Miskolc Mathematical Notes
Vol. 17 (2016), No. 1, pp. 255–266
HU e-ISSN 1787-2413
DOI: 10.18514/MMN.2016.748
MULTIPLE FRACTIONAL PART INTEGRALS AND EULER’S
CONSTANT
OVIDIU FURDUI
Received 27 December, 2013
Abstract. The paper is about calculating multiple fractional part integrals of the form
p
m
Z 1Z 1
Z 1Z 1Z 1
1
1
k
.x C y/
dxdy and
dxdyd´;
x Cy
x Cy C´
0 0
0 0 0
where fxg denotes the fractional part of x and k; m and p are nonnegative integers. We show
that these integrals can be expressed as series involving products of Riemann zeta function values and some binomial coefficients. We obtain, as particular cases of our results, new integral
representations of Euler’s constant as double and triple symmetric fractional part integrals.
2010 Mathematics Subject Classification: 26A42; 26B15; 40A30
Keywords: the Euler–Mascheroni constant, harmonic numbers, fractional part integrals, Riemann
zeta function
1. I NTRODUCTION
Let m, p and k be nonnegative integers and let Dk;p and Im denote the double
and triple symmetric integrals
p
Z 1Z 1
1
k
dxdy
(1.1)
Dk;p D
.x C y/
x Cy
0
0
and
Z
Im D
0
1Z 1Z 1
0
0
1
x Cy C´
m
dxdyd´;
(1.2)
where fxg denotes the fractional part of x and is therefore related to the floor function
by fxg D x bxc. The integral Im can be viewed as a three dimensional version of
R1
Havil’s integral 0 f1=xg dx D 1 ([3, pp. 109–111]), while the integral Dk;m
generalizes the double symmetric integral
Z 1Z 1
1
dxdy
x Cy
0
0
c 2016 Miskolc University Press
256
OVIDIU FURDUI
proposed by Furdui as a problem in [1] and solved by Qin in [4]. We also consider
the multiple integral
Z 1 Z 1
1
n
Mm
dx1 dxn ;
D
.x1 C x2 C C xn /m
x1 C C xn
0
0
where n 2 and m 0 are integers, and we show that this integral can be calculated
via a recurrence formula.
The goal of this paper is to calculate, in closed form, the three classes of integrals given above and to prove that these integrals can be expressed in terms of
series involving products of Riemann zeta function values and some binomial coefficients. We obtain, as particular cases of our results, new integral representations of
the Euler–Mascheroni constant in terms of double symmetric fractional part integrals. The organization of the paper is as follows: in the next section we calculate the
double integrals Dk;p ; in section 3 we concentrate on the evaluation of the integral
n via a recurrence formula.
Im and in section 4 we evaluate the integral Mm
Before we give the main results of this paper we need to collect a result from [2],
Theorem 1 below, which is about calculating the special class of single fractional part
integrals Vk;m defined by
k
Z 1
m 1
dx;
Vk;m D
x
x
0
where k and m are nonnegative integers.
Theorem 1. Let m 0 and let k 1 be integers. Then
k
Z 1
1
X
.m C j /Š
kŠ
m 1
Vk;m D
x
dx D
..m C j C 1/
x
.m C 1/Š
.k C j /Š
0
1/:
j D1
In particular, one has as a consequence of Theorem 1, the following corollary.
Corollary 1. a) Let m 1 be an integer. Then
Z 1 m
.2/ C .3/ C C .m C 1/
1
Vm;m D
x m dx D 1
:
x
mC1
0
b) Let m 1 be an integer. Then
mC1
Z 1
mC1
X .j /
m 1
x
dx D HmC1 VmC1;m D
;
x
j
0
j D2
where HmC1 denotes the .m C 1/th harmonic number.
c) Let m 0 be an integer. Then
m
Z 1
1
1
1
Vm;mC1 D
x mC1
dx D C
.2/
x
2 .m C 1/.m C 2/
0
mC1
X
i D1
!
i.i C 1/ :
MULTIPLE FRACTIONAL PART INTEGRALS
257
Proof. The first two parts of the corollary are proved in [2], so we need to prove
only part c) of the corollary. We have, in view of Theorem 1, that
1
Vm;mC1 D
mŠ X .m C 1 C j /Š
..m C j C 2/
.m C 2/Š
.m C j /Š
1/
j D1
1
X
1
D
.m C j C 1/ ..m C j C 2/
.m C 1/.m C 2/
1/ ;
j D1
and the result follows since
P1
i D1 i..i
C 1/
1/ D .2/.
2. A DOUBLE SYMMETRIC FRACTIONAL PART INTEGRAL
In this section we calculate the double integral Dk;p . We prove that the evaluation
of the double integral Dk;p reduces to the calculation of the single integral Vk;m . The
main result of this section is the following theorem.
Theorem 2. a) Let Dk;p be as in (1.1) and let k be a nonnegative integer. Then
1Z 1
Z
Dk;kC1 D
0
0
D 2 ln 2
kC1
1
dxdy
x Cy
.2/ C .3/ C C .k C 2/
:
k C2
.x C y/k
b) A new integral formula for Euler’s constant. Let k be a nonnegative integer. Then
Z
Dk;kC2 D
0
1Z 1
0
k
.x C y/
1
x Cy
kC2
dxdy D 1
ln 2 C HkC2
kC2
X
i D2
.i /
:
i
c) If p ¤ k C 1 and p ¤ k C 2 then
Dk;p D
2k
k
pC2
2
pC1
2k
k
pC2
1
C Vp;kC1 :
pC2
Proof. We have, based on the substitution x C y D t in the inner integral, that
p Z 1 Z xC1 p Z 1 Z 1
1
1
k
Dk;p D
.x C y/
dy dx D
tk
dt dx:
x Cy
t
0
0
0
x
We integrate by parts with
Z xC1 p
1
f .x/ D
tk
dt;
t
x
f 0 .x/ D .x C 1/k
1
x C1
p
xk
p
1
;
x
258
OVIDIU FURDUI
g 0 .x/ D 1; g.x/ D x, and we get that
p ˇxD1
ˇ
1
dt ˇˇ
t
x
xD0
p
p Z 1 1
1
dx
x .x C 1/k
xk
x
C
1
x
0
p
Z 2 p
Z 1
Z 1
1
1
D
tk
dt
x.x C 1/k p dx C
x kC1
dx
t
x
1
0
0
Z 2
Z 1
D
t k p dt
x.x C 1/k p dx C Vp;kC1 :
Z
Dk;p D x
xC1
tk
1
(2.1)
0
We used the fact that, for x > 0, one has that f1=.1 C x/g D 1=.1Cx/. We distinguish
here the following cases.
Case p D k C 1. We have, based on (2.1) and part a) of Corollary 1, that
Z 1
x
dt
dx C VkC1;kC1
Dk;kC1 D
0 x C1
1 t
D 2 ln 2 1 C VkC1;kC1
1
..2/ C .3/ C C .k C 2// :
D 2 ln 2
k C2
Z
2
Case p D k C 2. We have, in view of (2.1) and part b) of Corollary 1, that
D1
Z 1
x
dt
dx C VkC2;kC1
2
2
t
0 .x C 1/
ln 2 C VkC2;kC1
D1
ln 2 C HkC2
Z
Dk;kC2 D
1
2
kC2
X
i D2
1
.i /:
i
Case p ¤ k C 1 and p ¤ k C 2. Equality (2.1) implies that
ˇ
t k pC1 ˇˇt D2
Dk;p D
k p C 1 ˇ t D1
D
2k pC2
k
2
pC1
ˇ
.x C 1/k pC2 ˇˇxD1 .x C 1/k pC1
C
k p C 2 ˇxD0
k pC1
ˇxD1
ˇ
ˇ
CVp;kC1
ˇ
xD0
2k pC2
and the theorem is proved.
k
1
C Vp;kC1 ;
pC2
MULTIPLE FRACTIONAL PART INTEGRALS
259
Corollary 2. Let k 0 be an integer. Then
k
Z 1Z 1
1
k
Dk;k D
.x C y/
dxdy
x Cy
0
0
1
0
kC1
X
1
@.2/
i.i C 1/A :
D 1C
.k C 1/.k C 2/
i D1
Proof. We have, based on Theorem 2 with p D k, that Dk;k D 1=2 C Vk;kC1 ; and
the result follows from part c) of Corollary 1.
In particular one has that
Z 1Z 1
0
0
1
.x C y/
dxdy D 1
x Cy
.3/
3
and
Z
1Z 1
0
0
2
.x C y/
1
x Cy
2
dxdy D 1
.3/
6
.4/
:
4
Corollary 3. The following equality holds
p
Z 1Z 1
1
1
k
dxdy D DkC1;p :
x.x C y/
x Cy
2
0
0
Proof. We have, by symmetry reasons, that
p
p
Z 1Z 1
Z 1Z 1
1
1
k
k
dxdy D
dxdy;
x.x C y/
y.x C y/
x Cy
x Cy
0
0
0
0
and it follows that
Z 1Z 1
k
x.x C y/
0
0
1
x Cy
p
p
Z Z
1 1 1
1
kC1
dxdy D
.x C y/
dxdy
2 0 0
x Cy
1
D DkC1;p ;
2
and the corollary is proved.
We have that, for any nonnegative integer k, the following equality holds
PkC1
k
Z 1Z 1
1
1 .2/
i D1 i.i C 1/
x.x C y/k 1
dxdy D C
:
x
C
y
2
2.k
C
1/.k C 2/
0
0
260
OVIDIU FURDUI
3. A TRIPLE H AVIL INTEGRAL
In this section we calculate the integral Im , which we call the triple Havil integral,
and we prove that the evaluation of this class of integrals reduces to the calculation
of the double integral D1;m . The main result of this section is the following theorem.
Theorem 3. Let Im be as in (1.2). a) Then
I1 D
9 ln 3
2
6 ln 2
.3/
6
and
I2 D 6 ln 2
b) A cubic integral. We have
3
Z 1Z 1Z 1
1
ln 3
I3 D
dxdyd´ D
x Cy C´
2
0
0
0
3 ln 3
3 ln 2 5
C
2
3
.2/ C .3/
:
6
2
.2/
4
.3/
:
6
c) Let m 4 be an integer. Then
Im D
.7
m/22 m C m 4 33
.m 1/.m 2/.m 3/
m
1
C D1;m :
2
Proof. We have, based on the substitution x C y C ´ D t, that
Z 1 Z 1 Z xCyC1 m 1
dt dxdy
Im D
t
0
0
xCy
Z 1 Z 1 Z xCyC1 m 1
dt dx dy:
D
t
0
0
xCy
We calculate the double inner integral by parts with
m
Z xCyC1 m
1
1
0
f .x/ D
dt; f .x/ D
t
1Cx Cy
xCy
1
x Cy
m
;
g 0 .x/ D 1; g.x/ D x; and we get that
Z xCyC1 m ˇxD1
Z 1 Z xCyC1 m ˇ
1
1
dt dx D x
dt ˇˇ
t
t
0
xCy
xCy
xD0
m m Z 1 1
1
x
dx
1Cx Cy
x Cy
0
Z 2Cy
Z 1
1
x
D
dt
dx
m
m
1Cy t
0 .1 C x C y/
m
Z 1 1
C
x
dx:
x Cy
0
MULTIPLE FRACTIONAL PART INTEGRALS
261
Thus,
1 Z 2Cy
Z
Im D
0
Z
C
Z
D
0
1Cy
1Z 1
1
tm
Z
1Z 1
dt dy
0
0
x
dxdy
.1 C x C y/m
m
x
0
0
1 Z 2Cy
1Cy
t
1
dxdy
x Cy
Z 1Z
1
dt dy
m
0
0
(3.1)
1
x
1
dxdy C D1;m :
m
.1 C x C y/
2
We distinguish the following cases.
Case m D 1. We have, based on (3.1) and Corollary 2, that
Z 1 Z 1Z 1
x
1
2Cy
I1 D
ln
dy
dxdy C D1;1
1Cy
2
0
0
0 1Cx Cy
9 ln 3
.3/
D
6 ln 2
:
2
6
Case m D 2. A calculation, based on (3.1) and part a) of Theorem 2, shows that
Z 1Z 1
1
4
x
dxdy C D1;2
I2 D ln
2
3
.1
C
x
C
y/
2
0
0
.2/ C .3/
D 6 ln 2 3 ln 3
:
6
Case m D 3. We have, based on (3.1) and part b) of Theorem 2, that
Z 1Z 1
1
x
1
dxdy
C
D1;3
I3 D
3
6
2
0
0 .1 C x C y/
ln 3 3 ln 2 5 .2/ .3/
D
C
:
2
2
3 2
4
6
Case m 4. A calculation, based on (3.1), shows that
Im D
.7
and the theorem is proved.
m/ 22 m C m 4 33
.m 1/.m 2/.m 3/
m
1
C D1;m ;
2
Remark 1. It is worth mentioning that this integral is recorded in [5] as the second
part of Theorem 8. However, Qin and Lu have skipped the calculations of Im and
instead they calculated a double integral and explained how the triple integral can be
calculated by analogy. It turns out that the values of Im when m D 1; 2 and 3, as
given by Qin and Lu, are incorrect and by our method, which is different than the one
mentioned in [5], these values are corrected.
262
OVIDIU FURDUI
4. A SPECIAL CLASS OF INTEGRALS AND A RECURRENCE FORMULA
In this section we consider the multiple integral
Z 1 Z 1
1
n
m
dx1 dxn ;
Mm D
.x1 C x2 C C xn /
x1 C C xn
0
0
where n 2 and m 0 are integers, and we show that this integral can be calculated
via a recurrence formula. The main result of the section is the following theorem.
Theorem 4. Let n 2 and m 0 be integers. The following recurrence formulas
hold
Z 1 Z 1
1
n
n 1
M0 D
M
C
ln.1 C x1 C C xn 1 /dx1 dxn 1
n 1 1
0
0
Z 1 Z 1
ln.1 C x1 C C xn 2 /dx1 dxn 2
0
0
and, for m 1,
n
Mm
D
1
n
1
m
n 1
C
MmC1
1
Z
0
1
Z
0
1
m
Z
0
1
Z
0
1
.1 C x1 C C xn
1
.1 C x1 C C xn
m
1 / dx1 dxn 1
m
2 / dx1 dxn 2 :
Proof. We discuss only the case when m 1 since the case when m D 0 is handled
similarly. We have, in view of the substitution x1 C C xn D y, that
Z 1 Z 1
1
n
m
Mm D
.x1 C x2 C C xn /
dx1 dxn
x1 C C xn
0
0
Z 1 Z 1 Z 1Cx1 CCxn 1
m 1
D
y
dy dx1 dxn 1
y
0
0
x1 CCxn 1
Z 1 Z 1 Z 1 Z 1Cx1 CCxn 1
m 1
D
y
dy dxn 1 dx2 dx1 :
y
0
0
0
x1 CCxn 1
R 1Cx1 CCxn 1 R 1 R 1Cx1 CCxn 1 R x1 CCxn 1
Since x1 CCx
D 0C 1
, we have
0
n 1
Z 1Cx1 CCxn 1
1
ym
dy
y
x1 CCxn 1
Z 1
Z 1Cx1 CCxn 1
Z x1 CCxn 1
m 1
m 1
m 1
D
y
dy C
y
dy
y
dy
y
y
y
0
1
0
Z 1
Z x1 CCxn 1
1
.1 C x1 C C xn 1 /m 1
1
D
ym
dy C
ym
dy;
y
m
y
0
0
MULTIPLE FRACTIONAL PART INTEGRALS
and it follows that
Z 1 Z
0
Z
C
1Cx1 CCxn
1
1
y
dy dxn
y
m
x1 CCxn 1
1
.1 C x1 C C xn 1 /m
m
0
Z 1 Z x1 CCxn 1
0
0
We calculate the integral
Z 1 Z
0
Z
1
D
0
1
263
1
y
dy
y
m
1
dxn 1
m 1
y
dy dxn 1 :
y
x1 CCxn
0
1
1
y
dy dxn
y
m
1
by parts with
g 0 .xn
x1 CCxn
1
f .xn 1 / D
y
dy;
y
0
1
0
m
f .xn 1 / D .x1 C C xn 1 /
;
x1 C C xn 1
1 / D 1, g.xn 1 / D xn 1 , and we get that
Z 1 Z x1 CCxn 1
1
dy dxn 1
ym
y
0
0
ˇxn 1 D1
Z x1 CCxn 1
ˇ
m 1
D xn 1
y
dy ˇˇ
y
0
xn 1 D0
Z 1
1
m
dxn
xn 1 .x1 C C xn 1 /
x1 C C xn 1
0
Z 1Cx1 CCxn 2
m 1
D
y
dy
y
0
Z 1
1
m
dxn
xn 1 .x1 C C xn 1 /
x1 C C xn 1
0
Z 1
Z 1Cx1 CCxn 2
m 1
m 1
D
y
dy C
y
dy
y
y
0
1
Z 1
1
m
xn 1 .x1 C C xn 1 /
dxn
x1 C C xn 1
0
Z 1
.1 C x1 C x2 C C xn 2 /m 1
m 1
D
y
dy C
y
m
0
Z 1
1
xn 1 .x1 C C xn 1 /m
dxn
x1 C C xn 1
0
Z
1
m
1
1
1
1:
264
OVIDIU FURDUI
Thus,
1 Z 1Cx1 CCxn
Z
0
x1 CCxn
1
1
y
dy dxn
y
m
1
1
.1 C x1 C C xn 1 /m
dxn
m
0
.1 C x1 C C xn 2 /m
m
Z 1
m
C
xn 1 .x1 C C xn 1 /
Z
D
1
1
1
dxn 1 :
x1 C C xn 1
0
Integrating with respect to variables x1 ; : : : ; xn 2 , we get that
Z 1 Z 1
.1 C x1 C C xn 1 /m
n
Mm
dx1 dxn 1
D
m
0
0
Z 1 Z 1
.1 C x1 C C xn 2 /m
dx1 dxn 2
m
0
0
Z 1 Z 1
1
m
dx1 dxn
C
xn 1 .x1 C C xn 1 /
x1 C C xn 1
0
0
1:
By symmetry, for all i; j D 1; : : : ; n 1, one has that
Z 1 Z 1
1
m
xi .x1 C C xn 1 /
dx1 dxn 1
x1 C C xn 1
0
0
Z 1 Z 1
1
dx1 dxn
D
xj .x1 C C xn 1 /m
x1 C C xn 1
0
0
and hence
Z 1 Z
0
1
1
xn 1 .x1 C C xn 1 /
dx1 dxn 1
x1 C C xn 1
0
Z 1 Z 1
1
1
D
.x1 C C xn 1 /mC1
dx1 dxn
n 1 0
x1 C C xn 1
0
m
1;
1:
It follows that
n
Mm
D
1
n
1
m
n 1
MmC1
C
1
Z
0
1
Z
R1
0
1
m
Z
0
1
Z
0
1
.1 C x1 C C xn
1
.1 C x1 C C xn
m
1 / dx1 dxn 1
m
2 / dx1 dxn 2 :
R1
Integrals of the form 0 0 .1 C x1 C C xk /m dx1 dxk are calculated by using
the multinomial formula. The case when m D 0 follows by a similar argument. The
theorem is proved.
MULTIPLE FRACTIONAL PART INTEGRALS
265
Now we show how this recurrence formulas can be used for establishing the integral equalities
Z 1Z 1Z 1
1
9
.3/
3
M0 D
dxdyd´ D ln 3 6 ln 2
x Cy C´
2
6
0
0
0
and
M13 D
Z
0
1Z 1Z 1
0
0
.x C y C ´/
1
dxdyd´
x Cy C´
1
5
1 X
D C
.2 C j /.3 C j / ..j C 4/
6 48
1/ :
j D1
We have, based on Theorem 4 with m D 0 and n D 3, that
Z 1Z 1
Z 1
1 2
3
ln.1 C x C y/dxdy
ln.1 C x/dx
M0 D M1 C
2
0
0
0
1
1
D D1;1 C .9 ln 3 8 ln 2 3/ .2 ln 2 1/
2
2
9
.3/
D ln 3 6 ln 2
:
2
6
On the other hand,
M13
Z 1Z 1
Z 1
1 2
D M2 C
.1 C x C y/dxdy
.1 C x/dx
2
0
0
0
1 1 1
1
D D2;1 C D C V1;3
2
2 3 2
1
X
5
1
D C
.2 C j /.3 C j / ..j C 4/ 1/ ;
6 48
j D1
where the last equality follows from Theorem 1 with k D 1 and m D 3.
R EFERENCES
[1] O. Furdui, “Problem 150, Problems,” Missouri J. Math. Sci., vol. 16, no. 2, 2004.
[2] O. Furdui, “Exotic fractional part integrals and Euler’s constant,” Analysis, vol. 31, pp. 249–257,
2011, doi: 10.1524/anly.2011.1131.
[3] J. Havil, Gamma. Exploring Euler’s constant. Princeton and Oxford: Princeton University Press,
2003.
[4] H. Qin, “Problem 150, Solutions,” Missouri J. Math. Sci., vol. 17, no. 3, 2005.
[5] H. Qin and Y. Lu, “Integrals of fractional parts and some new identities on Bernoulli numbers,” Int.
J. Contemp. Math. Sciences, vol. 6, no. 15, pp. 745–761, 2011.
266
OVIDIU FURDUI
Author’s address
Ovidiu Furdui
Technical University of Cluj-Napoca, Department of Mathematics, Str. Memorandumului Nr. 28,
400114 Cluj-Napoca, Romania
E-mail address: [email protected]