PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 133, Number 5, Pages 1257–1265
S 0002-9939(04)07863-3
Article electronically published on December 15, 2004
EULER’S INTEGRALS AND MULTIPLE SINE FUNCTIONS
SHIN-YA KOYAMA AND NOBUSHIGE KUROKAWA
(Communicated by David E. Rohrlich)
Abstract. We show that Euler’s famous integrals whose integrands contain
the logarithm of the sine function are expressed via multiple sine functions.
1. Introduction
π
Euler studied the definite integrals 02 xn log(sin x)dx for n = 0 and 1. In [E1]
(1769), he proved the famous result
π2
π
(1)
log(sin x)dx = − log 2,
2
0
which is frequently explained as an example of tricky integrals in analysis courses.
A bit later, Euler [E2] (1772) stated that


π2
∞
1 1
π2
log 2
x log(sin x)dx =
−
2
n3
4
0
n=1
n:odd
=
(2)
7
π2
ζ(3) −
log 2.
16
8
Euler proved (1) by using
(3)
log(sin x) = −
∞
cos(2nx)
− log 2
n
n=1
for 0 < x < π. The actual integration is obvious since
π2
cos(2nx)dx = 0
0
for n = 1, 2, 3, · · · . Hence, the original proof of Euler is not tricky contrary to the
usual explanation. In the case of (2), Euler primarily wanted to calculate the value
ζ(3). He started from the divergent series expression
ζ(3) =
=
−4π 2 ζ (−2)
∞
4π 2
n2 log n
n=1
Received by the editors August 4, 2003.
2000 Mathematics Subject Classification. Primary 11M06.
c
2004
American Mathematical Society
Reverts to public domain 28 years from publication
1257
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1258
SHIN-YA KOYAMA AND NOBUSHIGE KUROKAWA
and by investigating it he reached (2). Thus his arguments are difficult to follow
and literally invalid. Moreover Euler did not prove the functional equation ζ(3) =
−4π 2 ζ (−2) conjectured by himself in [E3]. We notice that when we use (3) we can
give a secure calculation for (2):
π2
π
∞
1 2
π2
log 2
x log(sin x)dx = −
x cos(2nx)dx −
n 0
8
0
n=1
with integration by parts
0
π
2

1

 − 2
2n
x cos(2nx)dx =


0
if n is odd,
if n is even.
It might be remarkable that Euler missed this way.
In this paper we investigate definite integrals
x
θr−2 log(sin θ)dθ
0
for r = 2, 3, 4, · · · containing Euler’s case x = π/2 from the point of view of multiple
sine functions. Let
∞
x nr−1
xr−1
Sr (x) = e r−1
Pr
n
n=−∞
n=0
= e
xr−1
r−1
∞
Pr
n=1
x (−1)r−1 nr−1
Pr −
n
n
x
be the multiple sine function studied in [K1, K2, KK1, KK2, KOW, KW], where
ur
u2
+ ···+
Pr (u) = (1 − u) exp u +
.
2
r
For example,
∞ 1−
S2 (x) = ex
n=1
S3 (x) = e
x2
2
and
S4 (x) = e
x3
3
∞
n=1
1+
x
n
x
n
n
e2x ,
n2
2
x2
ex
1− 2
n
n=1
∞
1−
1+
x
n
x
n
n3
e
2n2 x+ 23 x3
.
Then we show the following result.
Theorem 1. For 0 ≤ x < π and for r = 2, 3, 4, ..., we have
x
x
π r−1
xr−1
log(sin x) −
log Sr
.
θr−2 log(sin θ)dθ =
r−1
r−1
π
0
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EULER’S INTEGRALS AND MULTIPLE SINE FUNCTIONS
1259
In particular we have:
Theorem 2. For r = 2, 3, 4, ...,
π2
1
π r−1
r−2
log Sr
θ
log(sin θ)dθ = −
.
r
−
1
2
0
Examples.
0
1
= −π log S2
2
n ∞ 1
2n − 1
= −π log e 2
e
,
2n + 1
n=1
π
2
log(sin θ)dθ
(4)
π
2
0
θ log(sin θ)dθ
=
(5)
=
1
π2
− log S3
2
2
n2 ∞
1
1
π2
1
,
e4
− log e 8
1− 2
2
4n
n=1
1
π3
θ log(sin θ)dθ = − log S4
3
2
0
n3
∞
1
2n − 1
π3
1
= − log e 24
exp n2 +
(6)
.
3
2n
+
1
12
n=1
√
1
from Euler’s
We notice that we have S2 12 = 2 and S3 12 = 2 4 exp − 7ζ(3)
2
8π
π
2
2
results (1) and (2)
combined with (4) and (5). We demonstrate a calculation of the
special value S4 12 from the product expression directly as follows.
Theorem 3.
S4
and
0
π
2
1
1
9ζ(3)
= 2 8 exp −
2
16π 2
θ2 log(sin θ)dθ =
π3
3π
ζ(3) −
log 2.
16
24
In our calculation a generalization of the Stirling formula is crucial.
2. Multiple sine functions
To make this paper self-contained we prove some basic properties of multiple
sine functions. For general background we refer to [KK1, KK2, KOW, M].
Proposition 1. For r = 2, 3, 4, ..., Sr (x) is a meromorphic function in x ∈ C and
it satisfies
Sr (x)
= πxr−1 cot(πx).
Sr (x)
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1260
SHIN-YA KOYAMA AND NOBUSHIGE KUROKAWA
Proof. The fact that Sr (x) is a meromorphic function in x ∈ C (and its order as a
meromorphic function being r) is seen from the product expression defining Sr (x).
Let us calculate the logarithmic derivative. From
∞ x x (−1)r−1 nr−1
xr−1
r−1
Pr −
Sr (x) = e
Pr
n
n
n=1
we have
log Sr (x)
=
=
∞
x
x xr−1
+
+ (−1)r−1 log Pr −
nr−1 log Pr
r − 1 n=1
n
n
∞
xr−1
x
x
+
+ (−1)r−1 log 1 +
nr−1 log 1 −
r − 1 n=1
n
n
x 1 x 2
1 x r
+
+ ··· +
+
n 2 n
r n
2
r −x
−x
−x
1
1
+(−1)r−1
+ ···+
.
+
n
2
n
r
n
Hence
∞
Sr (x)
1
xr−1
(−1)r−1
1
x
nr−1
= xr−2 +
+
+
+ 2 + ··· + r
Sr (x)
x−n
x+n
n n
n
n=1
x
xr−1
1
+(−1)r−1 − + 2 + · · · + (−1)r r
.
n n
n
Here
( x )r − 1
x
xr−1
1
+ 2 + ···+ r = n
n n
n
x−n
and
−
(−1)r ( nx )r − 1
1
x
xr−1
+ 2 + · · · + (−1)r r =
.
n n
n
x+n
Thus
Sr (x)
Sr (x)
=
r−2
x
+
∞
n
r−1
n=1
∞
( nx )r
( x )r
− n
x−n x+n
2xr
x2 − n2
n=1
=
xr−2 +
=
πxr−1 cot(πx),
where we used
∞
π cot(πx) =
1 2x
+
.
x n=1 x2 − n2
Proposition 2. For 0 ≤ x < 1 and for r = 2, 3, 4, ...,
x
log Sr (x) =
πtr−1 cot(πt)dt.
0
Proof. Since Sr (0) = 1, both sides are 0 at x = 0. Hence it is sufficient to remark
that the differentiations of both sides are πxr−1 cot(πx) from Proposition 1.
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EULER’S INTEGRALS AND MULTIPLE SINE FUNCTIONS
1261
3. Euler’s integrals
Using Proposition 2 we show Theorems 1 and 2.
Proof of Theorems 1 and 2. By integration by parts in
x
log Sr (x) =
πtr−1 cot(πt)dt
0
we have
log Sr (x)
=
=
x
log(sin πt) 0 −
x
(r − 1)tr−2 log(sin πt)dt
x
tr−2 log(sin πt)dt.
xr−1 log(sin πx) − (r − 1)
t
r−1
0
0
Hence changing the variable to θ = πt in the integral, we have
r − 1 πx r−2
log Sr (x) = xr−1 log(sin πx) − r−1
θ
log(sin θ)dθ.
π
0
This gives Theorem 1. Then, letting x = 1/2 we have Theorem 2.
Examples.
1
π
log(sin θ)dθ = − log 2 − π log S2
,
8
4
0
√
π
3
1
3
π
− π log S2
log(sin θ)dθ = log
,
3
2
3
0
π4
1
π2
π2
log S3
θ log(sin θ)dθ = − log 2 −
,
64
2
4
0
√
π3
π2
1
3 π2
log
−
log S3
θ log(sin θ)dθ =
.
18
2
2
3
0
π
4
4. A calculation of the special value
Proof of Theorem 3. Since
n3
∞
1
2n − 1
1
1
2
24
S4
exp n +
=e
,
2
2n + 1
12
n=1
we put
AN
1
2
= e
exp n +
12
n=1
N
3
3
1
N
+ (12 + · · · + N 2 ) +
(2n − 1)n −(n−1)
= exp
×
24
12
n=1
1
24
N
2n − 1
2n + 1
×(2N + 1)−N
n3
3
and show
1
8
lim AN = 2 exp
N →∞
1
9 9ζ(3)
8
ζ (−2) = 2 exp −
.
4
16π 2
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1262
SHIN-YA KOYAMA AND NOBUSHIGE KUROKAWA
Then the value of the integral follows from (6). We use the Stirling formula
N! =
N
n∼
√
1
2πN N + 2 e−N
n=1
and its generalization
N
nn ∼ exp (−ζ (−2)) N
2
N3
3
2
3
N
+ N2 + N
− N9 + 12
6
e
,
n=1
which follow from the Euler-Maclaurin summation formula for ζ (s):
ζ (s) = lim
N →∞
−
N
n−s log n +
n=1
N 1−s
N 1−s log N
−
1−s
(1 − s)2
1
s
1
+ N −s log N − N −s−1 log N + N −s−1
2
12
12
valid in Re(s) > −3. We refer to Hardy [H], Chap. XIII, for the Euler-Maclaurin
summation formula and its applications. Then, letting s = 0 and −2 we see
1
lim
log n −
N+
log N − N
N →∞
2
n=1
N!
lim log
1
N →∞
N N + 2 e−N
√
log 2π = −ζ (0) =
=
N
and
−ζ (−2) =
=
lim
N →∞
N
n log n −
n=1
N
lim log
N →∞
2
N
N3
3
N3
N2
N
+
+
3
2
6
2
n
n=1 n
N3
N
+N
6 e− 9 + 12
2
+ N2
N
N3
+
log N −
9
12
.
Using
N
(2n − 1)n
3
−(n−1)3
=
n=1
N
(2n − 1)3n
2
−3n+1
n=1
=
=
N
34
(2n−1)2
(2n − 1)
n=1
2N
N
n
n=1 n
2
(2n)2
n=1 (2n)
×
N
14
(2n − 1)
n=1
34
×
2N
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n=1 n
N
n=1 (2n)
14
EULER’S INTEGRALS AND MULTIPLE SINE FUNCTIONS
1263
and the (generalized) Stirling formulas
2N
nn ∼ e−ζ
2
(−2)
(2N )
8N 3
3
3
N
+2N 2 + N
− 8N
3
9 + 6
e
,
n=1
N
(2n)(2n) = 24(1
2
2
+···+N )
2
n=1
N
4
nn
2
n=1
∼ 2 3 N (N +1)(2N +1) e−ζ
2
2N
n∼
(−2)
N
N3
3
2
+ N2
3
N
+N
− N9 + 12
6
4
e
,
√
1
2π(2N )2N + 2 e−2N ,
n=1
N
(2n) = 2N
n=1
N
n∼
√
1
2πN N + 2 e−N 2N ,
n=1
we have
AN
∼
exp
×e 4 ζ
9
1
+
24
(−2)
N
N
N3
N2
N
+
+
3
2
6
NN
3
−N
4
2N
3
−N
4
e−
N3
3
+
N
12
−N
8
N
×N 4 2 4 + 8 e− 4 × (2N + 1)−N .
1
3
Hence, combining with
−N 3
(2N + 1)
−N 3
1
= (2N )
1+
2N
1
−N 3
3
= (2N )
exp −N log 1 +
2N
2
3 1
1
1
1
1
−N 3
3
−
∼ (2N )
exp −N
+
2N
2 2N
3 2N
−N 3
−N 3
= (2N )
N
1
N2
+
−
exp −
,
2
8
24
we obtain the desired result
1
8
lim AN = 2 exp
N →∞
9 ζ (−2) .
4
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1264
SHIN-YA KOYAMA AND NOBUSHIGE KUROKAWA
√
Remarks. (1) The fact that S2 ( 12 ) = 2 is also proved as Theorem 3 and we get
(1) again from (4). In fact:
∞
1 n
1 − 2n
1
1
S2
e
= e2
1
2
1 + 2n
n=1
n N 1
2n − 1
2
e
= lim e
N →∞
2n + 1
n=1
N
1
2
3
2N − 1
1
3
5
1
N
= lim e 2
···
e
N →∞
3
5
7
2N + 1
1 3 · 5 · · · (2N − 1)
N
= lim e 2
e
N →∞
(2N + 1)N
1
(2N )!
N
= lim e 2
1 Ne
N →∞
N !22N N N (1 + 2N
)
√
=
2
by the (usual) Stirling formula.
(2) The case of S3 (1/2) is similar by using the generalized Stirling’s formula:
n2 ∞
1
1
1
1
S3
e4
1− 2
= e8
2
4n
n=1
n2 N
1
1
1
8
= lim e
e4
1− 2
N →∞
4n
n=1
n2
N N
1
(2n − 1)(2n + 1)
= lim e 4 + 8
N →∞
(2n)2
n=1
1/2
2N
1
n2
2
n
n=1
(2N )! 2
N
(2N + 1)N
+ 18
4
= lim e
× 4(12 +···+N 2 )
4 ×
N →∞
2N N !
2
N
n2
n
n=1
1
7 ζ (−2) .
= 2 4 exp
2
Thus we obtain Euler’s formula
√ (2) via (5).
(3) Except for S2 (1/2) = 2 we do not know the algebraicity of Sr (1/2) for
r ≥ 2. In fact we cannot deny even the optimistic expectation Sr (1/2) ∈ 2Q .
Acknowledgement
We thank the referees for their suggestions to refine the presentation.
References
[E1]
[E2]
L. Euler “De summis serierum numeros Bernoullianos involventium” Novi commentarii
academiae scientiarum Petropolitanae 14 (1769) 129-167 [Opera Omnia I-15, pp.91-130].
L. Euler “Exercitationes analyticae” Novi commentarii academiae scientiarum Petropolitanae 17 (1772) 173-204 [Opera Omnia I-15, pp.131-167].
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
EULER’S INTEGRALS AND MULTIPLE SINE FUNCTIONS
1265
[E3]
L. Euler “Remarques sur un beau rapport entre les séries des puissances tant directes
que réciproques” Mémoires de l’académie des sciences de Berlin 17 (1761) 83-106 (Lu en
1749) [Opera Omnia I-15, pp. 70-90].
[H]
G.H. Hardy “Divergent Series” Oxford Univ. Press 1949. MR0030620 (11:25a)
[K1]
N. Kurokawa “Multiple sine functions and Selberg zeta functions” Proc. Japan Acad.
67A (1991) 61-64. MR1105522 (92d:11094)
[K2]
N. Kurokawa “Multiple zeta functions: an example” Adv. Stud. in Pure Math. 21 (1992)
219-226. MR1210791 (94f:11084)
[KK1] N. Kurokawa and S. Koyama “Multiple sine functions” Forum Math. 15 (2003) 839-876.
MR2010282 (2004g:11077)
[KK2] S. Koyama and N. Kurokawa “Kummer’s formula for multiple gamma functions” J. Ramanujan Math. Soc. 18 (2003) 87-107. MR1966530 (2004a:33003)
[KOW] N. Kurokawa, H. Ochiai and M. Wakayama “Multiple trigonometry and zeta functions”
J. Ramanujan Math. Soc. 17 (2002) 101-113. MR1913896 (2003f:11132)
[KW] N. Kurokawa and M. Wakayama “On ζ(3)” J. Ramanujan Math. Soc. 16 (2001) 205-214.
MR1863604 (2002i:11083)
[M]
Yu. I. Manin “Lectures on zeta functions and motives (according to Deninger and
Kurokawa)” Astérisque 228 (1995) 121-163. MR1330931 (96d:11076)
2-5-27 Hayabuchi, Tsuzuki-ku, Yokohama 224-0025, Japan
E-mail address: [email protected]
Department of Mathematics, Tokyo Institute of Technology, 152-8551, Tokyo, Japan
E-mail address: [email protected]
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