MELLIN TRANSFORM AND RIEMANN ZETA FUNCTION
CINDY TAGARIS
Definiton
∞
Z x+i∞
1
x−s g(s)dx.
2πi
0
c−i∞
Note:
Z ∞ We need to restrict s to values where the integral will converge. We have g(s) exists
|f (x)|xs−1 dx is bounded for some k > 0. And f (x) will have c > k.
if
Z
Define g(s) = M{f }(s) =
f (t)ts−1 dt and f is recoverable: f (x) =
0
Examples
Z
∞
(1) The Γ function is defined Γ(s) =
e−t ts−1 dt.
0
Immediately, we see that Γ(s) = M{f }(s) where f (t) = e−t .
Notice this analytically continues Γ to C except at s = 0, −1, −2, . . . .
Aside: Γ has some nice properties that we will be utilizing later:
√
1
Γ(s + 1) = Γ(s)
Γ
= π.
2
1
(2) Let f (x) = x
. And let Re(s) > 1.
e −1
Z ∞
1
dx
M{f }(s) =
xs−1 x
e
−1
0
Z ∞
∞
X
s−1
=
x
e−nx dx
0
=
n=1
∞ Z ∞
X
xs−1 e−nx dx
n=1 0
Set: nx = t dt = ndx
∞ Z ∞ s−1
X
t
1
=
e−t dt
n
n
n=1 0
Z
∞
∞
X
=
n−s
ts−1 e−t dt
0
n=1
= ζ(s)Γ(s).
1
2
CINDY TAGARIS
Properties
Note: There are many interesting properties of the Mellin transform, but we will only cover
a few of interest.
Z
Z
∞
(1) M{f (ax)}(s) =
f (ax)xs−1 dx = a−s
∞
M{f }(s)
0
−ct
0
Example: Let f (t) = e−t , c > 0, f (ct) = e .
We already know that M{f }(s) = Γ(s). So, we anticipate M{f (ct)}(s) = c−s Γ(s).
M{e
−ct
Z
∞
dt
t
0
u
ct ⇒ du =cdt
Z=
u s 1 c ∞
=
e−u
du
c
c
u
0
Z ∞
e−u us−1 du
= c−s
}(s) =
e−ct ts−1
0
= c−s Γ(s).
Z ∞
Z ∞
a
a s−1
(2) M{x f (x)}(s) =
f (x)x x dx =
f (x)x(a+s)−1 dx = M{f (x)}(s + a).
0
0
1
1
(3) M
f
(s) = M{f (x)}(1 − s).
x
x
Recall
The theta function: θ(t) =
∞
X
2
e−πtn
(Re(t) > 0) has properties:
−∞
1
1
(1) θ(t) = √ θ
. The proof will use Poisson and Fourier transforms.
t t
−c
−1
1
(2) θ(t) − t− 2 < e t : c > 0 ⇒ θ(t) t 2 .
MELLIN TRANSFORM AND RIEMANN ZETA FUNCTION
3
Idea:
We are going to relate θ and ζ. In particular, there is a relationship between the Riemann
Zeta function and the Mellin transform of the Theta function. This relationship (along
with the functional equation for the Theta function) will provide insight on the Riemann
Zeta function. In particular, we will get a functional equation and meromorphic continuuation for the Riemann Zeta function.
Theorem
Let ζ(s) =
∞
X
for Re(s) > 1. We have that ζ extends analytically onto C except for a
s
−s
simple pole at s = 1 with Residue of 1. And if we define Λ(s) = π 2 Γ
ζ(s), then we
2
obtain the functional equation: Λ(s) = Λ(s − 1). In particular, ζ(s) satisfies the functional
equation
n=1
π
−s
2
Γ
s
2
ζ(s) = π
−(1−s)
2
Γ
1−s
2
ζ(1 − s).
Proof
Define
Z
∞
s
t 2 (θ(t) − 1)
φ(s) =
1
Note that θ(t) − 1 = 2
−1
∞
X
dt
+
t
1
Z
s
t2
0
1 dt
.
θ(t) − √
t t
2
e−πn t → 0 at ∞ ⇒ the integral converges.
n=1
Also, since θ(t) t 2 , the second integral will converge, as well.
Let Re(s) > 1 and focus on the second integral:
Z
1
s
t 2 θ(t)
0
dt
−
t
Z
1
t
0
s−1
2
dt
=
t
Z
1
s
t 2 θ(t)
0
2
dt
−
.
t
s−1
4
CINDY TAGARIS
Z
φ(s) = 2
∞
∞X
1
n=1
∞
XZ
−πnt
e
dt
t
+
t
s
2
Z
0
1
∞
X
dt
t
+2
t
s
2
Z
n−1 0
1
s
2t
t 2 e−πn
dt
2
−
t
s−1
dt 2
2
+ +
t
s
1
−
s
n=1 0
s
Letting c → πn2 , s → , and using the Scaling property previously discussed:
2
∞
X
−s
1
s
1
1
φ(s) =
+ +
(πn2 ) 2 Γ
2
2
s 1−s
= 2
=
n=1
∞
X
π
−s
2
n
2
s
e−πn t t 2
−s
Γ
s
n=1
+
1
1
+
s 1−s
1
1
+
.
2
s 1−s
We already established the integrals making up φ were convegent so for all s such that
Re(s) > 1, we have that φ is entire with respect to s. Thus, we have meromorphic continuuation:
s
1
1
1
2
π
2 φ(s) − s − 1 − s
s
= ζ(s).
Γ
2
s
1
Moreover, π 2 , s , φ(s) are entire so the poles of ζ are simple and at s = 0, 1.
Γ
2
Consider the
spole
at s = 0 :
s
s s
s
We have Γ
s in the denominator. Replace Γ
s=2
Γ
= 2Γ
+ 1 6= 0
2
2
2
2
2
as s → 0. So, we have fixed this.
Next, consider the pole at s = 1 :
To find the residue, we consider
s
1
π2
1 1
1
π2
lim (s − 1) s − −
= = 1.
s→1
1
2 s 1−s
Γ
Γ
2
2
= π
−s
2
ζ(s)Γ
s
2
+
MELLIN TRANSFORM AND RIEMANN ZETA FUNCTION
5
Finally, we need to establish the functional equation:
1
1
1
1
1
. Since and
are invariant replacing s → 1 − s, we
Let Λ(s) = φ(s) − −
2
s
1−s
s
1−s
just need to show that φ(s) = φ(1 − s).
1
Recall the functional equation for θ. Then, switch the bounds on φ from t → . After a
t
little work, we obtain:
Z 1
Z ∞
√ dt
−s
−s
1
dt
1
φ(s) =
t2 θ
−1
− t
+
t2 θ
t
t
t
t
0
1
Z 1
Z
∞
√
√
√ d
−s
−s
d
=
t2
tθ(t) − 1
+
t2
tθ(t) − t
t
t
0
1
Z 1
Z ∞
1−s
1−s
1 dt
dt
t 2 θ(t) − √
=
t 2 (θ(t) − 1)
+
t
t t
0
1
= φ(1 − s).
This completes the proof of the functional equation and, thus, the theorem.
Remark
In a similar way, one can obtain analytic continuuation and functional equations for more
general series by inserting Dirichlet characters. Furthermore, the Mellin transform extends
to C easily, i.e. one could define the Mellin transform for the C case, as well, taking the
integral to i∞.