λx) dω
How do the Fourier transform and the multiplication operators interact?
6.1. Proposition. For a 6= 0
b
µd
a f = ||a|| µa−1 f .
Proof. Because
µd
a f (λ) =
Z
[µa f ](x)e−2πiλx dx
ZF
f (a−1 x)e−2πiλx dx
Z
f (y)e−2πiλay dy
= |a|
=
F
F
= |a|µa−1 fb(λ) .
b of a distribution Φ is defined by
The Fourier transform Φ
b f i = hΦ, fbi .
hΦ,
This, as an easy calculation will show, agrees with the definition of the Fourier transform on S(R). It can be
rewritten as a version of the Plancherel formula:
b µ−1 fbi .
hΦ, f i = hΦ,
Suppose χ to be a multiplicative character. The distribution Φ = Φχ is defined by
hΦχ , i =
Z
R
χ(x)f (x)
dω
||z||
The Gamma function
16
defined by convergence for certain χ and extended meromorphically. What is the Fourier transform of Φ?
Since µa Φχ = χ−1 (a)Φχ we have
b f i = hΦ,
b µa−1 f i
hµa Φ,
= hΦ, µ\
a−1 f i
= hΦ, ||a|| −1 µa fbi
= ||a|| −1 hµa−1 Φ, fbi
= ||a|| −1 χ(a)hΦ, fbi
b fi
= ||a|| −1 χ(a)hΦ,
b must be a scalar multiple γχ Φ where χ
so because of uniqueness Φ
e(a) = ||a|| χ−1 (a). To calculate the scalar
χ
e
γχ explicitly, we calculate first the Fourier transform of some particular functions.
THE REAL CASE .
The characters are of two kinds, depending on parity:x 7→ xm ||x|| s for m = 0, 1.
2
6.2. Lemma. The Fourier transform of e−πx is itself.
2
Proof. Let f (x) = e−πx . Then
fb(λ) =
Z
∞
2
e−2πiλx−πx dω
−∞
Z ∞
2
2
−πλ2
=e
eπλ −2πiλx−πx dx
−∞
Z ∞
2
2
= e−πλ
e−π(x−iλ) dx
−∞
Z ∞
2
2
= e−πλ
e−πx dx
−∞
=e
−πλ2
.
Let now χ(x) = |x|s , Φ = Φχ . Then
2
2
b e−πx i = hΦ, e−πx i
hΦ,
Z ∞
2 dx
=
||a|| s e−πx
|x|
−∞
Z ∞
2 dx
= π −s/2
|x|s e−x
|x|
−∞
= π −s/2 Γ(s/2)
= ζR (s) .
Since χ
e = |x|1−s :
6.3. Proposition. We have
where
b s,0 = γs Φ1−s,0
Φ
γs =
ζR (s)
.
ζR (1 − s)
The Gamma function
17
This formula isn’t quite right for values of s where Φs,0 or Φ1−s,0 have poles. The simplest way to formulate
things is to observe that Φs,0 /ζ(s) is entire, and that this formula says that the Fourier transform of Φs,0 /ζR (s)
is Φ1−s,0 /ζR (1 − s).
2
We can reason similarly for |x|s sgn(x) with xeπx .
2
2
6.4. Proposition. The Fourier transform of xe−πx is −iλe−πλ .
Proof. Differentiate the equation
Z
∞
2
2
e−πx e−2πiλx dx = e−πλ
−∞
with respect to λ.
Therefore
2
2
−πx
d
hΦ
i = hΦs,1 , −ixe−πx i
s,1 , xe
Z
2
= −i |x|s−1 sgn(x)xe−πx dx
ZR
2
= −i |x|s e−πx dx
R
= −iζR (1 + s)
2
hΦ1−s,1 , xe−πx i = ζR 1 + (1 − s)
and hence:
6.5. Proposition. We have
b s,1 = λs Φ1−s,1
Φ
where
λs = −i
LR (s)
,
LR (1 − s)
LR (s) = π −(s+1)/2 Γ
s+1
2
.
I conclude with a useful calculation, then I examine some special cases.
6.6. Proposition. Suppose Φ to be a tempered distribution on R. Then
b;
(a) the Fourier transform of Φ′ is 2πiλΦ
′
b
(b) the Fourier transform of xΦ is Φ /(−2πi).
Proof. First assume Φ to be in S(R). The first assertion follows from integration by parts, the second by
differentiating
Z ∞
b
Φ(x)e−2πiλx dx = Φ(λ)
−∞
with respect to λ. Proving the assertion for distributions follows from this simpler case.
The distributions defined by integrals
Z
R
are of particular importance.
6.7. Proposition. For n ≥ 0
xn f (x) dx,
Z
R
xn sgn(x)f (x) dx
The Gamma function
18
(n)
(a) the Fourier transform of xn is δ0 /(−2πi)n ;
(b) the transform of xn sgn(x) is
2 n!
Pf(1/xn+1 ) .
(2πi)n+1
As for the first, calculation shows that the transform of 1 is δ0 . But then by the previous lemma the transform
(n)
of xn is δ0 /(−2πi)n .
For the second, we can write xn sgn(x) as |x|n+1 sgnn+1 (x)/|x|, so its transform will be an eigendistribution
for |x|−n sgnn+1 = xn sgn(x), which means that it is a multiple of Pf(1/xn+1 ). To compute the constant, let’s
look at n = 0, where we want the Fourier transform of sgn(x) itself. Here
−i
2
=
π
2πi
2
hPf(1/x), xe−πx i = 1
2
hd
sgn, xe−πx i =
so that the transform of sgn is (2/2πi)Pf(1/x). Then
xn\
sgn(x) =
(n)
2
2n!
1
Pf(1/x)
=
Pf(1/xn+1 )
n
(−2πi) 2πi
(2πi)n+1
since Pf(1/xn )′ = −n Pf(1/xn+1 ).
THE COMPLEX CASE .
In this case the characters are of the form z n ||z|| s and z n ||z|| s with n ≥ 0. Let
fn (z) =
z n e−2π|| z||
z n e−2π|| z||
if n ≥ 0
if n < 0.
6.8. Lemma. The Fourier transform of fn is i|n| f−n .
Proof. I’ll first show this for f0 .
fb0 (λ) =
Z
e−2π|| z|| e−2πi(λz+λz |dz ∧ dz|
Z
−2π|| λ||
=e
e−2π|| z+λ|| |dz ∧ dz|
C
C
= f0 (λ) .
Note that the point of the extra factor of 2 in the measure |dz ∧ dz| balances out to make
Z
e
−2π|| z||
C
|dz ∧ dz| = 2π
Z
2
e−2πr 2r dr = 1 .
C
But now apply D = (−1/2π)∂/∂λ to this equation to get
[Dfb0 ](λ) = the Fourier transform of zf0 (z) evaluated at λ = iλf0 (λ)
and more generally
n
Let
[Dn fb0 ](λ) = Fourier transform of z n f0 (z) evaluated at λ = in λ f0 (λ) .
ΓC (s) = 2(2π)−s Γ(s) .
6.9. Proposition. The Fourier transform of Φχ is γ(χ)Φ||•|| χ−1 with
γ(χ) = (−i)|n| ·
if χ(z) = z n ||z|| s or zn ||z|| s with n ≥ 0.
Proof. Apply the Lemma.
ΓC (s)
ΓC (1 − s)
The Gamma function
19
7. The Laplace transform
Much of this section is taken from §IX.3 of [Reed-Simon:1972].
It begin with a new version of the Fourier transform. For f in S(R) and s in iR define
F
f (s) =
(7.1)
Z
f (x)e−sx dx
R
Of course Ff (2πiλ) = fb(λ). The inverse transform is
f (x) =
1
2πi
Z
f (s)esx ds .
F
RE (s)=0
The point is that in certain circumstances this Fourier transform will be the boundary value of a Laplace
transform holomorphic in the right half plane RE(s) > 0.
Tempered distributions with support on [0, ∞) annihilate all of S(−∞, 0) and may be identified with elements
in the continuous linear dual of S[0, ∞). Suppose Φ to have support on [0, ∞). Since the restriction of e−sx
to [0, ∞) lies in S[0, ∞) for RE(s) > 0, it makes sense to define the Laplace transform of Φ to be the function
L
(7.2)
Φ(s) = hΦ, e−sx i .
This is defined for s in the positive right half plane. It is easy to see that L Φ is continuous, and from that to
deduce by means of any one of several criteria that it is holomorphic. As we shall see, as RE(s) → 0 it has
the Fourier transform of Φ as limiting value.
Recall that S[0, ∞) is a Fréchet space defined by the norms
kf kn = sup xk f (ℓ) k,ℓ≤n
kf k = sup |f (x)| .
x≥0
Since Φ is a continuous linear function on S[0, ∞), there exists n such that
hΦ, f i ≪Φ,n kf kn .
(7.3)
If f (x) = e−sx then
f (n) (x) = (−s)n e−sx
and the maximum value of |xn e−sx | is
n n
en log n−n
(σ = RE(s)) .
es
σn
(evaluated at x = (n/s)). The function x log x − x is monotonic, and therefore we have proved:
=
7.4. Proposition. If Φ is a tempered distribution with support on [0, ∞) then
L
1
1+ n
Φ(s) ≪Φ (1 + |s|)
σ
m
(σ = RE(s))
for some m, n.
That is to say, on any right half plane RE(s) ≥ σ > 0 it is polynomially bounded, but it may not remain
bounded near the imaginary axis. For example, the Laplace transform of the Heaviside function
Y (x) =
is 1/s.
n
0 if x < 0
1 otherwise.
The Fourier transform of a tempered distribution is defined by duality, but if F is a summable function then
F is a continuous function for which (7.1) remains valid. This applies in particular to F with support in
[0, ∞). In these circumstances we can easily deduce a few properties of LF :
F
The Gamma function
20
(a) the Laplace transform of F extends continuously to the imaginary axis and FF (s) = LF (s) for RE (s) = 0;
(b) more generally, the Fourier transform of F (x)e−σx is the function t 7→ LF (σ + 2πit).
These properties remain valid in some sense for all distributions with support in [0, ∞). It will take a while
to justify this observation.
The first result is a partial converse to Proposition 7.4. Suppose ϕ(s) to be holomorphic in the region
RE(s) > 0 and satisfying the inequality
1
ϕ(s) ≪ϕ (1 + |s|)m 1 + n
σ
(7.5)
for some m, n. Then the function
ϕσ : t 7−→ ϕ(s + it)
is of moderate growth and hence defines a tempered distribution
Z
1
hϕσ , f i =
2πi
ϕ(σ + it)f (t) dt .
R
7.6. Theorem. In these circumstances
(a) the limit limσ→0 ϕσ exists and defines a tempered distribution ϕ0 ;
(b) if Φ is the inverse Fourier transform of ϕ0 then it has support on [0, ∞) and its Laplace transform is
ϕ(s).
I’ll start the proof in a moment.
Proposition 7.4 tells us that for any tempered distribution Φ with support on [0, ∞) the Laplace transform
L
Φ satisfies the hypotheses of Theorem 7.6. Hence:
7.7. Corollary. If Φ is a tempered distribution with support in [0, ∞) then for each σ > 0 the distribution
L
Φσ is the Fourier transform of e−σx Φ, and its limit as σ → 0 is FΦ.
Proof of Theorem 7.6 will take a while, and comes in a few steps. Assume ϕ given satisfying (7.5).
But now we can apply Corollary 1.5 to deduce what we want to know.
Step 1. Suppose given f in S(R), and consider
F (s) =
Z
ϕ(s + it)f (t) dt
R
for s real and small. It is to be shown that the limit of F (s) exists as s → 0, and that the limit depends
continuously on f . This is going to be an application of Corollary 1.5.
This involves derivatives of F . We have
Z
d
ϕ(s + it)f (t) dt
ds
ZR
1 d
ϕ(s + it)f (t) dt
=
R i dt
Z
= i ϕ(s + it)f ′ (t) dt
f ′ (s) =
R
and then
F (p) (x) = ip
Z
R
ϕ(s + it)f (p) (t) dt .
The Gamma function
21
Near s = 0 we have
ϕ(s + it) ≪
1 + tn
.
sm
Since f lies in S(R) we therefore have
m+1)(t)≪ s1m ·
ϕ(s + it)f
1
1+tN
for arbitrarily large N , hence also
1
.
sm
But this is exactly the hypothesis of Corollary 1.5, and we conclude that F (s) is continuous on [0, ∞), and
that the value of F (0) depends continuously on f . Therefore limσ→0 ϕσ exists and amounts to a tempered
distribution ϕ0 .
F m+1) (s) ≪
Step 2. Let Φ be the inverse Fourier transform of ϕ0 . It is next to be shown that Φ has support on [0, ∞),
which means that
whenever f is in Cc∞ with support on (−∞, 0].
hΦ, f i = 0
Let ψ(s) be the Laplace transform of f . It is holomorphic in all of C, and of uniform rapid decrease at infinity
in the right half plane. Furthermore
1
hΦ, f i =
2πi
Z
ϕ(s)ψ(−s) ds
RE (s)=σ
for any σ > 0. Using the uniform bounds and moving σ to ∞, we see that hΦ, f i = 0.
Step 3. The last step is to show that ϕσ is the Fourier transform of ϕ0 e−σx . This is now easy.
8. Another formula for Tate’s factor
The Fourier transform of the tempered distribution
h| • |s , f i =
(8.1)
Z
R
|x|s−1 f (x) dx
is a scalar multiple γ(s) of | • |1−s , and in a previous section we have seen Tate’s formula for it. But there is
also a second formula for it, and a comparison of the two is instructive.
8.2. Proposition. We have
γ(s) = 2(2π)−s cos(πs/2)Γ(s) .
It is easy enough to derive this formula from well known formulas. As we shall see in a moment, it is
equivalent to the combination of the reflection and duplication formulas for the Gamma function. But it also
implies these classical formulas, and sheds some light on them. In this section I’ll elucidate this.
b
Because fb(x) = f (−x), we see from Proposition 8.2 that
1 = ΓG (1 − s)ΓG (s)
= 2(2π)−s cos(πs/2)Γ(s) · 2(2π)s−1 cos(π(1 − s)/2)Γ(1 − s)
= 2 sin(πs/2) cos(πs/2) · π −1 Γ(s)Γ(1 − s)
= sin(πs)π −1 Γ(s)Γ(1 − s)
Γ(s)Γ(1 − s) = π/ sin(πs) .
The Gamma function
22
This is the reflection formula for Γ(s).
Since Tate’s formula and that of Proposition 8.2 must agree:
s
π
Γ
2
.
2(2π)−s cos(πs/2) Γ(s) =
1−s
π −(1−s)/2 Γ
2
−s/2
But by the reflection formula
Γ
1−s
2
π
1+s
π
Γ
=
=
.
2
sin π(1 − s)/2
cos(πs/2)
from which we deduce the duplication formula
−s
2(2π)
s 1 + s
Γ
cos(πs/2)π −1
cos(πs/2)Γ(s) = π
Γ
2
2
s
1+s
−s
−s/2
−(1+s)/2
2(2π) Γ(s) = π
Γ
·π
Γ
.
2
2
1/2−s
Remark. In number theory, the duplication formula plays a role in establishing that the L-function of an
induced representation is that of the inducing representation. This result for Γ(s) has an analogue for p-adic
fields known as the Hasse-Davenport relation, which is related to p-adic Gamma functions. The approach
here is of interest since it reinforces the similarity of R and Qp .
Proposition 8.2 follows from the reflection and duplication formulas, but it is natural to ask whether there is
a direct derivation, and this is a simple consequence of what I have said about the Laplace transform.
Proof of Proposition 8.2. The distribution (8.1) is the sum of two distributions, one with support on (−∞, 0]
and the other with support on [0, inf ty). Its Fourier transform is a corresponding sum. Each of these pieces
has a Laplace transform, one on the half plane RE(s) > 0, the other on RE (s) < 0, and the Fourier transform
is the sum of their limits. The formula thus reduces to:
8.3. Lemma. The Laplace transform of the distribution
hΦs , f i =
Z
∞
xs−1 f (x) dx
0
is |λ|−s Γ(s).
Proof of the Lemma. An easy calculation for s real, w which determines the formula everywhere since the
Laplace transform is holomorphic.
There is another curious question that arises here. Formally, the Fourier transform of |x|s−1 is given by the
formula
Z
G(λ) =
R
|x|s−1 e−2πiλx dx .
But the integrand converges near 0 if and only if RE(s) > 0, while it apparently converges near ∞ if and only
if RE(s) < 0. Still, it is natural to ask, can any meaning can be directly assigned to this expression? A peculiar
answer to my question is suggested by the kind of regularization found in [Casselman:1993]. Suppose c to
be imaginary. The integral
(8.4)
Z
R
f (x)|x|s−1 ecx dx
The Gamma function
23
makes sense for all f in S(R× ) and defines a continuous linear functional on this topological vector space.
Question (1) asks whether it can be extended to allow f (x) = 1. This suggests the more general problem of
extending it to all functions f (x) in C ∞ (P1 (R)). For this, express f as the sum of f0 and f∞ , the first with
support in some region |x| ≤ R, the second with support in some |x| ≥ 1/S . We must show how to evaluate
both
Z
(a)
f0 (x)|x|s−1 ecx dx
|x|≤R
and
Z
(b)
f∞ (x)|x|s−1 ecx dx .
|x|≥1/S
There is a difficulty because the first integral converges only for RE(s) > 0, the second only for RE (s) < 0. I
deal with these in turn.
(a) Suppose f (x) to be in Cc∞ (R), c an arbitrary complex number. The integral
Is =
Z
f (x)|x|s−1 ecx dx
|x|≤R
converges for RE(s) > 0, and varies holomorphically with s in that region. But f (x)ecx also lies in Cc∞ (R)
and by a familiar argument we have for RE (s) > 0
R
Z
1 R s
f (x)ecx |x|s
−
|x| (f (x)ecx )′ dx
s
s
0
0
Z
1 R s
=−
|x| (f (x)ecx )′ dx
s 0
Is =
This defines Is for RE(s) > −1, with a simple pole at s = 0, and we can continue in the same way to see that
it is defined and holomorphic for all s except for simple poles in −N.
(b) This is more interesting. I adapt the argument in the introduction to [Casselman-Hecht-Miličić:2000].
Suppose f (x) to have support in |x| ≥ 1/S > 0. Substituting y = 1/x and setting F (y) = f (1/y) the
integral we are looking at becomes
Z
|y|≤S
|y|−s−1 F (y)ec/y dy .
By assumption f (x) is smooth on P1 (R) is smooth, so F (y) lies in Cc∞ (R). I am assuming that c is imaginary,
and I now assume also that c 6= 0. The function ec/y oscillates wildly near the origin but remains bounded,
so it is at least plausible that integrating against a smooth function cancels out the oscillation. I’ll make this
precise, again integrating by parts.
Replace s − 1 by s for convenience. Set
ϕs = |x|s ec/x .
Since the derivative of |x|s is s|x|s /x,
ϕ′s = (s/x − c/x2 )ϕs
x2 ϕ′s = (sx − c)ϕs
1
ϕs = (sxϕs − x2 ϕ′s ) .
c
The Gamma function
24
Therefore
Z
Z
1 S
+
sxϕs f (x) dx
c −S
−S
Z
Z
S
1
1 S
1 S
= − ϕs (x)x2 f (x) −S +
ϕs (x)(x2 f (x))′ dx +
sxϕs f (x) dx
c
c −S
c −S
Z
1 S
ϕs (x) x2 f ′ (x) + (s + 2)xf (x) dx .
=
c −S
S
1
ϕs (x)f (x) dx = −
c
−S
Set
Z
S
ϕ′s (x)x2 f (x) dx
Df = x2 f ′ + (s + 2)xf .
Then Dn f vanishes of order n at 0, and repeated integration by parts expresses the integral as
n Z S
1
ϕs (x)[Dn f ](x) dx .
c
−S
This converges for RE (s) > −n − 1, and hence the integral defines a holomorphic function of s.
Part II. Classical formulas
9. The Beta function
The Gamma function appears in a wide variety of integration formulas. Many involve the Beta function
Γ(u)Γ(v)
.
Γ(u + v)
B(u, v) =
9.1. Proposition. We have
B(u, v) =
Z
π/2
cos2u−1 (θ) sin2v−1 (θ) dθ .
0
Proof. Start with
Γ(s) = 2
Z
∞
2
e−x x2s−1 dx .
0
Moving to two dimensions and switching to polar coordinates:
Γ(u)Γ(v) = 4
=4
=4
Z Z
Z Z
Z
2
e−s
−t2 2u−1 2v−1
s
t
ds dt
s≥0,t≥0
2
e−r r2(u+v)−1 cos2u−1 θ sin2v−1 θ dr dθ
r≥0,0≤θ≤π/2
2
e−r r2(u+v)−1 dr
r≥0
= Γ(u + v)B(u, v) .
9.2. Corollary. We have
Z
0
∞
1
tα
dt =
2
β
(1 + t )
2
Γ
α+1
2
Γ β− α+1
2
Γ(β)
The Gamma function
25
Proof. If we change variables in the Proposition to t = tan(θ) we get
θ = arctan(t)
dθ = dt/(1 + t2 )
p
cos(θ) = 1/ 1 + t2
p
sin(θ) = t/ 1 + t2
leading to
Z
∞
0
1
tα
dr =
2
β
(1 + t )
2
In particular
Γ2 (1/2) =
Z
∞
−∞
Γ
α+1
2
Γ β− α+1
2
.
Γ(β)
dr
= π,
1 + r2
Γ(1/2) =
√
π.
CAUCHY’S FORMULA .
One formula very useful in representation theory happens to be one of the very last
ones in the remarkable memoir [Cauchy:1830].
9.3. Proposition. For RE(α + β) > 1
Z
R
dx
π 21−α−β Γ(α + β − 1)
=
(1 + ix)α (1 − ix)β
Γ(α)Γ(β)
Proof I adapt an idea to be found in [Garrett:]—that of seeing this as a special case of the Plancherel formula
to the two functions
1
1
,
.
(1 + ix)α
(1 + ix)β
To do this, we must first find the Fourier transform of 1/(1 + ix)α .
9.4. Lemma. If f (x) = 1/(1 + ix)α then
fb(λ) =
if λ > 0
0
e2πλ (2π)α (−λ)α−1 /Γ(α) otherwise.
Proof of the Lemma. We get
Z
(1 + ix)−α e−2πixλ dx =
R
e2πλ
i
e2πλ
=
2πi
=
Z
RE (t)=1
Z
e−2πµ
2πi
t−α e−2πtλ dy (t = 1 + ix)
s−α e−sλ ds (s = 2πt)
RE (s)=2π
Z
RE (s)=2π
s−α esµ dy (µ = −λ)
To conclude the proof of the Proposition—the integral is the inverse Laplace transform of s−α evaluated at
−λ. We have seen this before—it is λα−1 /Γ(α).
The Gamma function
26
Therefore by the Plancherel formula our integral is
(2π)α+β
Γ(α)Γ(β)
Z
∞
1
1 2α+β π α+β
4π Γ(α)Γ(β) 2α+β−2 (2π)α+β−2
e−4πy y α+β−2 dy =
0
π 21−α−β Γ(α + β − 1)
=
.
Γ(α)Γ(β)
Z
∞
e−x xα+β−2 dy
0
10. The limit product formula
The exponential function e−t can be approximated by finite products.
Lemma. For any real t we have
n
t
.
1−
n→∞
n
e−t = lim
This can be seen most easily by taking logarithms since for 0 ≤ t < n
n
t
t
log 1 −
= n log 1 −
n
n
!
2
3
1 t
1 t
t
−
− ...
−
=n −
n
2 n
3 n
t2
t3
− 2 − ...
2n 3n
= −t − T
= −t −
where
T =
t2
t3
+ 2 + ....
2n 3n
which converges to 0 as n → ∞.
Another way of putting this is to define
ϕn (t) =
(
n
(1 − t/n)
0
0≤t≤n
t>n
and then define for each n an approximation Γn (s) to Γ(s):
Z
∞
ts−1 ϕn (t) dt
n
Z n
t
s−1
1−
=
t
dt
n
0
Γn (s) =
0
On the one hand, this can be explicitly calculated through repeated integration by parts:
Z
0
n
n
Z n
t
1 n−1 n−2
1
n! ns
ts−1 1 −
dt =
...
ts+n−1 dt =
n
s n(s + 1) n(s + 2)
n(s + n − 1) 0
s(s + 1) . . . (s + n)
On the other, since for all fixed t the limit of ϕn (t) as n → ∞ is equal to e−t , and both ϕn (t) and e−t are small
at ∞, this is at least plausible:
The Gamma function
27
Proposition. For any s with ℜ(s) > 1 the limit of Γn (s) as n → ∞ is equal to Γ(s). In other words, for any s
in C
n! ns
.
n→∞ s(s + 1) . . . (s + n)
Γ(s) = lim
The Euler constant γ is defined to be the limit
1
1 1
− log n.
γ = lim 1 + + + . . .
n→∞
2 3
n−1
The limit product formula implies immediately a limit formula for 1/Γ(s):
1
s
s
s
1+
... 1 +
n−s
= lim s 1 +
Γ(s) n→∞
1
2
n−1
but
n−s = e−s log n = e−s(1+ 2 + 3 +...+ n−1 )+sγn
1
1
1
where γn → γ . Therefore:
Proposition. The inverse Gamma function has the product expansion
∞ Y
1
s −s/n
1+
e
= seγs
Γ(s)
n
1
where γ is Euler’s constant.
The limit product formula also implies Legendre’s duplication formula:
s
s+1
1
s−1
Γ(s) = 2 Γ
Γ
Γ
2
2
2
Explicitly
s
2n+1 n! ns/2
= lim
n→∞ s(s + 2) . . . (s + 2n)
2
s+1
2n+1 n! ns+1/2
Γ
= lim
n→∞ (s + 1) . . . (s + 2n + 1)
2
Γ
so
s
s+1
2 Γ
Γ
Γ(s)
2
2
s
= lim 2s
n→∞
2n+1 n! n(s+1)/2
s(s + 1)(s + 2) . . . (s + 2n)
2n+1 n! ns/2
s(s + 2) . . . (s + 2n) (s + 1)(s + 3) . . . (s + 2n + 1)
(2n)! (2n)s
(n!)2 22n+2 n1/2
n→∞ (2n)! (s + 2n + 1)
(n!)2 22n+1
√
= lim
n→∞ (2n)! n
= lim
but this last does not depend on√
s, and is finite since the limit on the left hand side exists, so we may set
s = 1/2 to see that it is equal to 2 π .
The Gamma function
28
11. The reflection formula
The formula for the Beta function gives us
Γ(s)Γ(1 − s) =
=
Z
0
Z
0
1
us−1 (1 − u)−s du
∞
v s−1
dv (u = v/1 + v, v = (u/1 − u), du/(1 − u) = (1 + v)dv)
1+v
We can calculate this last integral by means of a contour integral in C. Let C be the path determined by these
four segments: (1) along the positive real axis, or just above it, from ǫ to R; (2) around the circle of radius R,
counter-clockwise, to the point just below R; (3) along and just below the real axis to ǫ; (4) around the circle
of radius ǫ, clockwise, to just above ǫ. We want to calculate the limit of the integral
Z
C
z s−1
dz
1+z
as ǫ → 0 and R → ∞.
On the one hand the integrals over the different components converge to
Z
0
∞
z s−1
dz + 0 − e2πis
1+z
Z
0
∞
z s−1
dz + 0 = (1 − e2πis )
1+z
Z
∞
0
z s−1
dz
1+z
But on the other there is exactly one pole inside the curves C , so the integral is also equal to −2πieπis .
Therefore
Z ∞ s−1
πis
Γ(s)Γ(1 − s) =
0
z
π
−2πie
=
dz =
2πis
1+z
1−e
sin πs
Incidentally, combined with the product formula for Γ(s) this gives the product formula for sin πs
sin πs = πs
∞ Y
s2
1− 2
n
1
12. The Euler-Maclaurin formula
Define a sequence of polynomials
B0 (x) = 1
B1 (x) = x − 1/2
B2 (x) = x2 − x + 1/6
...
recursively determined by
′
Bn+1
(x) = nBn (x),
Z
1
Bn (x) dx = 0 .
0
These are the Bernoulli polynomials. They determine in turn functions ψn by extension to all of R of period
1.
The following is a simple version of the much more interesting Euler-Maclaurin sum formula:
The Gamma function
29
Proposition. Suppose f to be a function on the interval [k, ℓ] which has continuous second derivatives. Then
f (k) + f (k + 1) + . . . f (ℓ − 1) =
Z
k
ℓ
f (x) dx −
where
R2 = −
1
2
Z
ℓ
1
1 ′
f (ℓ) − f (k) +
f (ℓ) − f ′ (k) + R2
2
12
f ′′ (x)ψ2 (x) dx .
k
The proof is very simple, a repetition of integration by parts. Suppose m to be an integer with f defined and
continuously differentiable on [m, m + 1]. Then since ψ0 = 1 and ψ1′ = ψ0
Z
m+1
f (x) dx =
m
Z
m+1
f (x)ψ0 (x) dx
m
m+1
= [f (x)ψ1 (x)]m
−
Z
m+1
m
1
= f (m) + f (m + 1) −
2
f ′ (x)ψ1 (x) dx
Z
m+1
f ′ (x)ψ1 (x) dx
m
since ψ1′ (x) = 1, and of course we look at the limit of ψ1 from above at m, the limit from below at m + 1.
Then we sum this equation over all the unit sub-intervals of [k, ℓ], using the periodicity of ψ1 .
Z
ℓ
k
f (x) dx = (1/2)f (k) + f (k + 1) + . . . + f (ℓ − 1) + (1/2)f (ℓ) −
Z
ℓ
f ′ (x)ψ1 (x) dx
k
We can rewrite this and apply integration by parts successively:
f (k) + f (k + 1) + . . . + f (ℓ − 1)
Z ℓ
Z ℓ
1
=
f (x) dx − f (ℓ) − f (k) +
f ′ (x)ψ1 (x) dx
2
k
k
Z ℓ
Z
1
1 ℓ ′′
1
′
′
=
f (x) dx − f (ℓ) − f (k) + ψ2 (ℓ)f (ℓ) − ψ2 (k)f (k) −
f (x)ψ2 (x) dx
2
2
2 k
k
Z ℓ
Z
1 ℓ ′′
1
1 ′
=
f (x) dx − f (ℓ) − f (k) +
f (ℓ) − f ′ (k) −
f (x)ψ2 (x) dx
2
12
2 k
k
The calculations can be continued to obtain an infinite asymptotic expansion involving the polynomials and
their constant terms, the Bernoulli numbers.
The Gamma function
30
13. Stirling’s formula
We know that the Gamma function can be evaluated as a limit product
(n − 1)! (n − 1)s
n→∞ s(s + 1) . . . (s + n − 1)
s
(n − 1)! ns
n−1
= lim
n→∞ s(s + 1) . . . (s + n − 1)
n
s
(n − 1)! n
= lim
n→∞ s(s + 1) . . . (s + n − 1)
Γ(s) = lim
We have proven this for s in the domain of convergence of the integral defining Γ(s), but in fact the limit
exists and defines an analytic function for all s except s = −n with n a non-negative integer, so that by the
principle of analytic continuation it must be valid wherever Γ(s) is defined. As a consequence
log Γ(s) = lim Sn−1 (1) − Sn (s) + s log n
n→∞
where
Sn (s) = log s + log(s + 1) + . . . + log(s + n − 1)
We can evaluate Sn (s) by the Euler-Maclaurin formula
f (0) + f (1) + . . . + f (n − 1)
Z
Z n
β2 ′
1 n (2)
1
f (n) − f ′ (0) −
f (x)ψ2 (x) dx
f (x) dx − f (n) − f (0) +
=
2
2
2 0
0
with
f (x) = log(s + x),
f ′ (x) =
1
,
s+x
f (2) (x) = −
1
(s + x)2
so
log s+ log(s + 1) + . . . + log(s + n − 1)
Z n
Z
1
1 n ψ2 (x)
1
1
1
=
log(s + x) dx − [log(s + n) − log s] +
+
dx
−
2
12 s + n s
2 0 (s + x)2
0
Z n
1
1
1
1
1
ψ2 (x)
s+n
= [x log x − x]s − [log(s + n) − log s] +
+
dx
−
2
12 s + n s
2 0 (s + x)2
Z n
1
1
1
ψ2 (x)
1
+
dx
−
= (s + n − 1/2) log(s + n) − (s − 1/2) log s − n +
12 s + n s
2 0 (s + x)2
and setting s = 1, n − 1 for n:
log 1 + log 2+ . . . + log n
Z
1 1
1 n ψ2 (x)
dx
−1 +
12 n
2 1
x2
Z
11
1
1 n ψ2 (x)
= (n − 1/2) log n − n +
dx
+
+
12 12n 2 1
x2
Z
1 ∞ ψ2 (x)
1
dx
−
= (n − 1/2) log n − n + C +
12n 2 n
x2
= (n − 1/2) log n − (n − 1) +
The Gamma function
31
where we define the constant
11 1
+
C=
12 2
Z
∞
1
ψ2 (x)
dx.
x2
Taking limits, therefore
log Γ(s) =
Z
1
1
1 ∞ ψ2 (x)
log s − s + C +
s−
dx.
−
2
12s 2 0 (s + x)2
This is valid for all s not on the negative real axis, and gives immediately the generalization of Stirling’s
formula
as s goes to infinity in any region
eC s s
Γ(s) ∼ √
s e
−π + δ ≤ arg(s) ≤ π − δ
since the remainder will have a uniform estimate in this region. The constant C can be evaluated by letting
t → ±∞ in the reflection formula. On the one hand
π
it sin πit
2πi
=−
it[e−πt − eπt ]
Γ(it)Γ(−it) = −
∼ 2πt−1 e−πt
while on the other
it C −it
e
eC it
−it
√
Γ(it)Γ(−it) ∼ √
e
−it
it e
2C
e
(i)it (−i)−it
=
t
e2C (it)(πi)/2 (−it)(−πi)/2
=
e
e
t
e2C −πt
=
e
t
I recall that
xy = ey log x
where log is given its principal value. This gives
C = log
√
2π
and finally the explicit version
Proposition. (Stirling’s asymptotic formula) As s goes to ∞ in the region
−π + δ ≤ arg(s) ≤ π − δ
we have the asymptotic estimate
Γ(s) ∼
r
2π s s
s e
The Gamma function
32
14. The volumes and areas of spheres
If we set s = 1 in the formula for ζR at the end of the last section, we get
Z ∞
2
1
=
e−πx dx .
π
Γ
2
−∞
The integral on the right cannot be evaluated as an improper integral, but there is a well known trick one can
use to evaluate the infinite integral. We move into two dimensions. We can shift to polar coordinates and get
Z
2 Z
Z
2
2
2
e−πx dx =
e−πx dx · e−πy dy
R
R
ZR
2
2
=
e−π(x +y ) dx dy
−1/2
R2
∞
=
=
=
=
=
Z
dr
Z
2π
2
e−πr r dθ
Z0 ∞
2πre−πr dr
Z0 ∞
e−πr (2πr) dr
Z0 ∞
Z0 ∞
0
2
√
2
2 πre−πr dr
2
e−s ds
0
so π
−1/2
√
Γ(1/2) = 1, and Γ(1/2) =
= 1,
π.
We can use this formula and the same trick to find a formula for the volumes of spheres in n dimensions. Let
Sn−1 be the volume of the unit sphere in Rn . Then
Z
e
−πx2
R
n
dx
=1
Z
2
e−πr dx1 . . . dxn
=
n
ZR∞
2
=
Sn−1 rn−1 e−πr dr
Z0 ∞
2 dr
Sn−1 rn e−πr
=
r
0
1 −n/2
Γ(n/2) .
= Sn−1 π
2
2π n/2
Sn−1 =
.
Γ(n/2)
For example, the area of the two-sphere in R3 is
2π 3/2
2π 3/2
=
= 4π .
Γ(3/2)
π/2
S2 =
The volume of the n-ball of radius R in Rn is
Vn (R) =
Z
0
R
Sn−1 rn−1 dr =
Sn−1 Rn
.
n
The Gamma function
33
Part III. References
15. References
1. W. A. Casselman, ‘Extended automorphic forms on the upper half plane’, Mathematische Annalen 296
(1993), 755–762.
2. ——, Henryk Hecht, and Dragan Miličić, ‘Bruhat filtrations and Whittaker vectors for real groups’, pp.
151–190 in [Doran:2000].
3. Robert S. Doran and V. S. Varadarajan (editors), The mathematical legacy of Harish-Chandra, Proceedings
of Symposia in Pure Mathematics 68, American Mathematical Society, 2000.
4. I. M. Gelfand, M. I. Graev, I. I. Pyatetskii-Shapiro, Representation theory and automorphic functions, W.
B. Saunders, 1969.
5. Jacques Hadamard, Lectures on Cauchy’s problem, Yale University Press, 1923.
Chapter 1 of Book III introduces ‘finite parts’ of integrals. This notion is necessary in order to interpret the
fundamental solutions to the wave equation in high dimensions.
6. Michael Reed and Barry Simon, Fourier analysis and self-adjointness, volume II of the series Methods
of Mathematical Physics, Academic Press, 1972.
7. Paul Sally and Mitchell Taibleson, ‘Special functions on locally compact fields’ Acta Mathematica 116
(1966), 279–309.
8. Laurent Schwartz, Méthodes mathématiques pour les sciences physiques, Hermann, 1965. Also available in English. This is very readable, although with few rigourous proofs.
9. John Tate, ‘Fourier analysis in number fields and Hecke’s zeta functions’, pp. 305–347 in in Algebraic
number theory, edited by J. W. S. Cassels and A. Fröhlich, Thompson, 1967. This is his Princeton Ph. D.
thesis.
10. André Weil, ‘Fonction zêta et distributions’, Séminaire Bourbaki, exposé 312, Juin 1966. Bourbaki