Mellin transforms and the functional equation of
the Riemann Zeta function
Marko R. Riedel
December 3, 2015
We collect several examples of using Mellin transforms and the functional
equation of the Riemann Zeta function to evaluate harmonic sums. These are
collected from my posts at math.stackexchange.com and have retained the question answer format used there.
The methods here are those of the papers by Szpankowski [Szp01] and Flajolet [FS96].
1
Contents
1 Evaluating Fourier series
3
2 Some sum formulae
2.1 First formula . .
2.2 Second formula .
2.3 Third formula . .
6
7
8
9
by
. .
. .
. .
Apery and Plouffe
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . .
3 A sum formula by Hardy and Ramanujan
10
4 Predecessor of sum formula by Cauchy and Ramanujan
12
5 A sum formula by Cauchy and Ramanujan
15
6 A functional equation relating two harmonic sums
19
7 Functional equation of the Hurwitz Zeta function
23
8 Two contrasting examples of fixed point scenarios
26
2
1
Evaluating Fourier series
Suppose we seek to show that
π
sin2 (n/q)
=
.
2
n
4q
n=1,3,5,...
X
As suggested we use
sin2 t =
1 − cos(2t)
2
to get
1 1 − cos(2n/q)
n2
2
n=1,3,5,...
X
which is
1
1
1 X
1 X
−
cos(2n/q).
2
2 n=1,3,5,... n
2 n=1,3,5,... n2
We will be using
1
=
s
n
n=1,3,5,...
X
1
1 − s ζ(s)
2
which gives for the sum
1
1 X
1
1
1 − 2 ζ(2) −
cos(2n/q)
2
2
2 n=1,3,5,... n2
or
1 X
π2
1
−
cos(2n/q).
16 2 n=1,3,5,... n2
Introduce S(x) given by
∞
S(x) =
X
1
1
cos(nx) =
cos((2k − 1)x)
2
n
(2k − 1)2
n=1,3,5,...
X
k=1
so that we are interested in S(2/q).
The sum term is harmonic and may be evaluated by inverting its Mellin
transform.
Recall the harmonic sum identity

 

X
X λk
 g ∗ (s)
λk g(µk x); s = 
M
µsk
k≥1
k≥1
where g ∗ (s) is the Mellin transform of g(x).
In the present case we have
λk =
1
,
(2k − 1)2
µk = (2k − 1)
3
and g(x) = cos(x).
We need the Mellin transform g ∗ (s) of g(x).
Now the Mellin transform of cos(x) was computed at math.stackexchange.com
question 479586 and found to be
Γ(s) cos(πs/2)
It follows that the Mellin transform Q(s) of the harmonic sum S(x) is given
by
Q(s) = Γ(s) cos(πs/2) 1 −
because
X λk
=
µsk
k≥1
1
2s+2
X
k≥1
ζ(s + 2)
1
1
=
(2k − 1)2 (2k − 1)s
1−
1
2s+2
ζ(s + 2)
for <(s) > −1.
The Mellin inversion integral here is
Z 1/2+i∞
1
Q(s)/xs ds
2πi 1/2−i∞
which we evaluate by shifting it to the left for an expansion about zero (the
abscissa <(s) = 1/2 is in the intersection of h−1, ∞i and h0, 1i from the cosine
transform).
The zeros of the cosine term at the negative odd integers cancel the poles of
the gamma function at those values. Additional cancelation is gained from the
trivial zeros of the zeta function term ζ(s + 2) at the even negative integers p
with p ≤ −4.
This leaves just two poles at s = 0 and s = 1 and we have
Ress=0 Q(s)/xs =
π2
8
and
π
Ress=−1 Q(s)/xs = − x
4
and therefore
π2
π
− x.
8
4
We will see that this is exact for x ∈ [0, π).
With q ≥ 1 we have 2/q ≤ 2 and we get for the initial sum the form
S(x) ∼
π2
1 π2
1π2
π
−
+
=
16 2 8
24q
4q
which is the claim we were trying to prove.
We still need to prove exactness on [0, π) to complete the argument.
Put s = σ + it with σ ≤ −3/2 where we seek to evaluate
1
2πi
Z
−3/2+i∞
Q(s)/xs ds
−3/2−i∞
4
by shifting it to the left.
Recall that with σ > 1 and for |t| → ∞ we have
|ζ(σ + it)| ∈ O(1).
Furthermore recall the functional equation of the Riemann Zeta function
πs 2
cos
Γ(s)ζ(s)
2s π s
2
ζ(1 − s) =
which we re-parameterize like so
ζ(s + 2) = 2 × (2π)
s+1
π(s + 1)
cos −
2
which is
ζ(s + 2) = −2 × (2π)s+1 sin(πs/2)
Γ(−s − 1)ζ(−s − 1)
Γ(1 − s)
ζ(−s − 1).
s(s + 1)
Substitute this into Q(s) to obtain
1
Γ(1 − s)
Γ(s) cos(πs/2) 1 − s+2 × −2 × (2π)s+1 sin(πs/2)
ζ(−s − 1).
2
s(s + 1)
Use the reflection formula for the Gamma function to obtain
π
1
1
ζ(−s − 1),
cos(πs/2) 1 − s+2 × −2 × (2π)s+1 sin(πs/2) ×
2
sin(πs) s(s + 1)
in other words we have
s+1
Q(s) = −π(2π)
1−
1
2s+2
ζ(−s − 1)
.
s(s + 1)
There are two components here, call them Q1 (s) and Q2 (s), which are
−π(2π)s+1
ζ(−s − 1)
s(s + 1)
and π(2π)s+1
1 ζ(−s − 1)
.
2s+2 s(s + 1)
We evaluate these with σ < −5/2. For the first component this implies (with
σ = −5/2 we have <(−s − 1) = 3/2)
|Q1 (s)/xs | ∼ 2π 2 (2π)σ x−σ |t|−2 .
or
|Q1 (s)/xs | ∼ 2π 2 (x/2/π)−σ |t|−2 .
We see from the term in |t| that the integral obviously converges. (This
much we knew already.) Moreover, when x ∈ (0, 2π) we have (x/2/π)−σ → 0 as
σ → −∞. The term in x does not depend on the variable t of the integral and
may be brought to the front. This means that the contribution from the left
5
side of the rectangular contour that we employ as we shift to the left vanishes
in the limit.
For the second component we get
|Q2 (s)/xs | ∼
π2
(π)σ x−σ |t|−2 .
2
or
π2
(x/π)−σ |t|−2 .
2
This is the same as the first only now we have convergence in (0, π).
Joining the bounds for Q1 (s) and Q2 (s) we have proved the exactness of the
formula for S(x) in the interval (0, π) obtained earlier.
As I have mentioned elsewhere there is a theorem hiding here, namely that
certain Fourier series can be evaluated by inverting their Mellin transforms
which is not terribly surprising and which the reader is invited to state and
prove.
This is math.stackexchange.com question 1153068.
|Q2 (s)/xs | ∼
2
Some sum formulae by Apery and Plouffe
Suppose we are trying to prove that
∞
∞
k=1
k=1
X
X
1
π3 √
1
3
√
√
−
ζ(3) =
2−2
2
24
k 3 (eπk 2 − 1)
k 3 (e2πk 2 − 1)
∞
∞
X
X
3
π √
1
1
√
√
ζ(5) =
2−4
+
5
πk
2
5
2πk
2 − 1)
2
270
k (e
− 1)
k (e
5
k=1
∞
k=1
∞
X
X
1
9
41π 7 √
1
√
√
ζ(7) =
2−8
−
7
πk
7
2πk
2
2 − 1)
2
37800
k (e
k (e
− 1)
k=1
k=1
Introduce the sum
S(x; α, p) =
X
n≥1
1
np (eαnx − 1)
√
with p √
a positive odd integer and α > 1, so that we seek e.g. 2S(1; π 2, 3) +
S(1; 2π 2, 3).
The sum term is harmonic and may be evaluated by inverting its Mellin
transform.
Recall the harmonic sum identity

 

X
X λk
 g ∗ (s)
M
λk g(µk x); s = 
µsk
k≥1
k≥1
6
where g ∗ (s) is the Mellin transform of g(x).
In the present case we have
λk =
1
,
kp
µk = k
1
.
eαx − 1
and g(x) =
We need the Mellin transform g ∗ (s) of g(x) which is
∞
Z
0
Z ∞
1
e−αx
s−1
xs−1 dx
x
dx
=
eαx − 1
1 − e−αx
0
Z ∞X
XZ
e−αqx xs−1 dx =
=
0
q≥1
q≥1
∞
e−αqx xs−1 dx
0
= Γ(s)
X
q≥1
1
1
= s Γ(s)ζ(s).
(αq)s
α
It follows that the Mellin transform Q(s) of the harmonic sum S(x; α, p) is
given by
Q(s) =
1
Γ(s)ζ(s)ζ(s + p)
αs
X λk
X 1 1
= ζ(s + p)
=
s
µk
kp ks
because
k≥1
k≥1
for <(s) > 1 − p.
The Mellin inversion integral here is
Z 3/2+i∞
1
Q(s)/xs ds
2πi 3/2−i∞
which we evaluate by shifting it to the left for an expansion about zero.
2.1
First formula
We take
Q(s) =
1
√
2+
1
2s
Γ(s)ζ(s)ζ(s + 3).
s
πs 2
We shift the Mellin inversion integral to the line s = −1, integrating right
through the pole at s = −1 picking up the following residues:
√
π3 2
Res(Q(s)/x ; s = 1) =
72x
s
and
and
3
Res(Q(s)/xs ; s = 0) = − ζ(3)
2
√
1
π 3 2x
s
Res(Q(s)/x ; s = −1) =
.
2
36
This almost concludes the proof of the first formula if we can show that the
integral on the line <(s) = −1 vanishes when x = 1. To accomplish this we must
show that the integrand is odd on this line.
7
Put s = −1 − it in the integrand to get
√ 1+it
π 1+it 2
2 + 21+it Γ(−1 − it)ζ(−1 − it)ζ(2 − it).
Now use the functional equation of the Riemann Zeta function in the following form:
πs 2
Γ(s)ζ(s)
ζ(1 − s) = s s cos
2 π
2
to obtain (with s = −1 − it)
√ 1+it
π 1+it 2
2 + 21+it ζ(2 + it)2−1−it π −1−it
1
2 cos
which is
√
1+it
2
1
2−it + 1 ζ(2 + it)
2 cos
π(1+it)
2
π(−1−it)
2
ζ(2 − it)
ζ(2 − it)
and finally yields
−
√ −1−it √ −1+it 1
2
+ 2
ζ(2 + it)ζ(2 − it).
sin(πit/2)
It is now possible to conclude by inspection: the zeta function terms and the
powers of the square root are even in t and the sine term is odd, so the whole
term is odd and the integral vanishes.
2.2
Second formula
We take
Q(s) =
1
√
s
πs 2
4−
1
2s
Γ(s)ζ(s)ζ(s + 5).
We shift the Mellin inversion integral to the line s = −2 (no pole on the line
this time) picking up the following residues:
√
π5 2
Res(Q(s)/x ; s = 1) =
540x
s
and
3
Res(Q(s)/xs ; s = 0) = − ζ(5)
2
and
√
π 5 2x
Res(Q(s)/x ; s = −1) =
.
540
It remains to verify that the integrand on the line <(s) = −2 is odd when
x = 1. Put s = −2 − it in the integrand to get
s
√ 2+it
4 − 22+it Γ(−2 − it)ζ(−2 − it)ζ(3 − it).
π 2+it 2
8
Applying the functional equation once again with s = −2 − it we obtain
√ 2+it
π 2+it 2
4 − 22+it ζ(3 + it)2−2−it π −2−it
1
2 cos
which is
√
2+it
2
1
2−it − 1 ζ(3 + it)
2 cos
which is in turn
√
2−it
2
−
√
2
2+it
1
2 cos
which finally yields
√ 2−it √ 2+it 2
− 2
−
π(2+it)
2
π(−2−it)
2
π(−2−it)
2
ζ(3 − it)
ζ(3 − it)
ζ(3 + it)ζ(3 − it)
1
ζ(3 + it)ζ(3 − it)
2 cos(πit/2)
The product of the zeta function terms is even, as is the cosine term. The
term in front is odd, so the integrand is odd as claimed.
2.3
Third formula
We take
Q(s) =
1
√
8+
1
2s
Γ(s)ζ(s)ζ(s + 7).
s
πs 2
We shift the Mellin inversion integral to the line s = −3, integrating right
through the pole at s = −3 picking up the following residues:
√
17π 7 2
Res(Q(s)/x ; s = 1) =
37800x
s
and
9
Res(Q(s)/xs ; s = 0) = − ζ(7)
2
and
Res(Q(s)/xs ; s = −1) =
√
π 7 2x
1134
√
1
π 7 2x3
Res(Q(s)/xs ; s = −3) = −
.
2
4050
and
This almost concludes the proof of this third formula if we can show that
the integral on the line <(s) = −3 vanishes when x = 1. To accomplish this we
must show once more that the integrand is odd on this line.
Put s = −3 − it in the integrand to get
√ 3+it
π 3+it 2
8 + 23+it Γ(−3 − it)ζ(−3 − it)ζ(4 − it).
By the functional equation we obtain with s = −3 − it
√ 3+it
π 3+it 2
8 + 23+it ζ(4 + it)2−3−it π −3−it
9
1
ζ(4 − it)
2 cos (π(−3 − it)/2)
which is
√
2
3+it
2−it + 1 ζ(4 + it)
1
ζ(4 − it)
2 cos (π(3 + it)/2)
which finally yields
√ 3−it √ 3+it 1
2
+ 2
ζ(4 + it)ζ(4 − it).
2 sin(πit/2)
This concludes it since the two zeta function terms together are even as is
the square root term while the sine term is odd, so their product is odd.
This is math.stackexchange.com question 157040.
3
A sum formula by Hardy and Ramanujan
Suppose we are trying to evaluate
X
coth(nπx) + x2 coth(nπ/x) /n3 .
n≥1
Put
S(x) = ζ(3) +
X −1 + coth(nπx)
n3
n≥1
and introduce the sum
T (x) =
X −1 + coth(nπx)
.
n3
n≥1
The sum term is harmonic and may be evaluated by inverting its Mellin
transform. We will construct a functional equation for T (x).
Recall the harmonic sum identity

 

X
X λk
 g ∗ (s)
M
λk g(µk x); s = 
µsk
k≥1
k≥1
∗
where g (s) is the Mellin transform of g(x).
In the present case we have
λk =
1
,
k3
µk = k
and g(x) = 2
e−2πx
.
1 − e−2πx
We need the Mellin transform g ∗ (s) of g(x) which is
Z ∞
e−2πx
2
xs−1 dx
−2πx
1
−
e
0
Z ∞X
XZ ∞
=2
e−2qπx xs−1 dx = 2
e−2qπx xs−1 dx
0
q≥1
q≥1
= 2Γ(s)
0
X
q≥1
10
1
2 1
= s s Γ(s)ζ(s).
s
(2πq)
2 π
It follows that the Mellin transform Q(s) of the harmonic sum T (x) is given
by
Q(s) =
2 1
Γ(s)ζ(s)ζ(s + 3)
2s π s
because
X λk
X 1 1
=
= ζ(s + 3)
s
µk
k3 ks
k≥1
k≥1
for <(s) > −2.
The Mellin inversion integral here is
1
2πi
Z
3/2+i∞
Q(s)/xs ds
3/2−i∞
which we evaluate by shifting it to the left for an expansion about zero.
Fortunately the trivial zeros of the two zeta function terms cancel the poles
of the gamma function term. Shifting to <(s) = −3 − 1/2 we get
T (x) =
π 3 x3
π3 x
π3
1
+ 4ζ 0 (−2)π 2 x2 +
− ζ(3) +
+
90
18
90x 2πi
Z
−7/2+i∞
Q(s)/xs ds.
−7/2−i∞
Substitute s = −2 − t in the remainder integral to get
1
−
2πi
Z
3/2−i∞
3/2+i∞
2
1
2−2−t
π −2−t
Γ(−2 − t)ζ(−2 − t)ζ(1 − t)xt+2 dt
which is
x2
2πi
Z
3/2+i∞
23+t π 2+t Γ(−2 − t)ζ(−2 − t)ζ(1 − t)xt dt.
3/2−i∞
In view of the desired functional equation we now use the functional equation
of the Riemann zeta function on Q(s) to prove that the integrand of the last
integral is in fact −Q(t).
Start with the functional equation
ζ(1 − s) =
πs 2
cos
Γ(s)ζ(s)
2s π s
2
and substitute this into Q(s) to obtain
Q(s) =
2 1 ζ(1 − s)2s π s
ζ(3 + s)
ζ(s + 3) =
ζ(1 − s).
2s π s 2 cos πs
cos πs
2
2
Apply the functional equation again (this time to ζ(s + 3)) to get
2
π(−2 − s)
1
cos
Γ(−2 − s)ζ(−2 − s)ζ(1 − s)
Q(s) =
2−2−s π −2−s
2
cos πs
2
11
Observe that
cos − πs
cos −π − πs
2
2
=
−
= −1
cos πs
cos πs
2
2
so we finally get
Q(s) = −23+s π 2+s Γ(−2 − s)ζ(−2 − s)ζ(1 − s),
thus proving the claim.
We have established the functional equation
T (x) =
π 3 x3
π3 x
π3
+ 4ζ 0 (−2)π 2 x2 +
− ζ(3) +
− x2 T (1/x).
90
18
90x
Finally returning to the sum that was the initial goal we see that it has the
value
ζ(3) + T (x) + x2 (ζ(3) + T (1/x))
or
ζ(3) + T (x) + x2 ζ(3) + x2 T (1/x).
Using the functional equation for T (x) this becomes
ζ(3) + T (x) + x2 ζ(3) +
π 3 x3
π3 x
π3
+ 4ζ 0 (−2)π 2 x2 +
− ζ(3) +
− T (x)
90
18
90x
which is
π3 x
π3
π 3 x3
+ 4ζ 0 (−2)π 2 x2 +
+
.
90
18
90x
In view of the fact that
x2 ζ(3) +
ζ(3) + 4ζ 0 (−2)π 2 = 0
this finally becomes
π 3 x3
π3 x
π3
π3
+
+
=
x4 + 5x2 + 1 .
90
18
90x
90x
This is math.stackexchange.com question 907480.
4
Predecessor of sum formula by Cauchy and
Ramanujan
Suppose we seek to show that
X coth(nπ)
19π 7
=
.
n7
56700
n≥1
Using
coth(x) =
ex + e−x
e−x
=
1
+
2
ex − e−x
ex − e−x
12
this is the same as
2
X 1
e−nπ
19π 7
=
−ζ(7)
+
.
n7 enπ − e−nπ
56700
n≥1
The sum term may be evaluated using harmonic summation techniques.
Since this method has not been presented I will detail this calculation here.
Put
X 1
e−nx
S(x) =
.
n7 enx − e−nx
n≥0
We will evaluate S(π) using a functional equation for S(x) that is obtained
by inverting its Mellin transform.
Recall the harmonic sum identity

 

X
X λk
 g ∗ (s)
M
λk g(µk x); s = 
µsk
k≥1
k≥1
where g ∗ (s) is the Mellin transform of g(x).
In the present case we have
λk =
1
,
k7
µk = k
and g(x) =
ex
e−x
.
− e−x
We need the Mellin transform g ∗ (s) of g(x) which is
Z
0
∞
Z ∞
e−x
e−2x
s−1
x
dx
=
xs−1 dx
ex − e−x
1 − e−2x
0
Z ∞X
XZ
−2x −2qx s−1
=
e
e
x dx =
0
q≥0
q≥0
∞
e−2(q+1)x xs−1 dx
0
= Γ(s)
X
q≥0
1
1
= s Γ(s)ζ(s)
2s (q + 1)s
2
with fundamental strip h1, ∞i.
It follows that the Mellin transform Q(s) of the harmonic sum S(x) is given
by
Q(s) = 2−s Γ(s)ζ(s)ζ(s + 7)
because
X λk
X 1 1
=
s
µk
k7 ks
k≥1
for <(s) > −6.
The Mellin inversion integral here is
1
2πi
Z
3/2+i∞
Q(s)/xs ds
3/2−i∞
13
k≥1
which we evaluate by shifting it to the left for an expansion about zero.
Fortunately the trivial zeros of the two zeta function terms cancel the poles
of the gamma function term. Shifting to <(s) = −7 − 1/2 we get
S(x) =
π8 1 1
π6 x
π 4 x3
π 2 x5
4
1
− ζ(7) +
−
+
+ ζ 0 (−6)x6 +
x7
18900 x 2
5670 8100
5670 45
18900
Z −15/2+i∞
1
+
Q(s)/xs ds.
2πi −15/2−i∞
We will turn this into the promised functional equation.
Substitute s = −6 − t in the remainder integral to get
Z 3/2−i∞
1
1
Γ(−6 − t)ζ(−6 − t)ζ(1 − t)xt+6 dt
−
2πi 3/2+i∞ 2−6−t
which is
x6
2πi
Z
3/2+i∞
26+t Γ(−6 − t)ζ(−6 − t)ζ(1 − t)xt dt
3/2−i∞
In view of the desired functional equation we now use the functional equation
of the Riemann zeta function on Q(s) to prove that the integrand of the last
integral is in fact −Q(t)/π 6+2t .
Start with the functional equation
πs 2
ζ(1 − s) = s s cos
Γ(s)ζ(s)
2 π
2
and substitute this into Q(s) to obtain
Q(s) = 2−s
1 ζ(s + 7)
ζ(1 − s)2s π s
ζ(s + 7) = π s
ζ(1 − s).
2
2 cos πs
cos πs
2
2
Apply the functional equation again (this time to ζ(s + 7)) to get
1 πs
π(−6 − s)
2
−6−s −6−s cos
Q(s) =
Γ(−6 − s)ζ(−6 − s)ζ(1 − s)
2 cos πs
2
π
2
2
Observe that
cos −3π − πs
cos − πs
2
2
= −1
=−
cos πs
cos πs
2
2
so we finally get
Q(s) = −26+s π 6+2s Γ(−6 − s)ζ(−6 − s)ζ(1 − s),
thus proving the claim.
Return to the remainder integral and re-write it as follows:
Z
(x/π)6 3/2+i∞ 6+t 6+2t
2 π
Γ(−6 − t)ζ(−6 − t)ζ(1 − t)(x/π 2 )t dt.
2πi
3/2−i∞
14
so that the fact of it being a multiple of the defining integral of S(x) becomes
readily apparent.
Using the fact that 4/45 × ζ 0 (−6) = −1/2 × ζ(7)/π 6 we have established the
functional equation
S(x) =
x7
π8 1 1
π6 x
π 4 x3
π 2 x5
1
− ζ(7) +
−
+
− ζ(7) 6 x6 +
18900 x 2
5670 8100
5670
2π
18900
x6
− 6 S(π 2 /x).
π
Now the value x = π is obviously special here and we get
1
1
1
1
1
7
S(π) = π
+
−
+
+
− ζ(7) − S(π)
18900 5670 8100 5670 18900
which gives
2S(π) = π 7
19
− ζ(7)
56700
as was to be shown.
This is math.stackexchange.com question 991066.
5
A sum formula by Cauchy and Ramanujan
We are trying to show that
2p+2
X coth mπ
X B2q
B4p+4−2q
4p+3
=
(2π)
(−1)q+1
,
4p+3
m
(2q)!
(4p
+ 4 − 2q)!
m∈Z
q=0
m6=0
Note that this sum is in fact
2
X coth mπ
.
m4p+3
m≥1
The sum term
Tp (x) = 2
X coth mx
m4p+3
m≥1
is harmonic and may be evaluated by inverting its Mellin transform.
Observe that
coth x =
ex + e−x
e−x
e−2x
=1+2 x
=1+2
.
x
−x
−x
e −e
e −e
1 − e−2x
This yields
Tp (x) = 2ζ(4p + 3) + 4
X
m≥1
15
1
m4p+3
e−2mx
.
1 − e−2mx
We will now work with
S(x) =
X
m≥1
e−2mx
m4p+3 1 − e−2mx
1
and establish a functional equation for S(x) that has x = π as a fixed point.
Recall the harmonic sum identity


 
X
X λk
 g ∗ (s)
M
λk g(µk x); s = 
µsk
k≥1
k≥1
∗
where g (s) is the Mellin transform of g(x).
In the present case we have
λk =
1
k 4p+3
,
µk = k
and g(x) =
exp(−2x)
.
1 − exp(−2x)
We need the Mellin transform g ∗ (s) of g(x) which we compute as follows:
Z ∞
Z ∞X
exp(−2x)
s−1
x dx =
exp(−2qx) xs−1 dx
1 − exp(−2x)
0
0
q≥1
XZ ∞
X 1
1
= s Γ(s)ζ(s).
=
exp(−2qx) xs−1 dx = Γ(s)
s qs
2
2
0
q≥1
q≥1
with fundamental strip h1, ∞i.
Hence the Mellin transform Q(s) of S(x) is given by
X λk
1
= ζ(s + 4p + 3)
Q(s) = s Γ(s)ζ(s)ζ(s + 4p + 3) because
2
µsk
k≥1
where <(s + 4p + 3) > 1 or <(s) > −4p − 2.
Intersecting the fundamental strip and the half-plane from the zeta function
term we find that the Mellin inversion integral for an expansion about zero is
Z 3/2+i∞
1
Q(s)/xs ds
2πi 3/2−i∞
which we evaluate in the left half-plane <(s) < 3/2.
The plain zeta function term cancels the poles of the gamma function term at
even negative integers 2q ≤ −2 and the compound zeta function term the poles
at odd negative integers 2q + 1 ≤ −4p − 5. We are left with the contributions
from s = 1, s = 0 and s = −1 which are
1
ζ(4p + 4)
2x
1
Res(Q(s)/xs ; s = 0) = − ζ(4p + 3)
2
1
s
Res(Q(s)/x ; s = −1) = xζ(4p + 2).
6
Res(Q(s)/xs ; s = 1) =
16
The remaining contributions are
2p+1
X
Res(Q(s)/xs ; s = −2q − 1)
q=1
=
2p+1
X
22q+1 x2q+1
q=1
=
2p+1
X
22q+1 x2q+1
q=1
=
2p+1
X
1
B2q+2
B4p+2−2q (2π)4p+2−2q
(−1)2p+1−q+1
(2q + 1)! 2q + 2
2(4p + 2 − 2q)!
22q+1 x2q+1
q=1
=
2p+2
X
B2q+2
B4p+2−2q (2π)4p+2−2q
(−1)−q
(2q + 2)!
2(4p + 2 − 2q)!
22q−1 x2q−1
q=2
4p+2
=2
(−1)2q+1
ζ(−2q − 1)ζ(4p + 2 − 2q)
(2q + 1)!
2p+2
X
x2q−1
q=2
4p+2
=2
B4p+4−2q (2π)4p+4−2q
B2q
(−1)q+1
(2q)!
2(4p + 4 − 2q)!
B2q
B4p+4−2q π 4p+4−2q
(−1)q+1
(2q)!
(4p + 4 − 2q)!
2p+2
X
x2q−1 π 4p+4−2q
q=2
B2q
B4p+4−2q
(−1)q+1
(2q)!
(4p + 4 − 2q)!
and the one from the pole of the compound zeta function term at s = −4p−2
which we’ll do in a moment.
Now some algebra shows that setting q = 0 and q = 1 in the sum produces
precisely the values that we obtained earlier for the poles at s = 1 and s = −1
so we may extend the sum to start at zero, keeping only the residue from the
pole at s = 0 to get
24p+2
2p+2
X
x2q−1 π 4p+4−2q
q=0
B2q
B4p+4−2q
(−1)q+1
.
(2q)!
(4p + 4 − 2q)!
We have thus established that
1
S(x) = − ζ(4p + 3) + Res(Q(s)/xs ; s = −4p − 2)
2
2p+2
X
B2q
B4p+4−2q
4p+2
+2
x2q−1 π 4p+4−2q
(−1)q+1
(2q)!
(4p
+ 4 − 2q)!
q=0
Z −4p−4+i∞
1
+
Q(s)/xs ds.
2πi −4p−4−i∞
To treat the integral recall the duplication formula of the gamma function:
1 s−1 s s+1
√
Γ(s) =
2 Γ
Γ
.
2
2
π
17
which yields for the integral
Z −4p−4+i∞
s s + 1
1
√ Γ
Γ
ζ(s)ζ(s + 4p + 3)/xs ds.
2
2
2
π
−4p−4−i∞
Furthermore observe the following variant of the functional equation of the
Riemann zeta function:
s
1−s
s−1/2
ζ(s) = π
Γ
Γ
ζ(1 − s)
2
2
which gives for the integral
Z
−4p−4+i∞
−4p−4−i∞
s+1
1−s
π s−1
Γ
Γ
ζ(1 − s)ζ(s + 4p + 3)/xs ds
2
2
2
Z −4p−4+i∞ s
1
π
ζ(1 − s)ζ(s + 4p + 3)/xs ds.
=
2
sin(π/2(s
+ 1))
−4p−4−i∞
Now put s = −4p − 2 − u to get
Z
2+i∞
2−i∞
π −4p−2−u
1
2
sin(π/2(−4p − 2 − u + 1))
× ζ(u + 4p + 3)ζ(1 − u)/x−4p−2−u du
Z
x4p+2 2+i∞ π −u
1
= 4p+2
π
2 sin(π/2(−(u + 1)))
2−i∞
× ζ(u + 4p + 3)ζ(1 − u)/x−u du
Z
x4p+2 2+i∞ π −u
1
= − 4p+2
π
2
sin(π/2(u
+ 1))
2−i∞
× ζ(u + 4p + 3)ζ(1 − u)/x−u du.
We have established the functional equation
1
S(x) = − ζ(4p + 3) + Res(Q(s)/xs ; s = −4p − 2)
2
2p+2
x 4p+2 π 2 X
B4p+4−2q
4p+2
2q−1 4p+4−2q B2q
q+1
+2
x
π
(−1)
−
S
.
(2q)!
(4p + 4 − 2q)!
π
x
q=0
Setting x = π we have
1
S(π) = − ζ(4p + 3) + Res(Q(s)/π s ; s = −4p − 2)
2
2p+2
π 4p+2
X B2q
B4p+4−2q
+ 24p+2 π 4p+3
(−1)q+1
−
S (π)
(2q)!
(4p + 4 − 2q)!
π
q=0
18
and hence
1
1
S(π) = − ζ(4p + 3) + Res(Q(s)/π s ; s = −4p − 2)
4
2
2p+2
X B2q
B4p+4−2q
+ 24p+1 π 4p+3
(−1)q+1
.
(2q)!
(4p
+ 4 − 2q)!
q=0
To conclude we treat the residue that we have defered until now. Recall the
alternate form of Q(s)/π s ,
1
πs
ζ(1 − s)ζ(s + 4p + 3)/π s
2 sin(π/2(s + 1))
1
1
ζ(1 − s)ζ(s + 4p + 3).
=
2 sin(π/2(s + 1))
It follows that the residue at s = −4p − 2 is
1
1
1
1
1
ζ(4p + 3) =
ζ(4p + 3) = − ζ(4p + 3).
2 sin(π/2(−4p − 1))
2 sin(−π/2)
2
This finally yields for S(π)
2p+2
X B2q
1
B4p+4−2q
4p+1 4p+3
S(π) = − ζ(4p + 3) + 2
π
(−1)q+1
.
2
(2q)!
(4p
+ 4 − 2q)!
q=0
Computing Tp (π) we thus obtain
2p+2
X B2q
B4p+4−2q
1
4p+3 4p+3
π
(−1)q+1
2ζ(4p + 3) − 4 × ζ(4p + 3) + 2
2
(2q)!
(4p
+ 4 − 2q)!
q=0
or
(2π)4p+3
2p+2
X
q=0
B2q
B4p+4−2q
(−1)q+1
,
(2q)!
(4p + 4 − 2q)!
QED.
This is math.stackexchange.com question 1293232.
6
A functional equation relating two harmonic
sums
Suppose we seek a functional equation for
S(x) =
X
k≥1
1
1
.
(2k − 1) exp(x(2k − 1)) − exp(−x(2k − 1))
19
The sum S(x) is harmonic and may be evaluated by inverting its Mellin
transform.
Recall the harmonic sum identity


 
X
X λk
 g ∗ (s)
M
λk g(µk x); s = 
µsk
k≥1
k≥1
where g ∗ (s) is the Mellin transform of g(x).
In the present case we have
λk =
1
,
(2k − 1)
µk = 2k − 1
and g(x) =
1
.
exp(x) − exp(−x)
We need the Mellin transform g ∗ (s) of g(x) which is computed as follows:
Z ∞
Z ∞
1
exp(−x)
g ∗ (s) =
xs−1 dx =
xs−1 dx
exp(x) − exp(−x)
1 − exp(−2x)
0
0
Z ∞X
X
1
1
s−1
=
exp(−(2q + 1)x)x dx =
Γ(s) = 1 − s Γ(s)ζ(s).
(2q + 1)s
2
0
q≥0
q≥0
Hence the Mellin transform Q(s) of S(x) is given by
1
1
Q(s) = 1 − s+1
1 − s Γ(s)ζ(s)ζ(s + 1)
2
2
X λk
1
= 1 − s+1 ζ(s + 1)
because
µsk
2
k≥1
where <(s) > 1. Intersecting the fundamental strip and the half-plane from the
zeta function term we find that the Mellin inversion integral for an expansion
about zero is
Z 3/2+i∞
1
Q(s)/xs ds
2πi 3/2−i∞
which we evaluate in the left half-plane <(s) < 3/2.
The two zeta function terms cancel the poles of the gamma function term
and we are left with just
π2
and
16x
1
Res(Q(s)/xs ; s = 0) = − log 2.
4
Res(Q(s)/xs ; s = 1) =
This shows that
S(x) =
π2
1
1
− log 2 +
16x 4
2πi
20
Z
−1/2+i∞
−1/2−i∞
Q(s)/xs ds.
(1)
(2)
To treat the integral recall the duplication formula of the gamma function:
1 s−1 s s+1
Γ(s) = √ 2 Γ
Γ
.
2
2
π
which yields for Q(s)
1
1 s−1 s s+1
1
1− s √ 2 Γ
Γ
ζ(s)ζ(s + 1)
1 − s+1
2
2
2
2
π
Furthermore observe the following variant of the functional equation of the
Riemann zeta function:
s
1−s
s−1/2
ζ(s) = π
Γ
Γ
ζ(1 − s)
2
2
which gives for Q(s)
1
1
1−s
1
s+1
1 − s √ 2s−1 π s−1/2 Γ
Γ
ζ(1 − s)ζ(s + 1)
1 − s+1
2
2
2
2
π
1
1
1
π
= 1 − s+1
1 − s √ 2s−1 π s−1/2
ζ(1 − s)ζ(s + 1)
2
2
sin(π(s + 1)/2)
π
1
1
πs
= 1 − s+1
ζ(1 − s)ζ(s + 1).
1 − s 2s−1
2
2
sin(π(s + 1)/2)
Now put s = −u in the remainder integral to get
1
2πi
Z
1/2+i∞
1/2−i∞
2u
π −u
(1 − 2u ) 2−u−1
ζ(1 + u)ζ(1 − u)xu du
1−
2
sin(π(−u + 1)/2)
=
1
2πi
Z
1/2+i∞
1−
1/2−i∞
2u
2
(1 − 2u ) 2−u−1
πu
×
ζ(1 + u)ζ(1 − u)(x/π 2 )u du.
sin(π(−u + 1)/2)
We may shift this to 3/2 as there is no pole at u = 1.
Now
sin(π(−u + 1)/2) = sin(π(−u − 1)/2 + π)
= − sin(π(−u − 1)/2) = sin(π(u + 1)/2)
and furthermore
2u
1
2u
1
1
1
u
−u−1
u−2
1−
(1 − 2 ) 2
=
1−
−1 =2
−1
−1
2
2
2
2u
2u−1
2u
1
1
u−2
=2
1 − u−1
1− u
2
2
21
1
1
u−2 1
1
−
−
3
×
2
1
−
2u+1
2u
2u+1
2u
1
1
3
1
1
1
= 2u−1 1 − u+1
1 − u − 2u−1 1 − u
.
2
2
2
4
2
2u
= 2u−2 1 −
1
We have shown that
π2
1
1
− log 2 + S(π 2 /x)
16x 4
2
Z 3/2+i∞ 1
3 1
1 − u Γ(u)ζ(u)ζ(u + 1)(x/π 2 /2)u du
−
4 2πi 3/2−i∞
2
S(x) =
or alternatively
S(x) =
π2
1
1
3
− log 2 + S(π 2 /x) − T (2π 2 /x)
16x 4
2
4
where
T (x) =
X1
1
k exp(kx) − exp(−kx)
k≥1
with functional equation
T (x) =
1
1
π2
x − log 2 +
− T (2π 2 /x).
24
2
12x
which finally yields
S(x) =
1
1
3
1
S(π 2 /x) − x + log 2 + T (x).
2
32
8
4
Using sinh with
S(x) =
X
k≥1
1
1
(2k − 1) sinh((2k − 1)x)
and T (x) =
X1
1
k sinh(kx)
k≥1
we obtain the functional equation
S(x) =
1
1
3
1
S(π 2 /x) − x + log 2 + T (x).
2
16
4
4
We also have
√
T ( 2π) =
√
√
√
2π 1
π 2
2π 1
− log 2 +
=
− log 2.
24
2
24
12
2
This was math.stackexchange.com problem 1417849.
22
7
Functional equation of the Hurwitz Zeta function
Suppose we seek to show that
∞
X
1
π
(−1)k
=
.
2k + 1 exp((2k + 1)π/2) + exp(−(2k + 1)π/2)
16
k=0
The sum term
S(x) =
X (−1)k+1
k≥1
1
2k − 1 exp(x(2k − 1)) + exp(−x(2k − 1))
is harmonic and may be evaluated by inverting its Mellin transform. We are
interested in S(π/2).
Recall the harmonic sum identity

 

X
X λk
 g ∗ (s)
M
λk g(µk x); s = 
µsk
k≥1
k≥1
where g ∗ (s) = M(g(x); s) is the Mellin transform of g(x).
In the present case we have
λk =
(−1)k+1
,
2k − 1
µk = 2k − 1
and g(x) =
1
.
exp(x) + exp(−x)
We need the Mellin transform g ∗ (s) of g(x) which is computed as follows:
Z ∞
Z ∞
1
exp(−x)
∗
s−1
g (s) =
x dx =
xs−1 dx
exp(x)
+
exp(−x)
1
+
exp(−2x)
0
0
Z ∞X
X (−1)q
=
= Γ(s)β(s)
(−1)q exp(−(2q + 1)x)xs−1 dx = Γ(s)
(2q + 1)s
0
q≥0
where
q≥0
1
3
β(s) = 4−s ζ s,
− ζ s,
.
4
4
Note that β(s) does not have a pole at s = 1 and β(1) =
Hence the Mellin transform Q(s) of S(x) is given by
Q(s) = Γ(s)β(s)β(s + 1)
because
π
4.
X λk
= β(s + 1)
µsk
k≥1
where <(s) > 0. Intersecting the fundamental strip and the half-plane from the
zeta function term we find that the Mellin inversion integral for an expansion
about zero is
Z 1/2+i∞
1
Q(s)/xs ds
2πi 1/2−i∞
23
which we evaluate in the left half-plane <(s) < 1/2.
The two beta function terms cancel the poles of the gamma function term
and we are left with just
Res(Q(s)/xs ; s = 0) =
π
.
8
This shows that
S(x) =
π
1
+
8
2πi
−1/2+i∞
Z
Q(s)/xs ds.
−1/2−i∞
To treat the integral recall the duplication formula of the gamma function:
s s + 1
1
Γ(s) = √ 2s−1 Γ
Γ
.
2
2
π
which yields for Q(s)
s
1
√ 2s−1 Γ
Γ
2
π
s+1
2
β(s)β(s + 1)
Furthermore observe the following variant of the functional equation of the
Riemann zeta function adapted to the beta function:
s+1
s
Γ
β(s) = 21−2s π s−1/2 Γ 1 −
β(1 − s)
2
2
which gives for Q(s)
s
1
s
√ 2s−1 Γ
21−2s π s−1/2 Γ 1 −
β(1 − s)β(s + 1)
2
2
π
= 2−s π s−1
= 2−s
π
β(1 − s)β(s + 1)
sin(πs/2)
πs
β(1 − s)β(s + 1).
sin(πs/2)
Now put s = −u in the remainder integral to get
Z 1/2+i∞
π −u
1
2u
ζ(1 + u)ζ(1 − u)xu du
2πi 1/2−i∞
sin(−πu/2)
Z 1/2+i∞
1
πu
2−u
ζ(1 + u)ζ(1 − u)(4x/π 2 )u du.
=−
2πi 1/2−i∞
sin(πu/2)
We have shown that
S(x) =
π
− S(π 2 /4/x).
8
24
In particular we get
S(π/2) =
π
− S(π/2)
8
or
S(π/2) =
π
16
as claimed.
Addendum. The functional equation for β(s) can be derived from the
functional equation of the Hurwitz Zeta function:
n 2Γ(s) X
πs 2πkm
k
m
=
cos
−
ζ
s,
.
ζ 1 − s,
n
(2πn)s
2
n
n
k=1
where 1 ≤ m ≤ n.
This yields
2Γ(s)
[cos(πs/2 − π/2)ζ(s, 1/4) + cos(πs/2 − π)ζ(s, 1/2)
23s π s
+ cos(πs/2 − 3π/2)ζ(s, 3/4) + cos(πs/2 − 2π)ζ(s)]
2Γ(s)
= 3s s [sin(πs/2)ζ(s, 1/4) − cos(πs/2)ζ(s, 1/2)
2 π
− sin(πs/2)ζ(s, 3/4) + cos(πs/2)ζ(s)] .
ζ(1 − s, 1/4) =
Similarly
2Γ(s)
[cos(πs/2 − 3π/2)ζ(s, 1/4) + cos(πs/2 − 3π)ζ(s, 1/2)
23s π s
+ cos(πs/2 − 9π/2)ζ(s, 3/4) + cos(πs/2 − 6π)ζ(s)]
2Γ(s)
= 3s s [− sin(πs/2)ζ(s, 1/4) − cos(πs/2)ζ(s, 1/2)
2 π
+ sin(πs/2)ζ(s, 3/4) + cos(πs/2)ζ(s)] .
ζ(1 − s, 3/4) =
Subtract to obtain
41−s β(1 − s) =
2Γ(s)
(2 sin(πs/2)ζ(s, 1/4) − 2 sin(πs/2)ζ(s, 3/4))
23s π s
or
41−s β(1 − s) =
2Γ(s)
2 sin(πs/2)4s β(s)
23s π s
which is
Γ(s)
β(1 − s) = 2s s sin(πs/2)β(s)
π
s s + 1
1
2s−1
= s+1/2 2
Γ
sin(πs/2)β(s)
Γ
2
2
π
1
s −1
s+1
π
= s+1/2 22s−1
Γ 1−
Γ
sin(πs/2)β(s)
sin(πs/2)
2
2
π
which yields
s
s+1
β(1 − s)Γ 1 −
= π 1/2−s 22s−1 β(s)Γ
2
2
which is the desired result.
This was math.stackexchange.com problem 1447489.
25
8
Two contrasting examples of fixed point scenarios
Suppose we seek a functional equation for the sum term
S(x) =
X
k≥1
k5
exp(kx) − 1
which is harmonic and may be evaluated by inverting its Mellin transform.
We are interested in possible fixed points of the functional equation especially
S(2π).
Recall the harmonic sum identity


 
X
X λk
 g ∗ (s)
M
λk g(µk x); s = 
µsk
k≥1
k≥1
where g ∗ (s) = M (g(x); s) is the Mellin transform of g(x).
In the present case we have
λk = k 5 ,
µk = k
and g(x) =
1
.
exp(x) − 1
We need the Mellin transform g ∗ (s) of g(x) which is computed as follows:
Z ∞
Z ∞
1
exp(−x)
g ∗ (s) =
xs−1 dx =
xs−1 dx
exp(x)
−
1
1
−
exp(−x)
0
0
Z ∞X
X 1
=
exp(−qx)xs−1 dx =
Γ(s) = Γ(s)ζ(s).
qs
0
q≥1
q≥1
Hence the Mellin transform Q(s) of S(x) is given by
Q(s) = Γ(s)ζ(s)ζ(s − 5)
because
X λk
X k5
= ζ(s − 5)
=
s
µk
ks
k≥1
k≥1
where <(s) > 6. Intersecting the fundamental strip and the half-plane from the
zeta function term we find that the Mellin inversion integral for an expansion
about zero is
Z 13/2+i∞
1
Q(s)/xs ds
2πi 13/2−i∞
which we evaluate in the left half-plane <(s) < 13/2.
The two zeta function terms cancel the poles of the gamma function term
and we are left with just
Res(Q(s)/xs ; s = 6) =
8π 6
63x6
and
26
Res(Q(s)/xs ; s = 0) =
1
.
504
This shows that
8π 6
1
1
S(x) =
+
+
15x6
504 2πi
Z
−1/2+i∞
Q(s)/xs ds.
−1/2−i∞
To treat the integral recall the duplication formula of the gamma function:
s s + 1
1
Γ
.
Γ(s) = √ 2s−1 Γ
2
2
π
which yields for Q(s)
s
1
√ 2s−1 Γ
Γ
2
π
s+1
2
ζ(s)ζ(s − 5)
Furthermore observe the following variant of the functional equation of the
Riemann zeta function:
s
1−s
Γ
ζ(s) = π s−1/2 Γ
ζ(1 − s)
2
2
which gives for Q(s)
1
√ 2s−1 π s−1/2 Γ
π
s+1
2
1−s
Γ
ζ(1 − s)ζ(s − 5)
2
1
π
= √ 2s−1 π s−1/2
ζ(1 − s)ζ(s − 5)
sin(π(s + 1)/2)
π
= 2s−1
πs
ζ(1 − s)ζ(s − 5).
sin(π(s + 1)/2)
Now put s = 6 − u in the remainder integral to get
Z 13/2+i∞
1 1
π 6−u
5−u
ζ(u − 5)ζ(1 − u)xu du
2
x6 2πi 13/2−i∞
sin(π(7 − u)/2)
Z 9/2+i∞
64π 6 1
πu
= 6
2u−1
ζ(u − 5)ζ(1 − u)(x/π 2 /22 )u du.
x 2πi 9/2−i∞
sin(π(7 − u)/2)
Now
sin(π(7 − u)/2) = sin(π(−u − 1)/2 + 4π)
= sin(π(−u − 1)/2) = − sin(π(u + 1)/2).
We have shown that
S(x) =
64π 6
8π 6
1
+
−
S(4π 2 /x).
63x6
504
x6
In particular we get
S(2π) =
1
1
+
− S(2π)
63 × 8 504
27
or
1
.
504
Remark. Unfortunately this method does not work for
S(2π) =
S(x) =
X
k≥1
k3
exp(kx) − 1
We get the functional equation
S(x) =
π4
1
16π 4
−
S(4π 2 /x).
+
15x4
240
x4
which yields
1
1
−
+ S(2π)
15 × 16 240
which holds without providing any data about the value itself.
This was math.stackexchange.com problem 1482918.
S(2π) =
References
[FS96] Philippe Flajolet and Robert Sedgewick. The average case analysis of
algorithms : Mellin transform asymptotics. Technical Report RR-2956,
INRIA, Rocquencourt, 1996.
[Szp01] Wojciech Szpankowski. Mellin transform and its applications. In Average Case Analysis of Algorithms on Sequences, pages 398–441. John
Wiley and Sons, Inc., 2001.
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