Int. J. Contemp. Math. Sciences, Vol. 6, 2011, no. 15, 745 - 761
Integrals of Fractional Parts and Some
New Identities on Bernoulli Numbers
Huizeng Qin
Institute of Applied Mathematics
Shandong University of Technology
Zibo, Shandong, P. R. China
Youmin Lu
Department of Mathematics and Computer Science
Bloomsburg University, Bloomsburg, PA 17815, USA
Abstract
1
In this paper, we calculate the values of the integrals 0 { x1 }m dx,
1
1
1
1
{ x+y
}m dxdy,
{ x+y+z
}m dxdydz and 0 { x1 }m { 1−x
}n dx,
0≤x,y≤1
0≤x,y,z≤1
where m and n are positive integers and {u} is the fractional part of u,
and express their values in terms of Euler’s constant and Riemann-Zeta
function. We also obtain a set of identities involving the Bernoulli and
Harmonic numbers.
Keywords: fractional parts, new identities, Euler’s constant, Bernoulli
numbers, Harmonic numbers
1
Introduction
1
In [2], Havil studied the integral of 0 { x1 }dx and expressed its value in terms
11 1
}dxdy. Reof Euler’s constant. In [3], Qin calculated the value of 0 0 { x+y
cently, Furdui [5] posted a problem for calculating the values of the integrals
746
Huizeng Qin and Youmin Lu
1
1 1 4
1 3
{
}
and
{ } dx and some results have been published. In this paper,
x
0
0 x
we generalize Furdui’s question and calculate the values of the integrals
1
1
{ }m dx,
x
(1.1)
1 m
} dxdy
x+y
(1.2)
1
}m dxdydz,
x+y+z
(1.3)
0
{
0≤x,y≤1
{
0≤x,y,z≤1
and
0
1
1
1
}dx,
{ }m {
x
1−x
(1.4)
where {u} represents the fractional part of u, for all positive integers m and
n.During the process of our calculation, we have also obtained infinitely many
identities involving Bernoulli numbers and Harmonic numbers. With our limited knowledge, we believe that some of these identities are new.
2
Calculation of Integral (1.1)
Theorem 1
0
1
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
1 − γ, if m = 1,
[ m−2
]
2
(−1)k (m)2k+1 ζ(2k+1)
1 m
ln(2π) −
−
−
22k+1 π 2k
{ } dx =
k=1
⎪
x
m−1
⎪
[ 2 ]
⎪
⎪
(−1)k−1 (m)2k ζ (2k)
⎪
⎪
+2
, for any integer m > 1.
⎩
(2π)2k
m
γ
2
1
m−1
k=1
where (m)k =
m!
(m−k)!
and ζ(x) is the Riemann-Zeta function.
747
Integrals of fractional parts
When m = 3 and 4,this theorem answers Furdui’s question [5]. In fact, we
have
1 2
1
dx = ln(2π) − γ − 1,
x
0
1 3
1
1 3ζ (2)
3
dx = ln(2π) − γ − +
,
x
2
2
π2
0
1 4
1
1 3ζ(3) 6ζ (2)
dx = ln(2π) − 2γ − +
+
.
x
3
π2
π2
0
These results improve the answer given by Janous[5] which are in terms of
infinite series of Riemann-Zeta functions. We first prove a sequence of lemmas.
1
Lemma 2 For any integer m > 0, 0 { x1 }m dx = Im where
Im =
∞
m−1
(m
n=1
Proof.
1
0
1
(−n)k−1
n+1
+ (−1)m−1 mnm−1 ln
−
).
m−k
n
n+1
k=1
(2.1)
m
∞
∞ n+1
1
{u}m
(u − n)m
dx =
du
=
du
x
u2
u2
1
n=1 n
∞ 1
um
=
du
2
(u
+
n)
0
n=1
1 m−1
∞ 1
u
du
−
=
m
u+n
n+1
0
n=1
∞
1 m−1
(−n)m−1
+
u+n
1
du −
m
(−n) u
=
n+1
0
n=1
k=1
∞
m−1
(−n)k−1
1
n+1
+ (−n)m−1 m ln
−
.
=
m
m−k
n
n+1
n=1
k=1
k−1 m−k−1
To prepare for the proof of the other lemmas, we list several formulas
related to zeta function ζ(x), gamma function Γ(x), Bernoulli numbers Bk
and harmonic numbers Hk [1].
ζ(2m) = (−1)m−1
(2π)2m B2m
for integer m > 0.
2(2m)!
(2.2)
748
Huizeng Qin and Youmin Lu
∞
0
∞
0
Γ(α)ζ(α)
y α−1
dy =
, for Re α > 1.
2πy
e −1
(2π)α
(2.3)
Γ (α)ζ(α) + Γ(α)ζ (α) − ln(2π)Γ(α)ζ(α)
y α−1 ln y
dy =
, for Re α > 1.
e2πy − 1
(2π)α
(2.4)
1
Γ (m)
= Hm−1 − γ, Hm =
for integer m > 0.
Γ(m)
k
k=1
m
(2.5)
Combining (2.1)-(2.4), we can get the following formulas.
∞
0
(−1)m−1 B2m
y 2m−1
dy
=
, for integer m > 0.
e2πy − 1
4m
(2.6)
y α−1 ln y
H2m−1 − γ − ln(2π) (2m − 1)!ζ (2m)
m−1
dy
=
(−1)
+
B
,
2m
e2πy − 1
4m
(2π)2m
0
for integer m > 0.
(2.7)
∞
The Abel-Plana formula [1], regular Taylor expansion and formula (2.6)
can be used to prove the following lemma..
Lemma 3 Let a and n be two non-negative integers with a < n and S =
{z|a ≤ Re z ≤ n}. Assume that f (z) is a function defined in S satisfying the
following conditions.
1). f (t) is real over a ≤ t ≤ n;
2). f (z) is continuous throughout S and holomorphic in the interior of S;
3). f (z) = o(e2π| Im z| ) as Im z → ±∞ in S uniformly with respect to Re z.
Then,
n
k=a
f (k) =
a
n
f (n) + f (a)
−2
f (x)dx +
2
0
∞
Im f (a + iy)
dy
e2πy − 1
(2.8)
M
2(−1)M ∞ y 2M Im f (2M) (n + iθn y)
B2j (2j−1)
f
dy,
(n) +
+
(2j)!
(2M)! 0
e2πy − 1
j=1
where M is a positive integer and 0 < θn < 1.
749
Integrals of fractional parts
By using the Leibniz rule for differentiation, one may obtain the following
lemma.
Lemma 4 For any integers k and m with m ≥ k ≥ 0,
where Cm,k
dk m
(x ln x) = (m)k xm−k ln x + Cm,k xm−k ,
dxk
k
m
(−1)j−1
.
= k!
j
k−j
j=1
(2.10)
Using this lemma and mathematical induction, one can also get next lemma.
Lemma 5 For any integers m and k with m ≥ k ≥ 0,
1
dk m
m−k
(x
ln
x)
=
(m)
x
(ln
x
+
).
k
dxk
m−j
j=0
k−1
In particular, Cm,m =
0, if m = 0,
and Cm,m−1 = m!(Hm − 1).
m!Hm , if m > 0,
Now, we combine the lemmas above to obtain a few summation formulas.
Lemma 6 For any integers m > 0, N > 0 and i = 0 or 1,the following
formulas are true.
N
k=1
k 2m−i =
N 2m−i
N 2m+1−i
+
2m + 1 − i
2
(2.11)
m−i
1
+
C 2j−1+i B2m−2j+2−2i N 2j−1+i ;
2m + 1 − i j=1 2m+1−i
N
k=1
k
2m−1
N 2m ln N
N 2m−1 ln N
1 2j−2
ln k =
C
B2m−2j+2 N 2j−2 ln N
+
+
2m
2
2m j=1 2m
m
N 2m C2m−1,2m−2j−1 B2m−2j 2j γ + ln(2π)
N +
B2m
+
−
4m2
(2m − 2j)!
2m
j=1
m−1
+2(−1)m
(2m − 1)!ζ (2m)
1
+ O( );
2m
(2π)
N
(2.12)
750
N
Huizeng Qin and Youmin Lu
k
k=1
2m
m
N 2m ln N
1 2j−1
N 2m+1 ln N
+
+
ln k =
C
B2m−2j+2 N 2j−1 ln N
2m + 1
2
2m + 1 j=1 2m+1
C2m,2m−2j+1 B2m−2j+2
N 2m+1
N 2j−1
+
−
(2m + 1)2 j=1
(2m − 2j + 2)!
m
(2m)!ζ(2m + 1)
1
).
+
O(
22m+1 π 2m
N
Proof. Let f (x) = x2m−1 and M = m. Then, (2.8) becomes
N
∞ 2m−1
N
N 2m−1
y
2m−1
2m−1
m
k
=
x
dx +
+ 2(−1)
dy
2πy
2
e −1
0
0
k=0
+(−1)m−1
+
(2.13)
m
B2j (2m − 1)2j−1 N 2m−2j
j=1
(2j)!
m
1 2m
N 2m N 2m−1 B2m
+
−
+
=
B2m−2j N 2j
2m
2
2m
2m j=1
2j
m−1
N 2m N 2m−1
1 2m
=
B2m−2j N 2j .
+
+
2m
2
2m j=1
2j
If we let f (x) = x2m and M = m+1 in (2.8), formula
(2.11) for i = 0 can be ob
2m−1
x
ln x, for 0 < x ≤ N,
tained similarly. To prove (2.12), we let f (x) =
Then,
0, for x = 0.
f (x) satisfies all conditions in Lemma 3 with a = 0 and n = N and, with
M = m, (2.8) can be written as
N
k
2m−1
ln k =
k=1
N
f (k)
k=0
N
=
0
2m−1
x
(2.14)
N 2m−1 ln N
+ 2(−1)m
ln xdx +
2
m
B2j 2m−1
1
+
(x
ln x)(2j−1) |x=N + O( ).
(2j)!
N
j=1
0
∞
y 2m−1 ln y
dy
e2πy − 1
Applying (2.7) and (2.10) to this expression, one can get formula (2.12). The
proof of (2.13) is similar.
Now, we are ready to complete the proof of Theorem 1. By using (2.1), we
first have
∞
n
1
1
n+1
−
) = lim (ln n −
+ 1) = 1 − γ.
I1 =
(ln
n→∞
n
n+1
k
n=1
k=1
751
Integrals of fractional parts
For m > 1, one may apply the Binomial Theorem to nm−1 = (n + 1 − 1)m−1
and rewrite (2.1) into
m−1
m−2
N
−1
(−n)k−1
m−1
k
(n + 1)k ln(n + 1)
+m
(m
(−1)
Im = lim
N →∞
m−k
k
n=1
k=1
k=1
+m ln(n + 1) + (−1)m−1 m(n + 1)m−1 ln(n + 1) + (−1)m mnm−1 ln n −
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
= lim ⎜
N →∞ ⎜
⎜
⎜
⎜
⎜
⎜
⎝
[ m−1
[ m−2
2 ]
2 ]
m(N−1)
N 2k−1
N 2k
+
m
−
m
m−1
m−2k
m−2k−1
k=1
k=1 [ m−1
[ m−1
N
N
2 ]
2 ]
m−1
1
2k−1
−m
n
−
m
n2k−1 ln n
m−2k
2k − 1 n=1
n=1
k=1
k=1
[ m−2
[ m−2
N
N
2 ]
2 ]
m
−
1
1
2k
n
+
m
n2k ln n
+m
m−2k−1
2k
n=1
n=1
k=1
k=1
N
1
+1
+m ln N! + (−1)m−1 mN m−1 ln N −
n
1
)
n+1
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟.
⎟
⎟
⎟
⎟
⎟
⎟
⎠
n=1
(2.15)
Applying Lemma 6, Stirling’s formula [4] and the formula
n=1
O( N1 ) to (2.15), we get
Im =
lim (
N →∞
m−1
m−1
k=1
k=0
ak N k +
N
1
bk N k ln N − (γ + ln(2π))
m−1 [
2 ]
k=1
n
= ln N + γ +
m
2k
B2k
m−1 [
2 ]
m−1
(2k − 1)!ζ (2k)
2(−1)k
−m
(2π)2k
2k − 1
k=1
[ m−2
2 ]
m−1
1
1
(2k)!ζ(2k + 1) m
(−1)k−1
+ O( ),
+m
+ ln(2π) − γ −
2k+1
2k
2
π
2
m−1
N
2k
k=1
where
ak
[ m−k−1
] 2
m
−
1
m
m
Ck+2j−1,2j−1
+
+
= (−1)k−1 (
(m
2(m − k − 1) k(m − k)
(2j)!
k + 2j − 1
j=1
k + 2j
1 m
m
−
+
)B2j ),
k
(m − k − 2j)(k + 2j)
k
k
752
Huizeng Qin and Youmin Lu
for k = 1, 2, 3, ..., m − 1 and
⎛
⎞
[ m−k−1
]
2
m ⎜ m−k
m − k⎟
B2j + 1 −
bk = (−1)k−1
⎝
⎠ for k = 0, 1, 2..., m − 1.
2
k
2j
j=1
Since Im is bounded and all ak and bk are independent of N, we have ak = 0
for all k = 1, 2, ..., m − 1 and bk = 0 for all k = 0, 1, 2, ..., m − 1. Therefore,
m−1 m−1 [
[
2 ]
2 ]
m
m−1
k (2k − 1)!ζ (2k)
B2k − 2m
(−1)
Im = −(γ + ln(2π))
(2π)2k
2k
2k − 1
j=1
j=1
m−2 [
2 ]
m−1
1
(2k)!ζ(2k + 1) m
ln(2π)
−
γ
−
.
+m
+
(−1)k−1
2k+1 π 2k
2
2
m
−
1
2k
j=1
Using the property b1 = 0, we complete the proof.
Equations ak = 0 for k = 1, 2, ..., m − 1 and bk = 0 for k = 0, 1, 2, ..., m − 1
give us infinitely many identities involving the Bernoulli and Harmonic numbers. In particular, we get the following results.
Theorem 7 For any integer m ≥ 2, the following equations are true.
m−1 [
2 ]
m
m
− 1,
(2.16)
B2k =
2
2k
k=1
m−1
[
2 ]
(
k=1
1
+ H2k
m − 2k
m
2k
)B2k =
1
1
m
−
−
.
2
m 2(m − 1)
(2.17)
Identity (2.16) is a direct result of b0 = 0. Identity (2.17) can be obtained
by replacing m by m − 1 and using (2.16) in a1 = 0.
3
Calculation of (1.2), (1.3) and (1.4)
Theorem 8 For any integer m > 0, we have
⎧
2
2 ln 2 − π12 , if m = 1,
⎪
m
⎨
1
2
3
dxdy =
− π12 − ln 2 − γ, if m = 2,
2
⎪
x+y
⎩ 1 π2
m−3+22−m
− 12 + (m−1)(m−2)
+ m2 Im−1 , if m > 2;
0≤x,y≤1
2
(3.1)
753
Integrals of fractional parts
and
0≤x,y,z≤1
1
x+y+z
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
m
dxdydz =
⎪
⎪
⎪
⎪
⎪
⎪
⎩
9
2
ln 3 − 13
− 19
ln 2 − ζ(3)
, m = 1,
24
4
3
2
ζ(3)
53
+ 4 ln 2 − 3 ln 3 − 3 − π12 , m = 2,
24
2
269
+ 12 ln 3 − 34 ln 2 − ζ(3)
− π6 − 3γ
, m = 3,
96
3
2
h(m)
6m+8−8ζ(3)−mπ2
+
(m−1)(m−2)(m−3)
24
+ m(m−1)
Im−2 , m > 3.
4
(3.2)
where h(m) = 2−1 (m2 − 4m + 2) − 2−m−1 (3m2 − 15m + 10) + 2−m (m2 − 9m +
26) − 3−m 27.
Proof. For the double integral, we first integrate it along the line L :
√
x + y = 2t, and then the line passing through the origin and perpendicular
to L. Since 0 ≤ x, y ≤ 1 is a square, we need to break the integral into two
√
√
parts: one for t changing from 0 to 22 and the other for t changing from 22
√
to 2. For the first part, we have
0≤x,y≤1,x+y≤1
1
x+y
m
dxdy =
0
√
= 2
0
∞
=
1
=
∞ n=1
1
0
x+y= 2t
√
2
2
m
1 π
−
+ Im−1 .
2 12
2
1
x+y
∞
1
0
m
dsdt
m
tdt =
{u}
du =
u3
1
1
m
= −
+
2 n=1 (n + 1)2
2 n=1
=
1
√
2t
m
um
du
(u + n)3
∞
2
√
2
2
1
0
∞
n+1
n=1
n
um−1
du
(u + n)2
1
u
du
u
(u − n)m
du
u3
754
Huizeng Qin and Youmin Lu
1
For 0 ≤ x, y ≤ 1 and x + y ≥ 1, 0 < x+y
< 1and
have
m
1
dxdy =
x+y
0≤x,y≤1,x+y≥1
1
x+y
(
0≤x,y≤1,x+y≥1
√2 =
=
√
2
2
√
√
x+y= 2t
√
2
2
√
2
=
1
.
x+y
Therefore, we
1 m
) dxdy
x+y
dsdt
(x + y)m
√
2
2 − 2t
2−u
√
dt =
du
um
( 2t)m
1
2
⎧
⎪
⎨ 2 ln 2 − 1, m = 1
1 − ln 2, m = 2
=
⎪
⎩ m−3+22−m
, m > 2.
(m−1)(m−2)
Combining the results, we get (3.1). The calculation of (3.2) is similar. The
√
only difference is to perform the integration over the plane P : x + y + z = 3t
and then along the line passing through the origin and perpendicular to plane
P. The interval for t needs to be broken into three parts: one for t changing
√
√
√
from 0 to 33 , the second for t changing from 33 to 2 3 3 and the other for t
√
√
changing from 2 3 3 to 3.
Theorem 9 For any integers m > 0 and n > 0, we have
0
1
m n
1
1
dx = I(m, n) + I(n, m),
x
1−x
(3.3)
where
I(n, m) =
m
n
n−i
(−1)n−i+j−1 i l
Cn (−1)l Cn−i
R(n − m + j − l)
i
+
j
−
m
j=1 i=0,i=m−j
l=1
m−2
n−m+j−1
jCnm−j
+
j=1,0≤m−j≤n
n−2
+(−1)n+m−1 mn
l=0
n−m+j−k
(−1)j−k−1Cn−m+j
J(k)
k=0
l
(−1)n−l−1 Cn−1
J(l) + (−1)m−1 γ
n
n
(−1)m+n−i i i
(−1)m+n−i i i−1
Cn (2 − 1) −
Cn (2 − 1)
−m
i
i
−
1
i=1
i=2
755
Integrals of fractional parts
+(−1)m H(n − 1) +
m(−1)m−1 (−1)m−1 n (−1)m+n
+
−
+ (n − m)(−1)m+n ln 2,
n
n−1
2
H(k) =
and
⎧
⎪
⎨
J(n) =
⎧
⎪
⎨
1, k > 0
, R(n) =
⎪
0, k ≤ 0
⎩
γ ln(2π)
Bn
2n
+ 2(−1)
(−1)
⎪
⎩
n+1
2
n
−1 n!ζ(n+1)
2
2n+1 π n
ln(2π)
,
2
0, n ≥ 0
,
γ, n = −1
ζ(−n), n < −1
n!ζ (n)
, if
(2π)n
n > 0 is odd
, if n > 0 is even
if n = 0.
In particular,
⎧
− ln(2π) + γ + 32 , if n = 1
⎪
⎪
⎪
⎪
n−1
n−k
⎪
k+j n−j−k
(−1)n+i−1 j i
⎪
k
⎪
(−1)
R(−k)
Cn
Cn−k−j
⎪
i+j−n
⎪
⎪
j=1
i=0
k=1
⎪
⎨ n−3
n−l−2
n−2
j
l
I(n, n) =
− Cnl J(l)
(−1)j (j + l)Cn−1
+ n2 (−1)n−l Cn−1
J(l)
⎪
⎪
j=0
l=0
l=0
⎪
⎪
n
n
i
⎪
(−1)i i
⎪
i
i
⎪
C
(2
−
1)
−
Cn (2i−1 − 1)
+(−1)n−1 γ − n (−1)
⎪
n
i
i
⎪
⎪
i=0
l=2
⎪
n−1 n
⎩
− 12 , if n > 1.
+(−1)n H(n − 1) + (−1)n−1 + (−1)
n−1
Proof.
n
n
n
∞ m
1 m ∞ k+1
1
1
u
1
u}
(u − k)m
dx =
du =
du
x
1−x
u2
u−1
u2
u−1
0
1
k
k=1
n
1
∞ 1
m
1
u du
um
=
+
du.
n
2
2
(u + k) (u + k + 1)
u
0 (u + 1)
k=1 0
If we denote I(m, n) =
∞ 1
k=1
0
1
um
(u + 1)2
um du
,
0 (u+k)n (u+k+1)2
then,
n
∞
∞ k+1
1
{v}n dv
(v − k)n dv
du =
=
u
v m (v + 1)2
v m (v + 1)2
1
k=1 k
∞ 1
v n dv
=
= I(n, m).
m
2
0 (v + k) (v + k + 1)
k=1
756
Huizeng Qin and Youmin Lu
Thus, we need to calculate for I(n, m). Using partial fractions and the Binomial
Theorem over un = (u + k − k)n and un = (u + k + 1 − k − 1)n , we first have
u
(u +
n
k)m (u
= un
1)2
+k+
m
(−1)j−1 j(u + k)j−m−1 +
j=1
m
n
m
m
(−1)
(−1) m
+
u + k + 1 (u + k + 1)2
(−1)n−i+j−1jk n−i Cni (u + k)i+j−m−1
=
j=1 i=0
n
+m
(−1)m+n−i (k + 1)n−i Cni (u + k + 1)i−1
i=0
+
n
(−1)m+n−i (k + 1)n−i Cni (u + k + 1)i−2 .
i=0
Plugging it back to the definition of I(n, m), one has
I(n, m) =
∞ 1
0
k=2
m n
(
(−1)n−i+j−1 j(k − 1)n−i Cni (u + k − 1)i+j−m−1
j=1 i=0
n
+m (−1)m+n−i k n−i Cni (u + k)i−1
i=0
+
n
(−1)m+n−i k n−i Cni (u + k)i−2 )du
i=0
n+m
= (−1)
n
lim
N
−1
N →∞
(
n
m
Cni
k=1 i=0 j=1,j=m−i
j(−1)m+i+j−1 k n−i (k + 1)i+j−m
i+j−m
m
j(−1)m+i+j−1 k n+j−m
−
Cni
i+j −m
i=0 j=1,j=m−i
min{m−1,n}
+
Cni (m − i)k n−i ln k
i=0
min{m−1,n}
n
(−1)i
i
n−i
(k + 1)n−i (k
−
Cn (m − i)k ln(k + 1) + m Cni
i
i=0
i=1
n
n
(−1)i
(−1)i
(k + 1)n +
(k + 1)n−i (k + 2)i−1
−m Cni
Cni
i
i−1
i=1
i=2
n
(−1)i Cni
1
n−1
n−1
−
i=2
i−1
(k + 1)
+ (k + 1)
[
k+2
+ 2)i
+ (mk + m − n) ln
k+2
]).
k+1
757
Integrals of fractional parts
Using the Binomial Theorem, we can rewrite it into
n+m
I(n, m) = (−1)
n
−
N
−1
n
m
n−i
j(−1)m−l+i+j−1 (k + 1)n−l+j−m
l
lim
(
Cni Cn−i
N →∞
i+j−m
k=1 i=0 j=1,j=m−i l=0
m
j(−1)m+i+j−1 k n+j−m
Cni
i+j −m
i=0 j=1,j=m−i
min{m−1,n}
−
Cni (m
i=0
n
i=1
+
+
n
i=2
n−1
i
(k + 1)
l=0
i−1
i
i−1
l
Ci−1
(k
n−i+l
Cni (m − i)k n−i ln k
i=0
+ 1)
n
(−1)i
− m Cni
(k + 1)n
i
i=1
n−i+l
−
n
i=2
l=0
n
l
(−1)l Cn−1
(k + 2)n−2−l + m
−n
l=0
i
i
(−1)
Cni
l=1
n−1
+
n−i
l
− i) (−1)l Cn−i
(k + 1)n−i−l ln(k + 1)
(−1)
Cni
+m
min{m−1,n}
Cni
(−1)i
(k + 1)n−1
i−1
(−1)l Cnl (k + 2)n−l ln(k + 2)
l=0
l
(−1)l Cn−1
(k + 2)n−l−1 ln(k + 2) − m(k + 1)n ln(k + 1)
l=0
+n(k + 1)n−1 ln(k + 1))
758
Huizeng Qin and Youmin Lu
Hence,
n+m
I(n, m) = (−1)
n
−
n
m
n−i
N
j(−1)m−l+i+j−1 n−l=J−M
i l
lim (
Cn Cn−i
k
N →∞
i
+
j
−
m
i=0 j=1,j=m−i l=0
k=1
min{m−1,n}
m
j(−1)m+i+j−1 k n+j−m
Cni
i+j −m
i=0 j=1,j=m−i
min{m−1,n}
−
n−i
Cni (m − i)
i=0
l
(−1)l Cn−i
l=0
+
Cni (m
N
− i) k n−i ln k
i=0
k=1
N
k n−i−l ln k
k=1
n
i
n
N
N
(−1)i n−i+l
(−1)i n
+m Cni
k
− m Cni
k
i
i
i=1
i=1
l=0 k=1
k=1
+
n
Cni
i=2
+
n−1
i−1
n
N
N
(−1)i l n−i+l i (−1)i n−1
Ci−1 k
−
Cn
k
i − 1 l=0
i
−
1
i=2
k=1
k=1
l
(−1)
l=1
n−1
−n
l
Cn−1
l
(−1)
N
k=1
N
l
Cn−1
l=0
+n(k + 1)
k
n−2−l
l=0
k
n−l−1
k=1
n−1
N
n
l l
+ m (−1) Cn k n−l ln k + Qn,m (N)
k=1
N
ln k − m
N
k ln k + n k n−1 ln k + P n, m,
n
k=1
k=1
ln(k + 1))
where
Pn,m
n
= −m Cni
m
n−i
j(−1)m−l+i+j−1
l
Cn−i
i+j−m
n
(−1)i i i
−m
Cn 2
i
i=1
j=1,j=m−i l=0
n
n
(−1)i i (−1)i i i−1 (−1)i i
Cn −
C 2 +
C
+m
i
i−1 n
i−1 n
i=1
i=2
i=2
n−1
n−1
n
l i
n−2−l
l i
− (−1) Cn−1 2
−
(−1) Cn−1 − m (−1)l Cni 2n−l
l=0
l=0
l=0
n−1
i
2n−1−l ln 2,
+ (−1)l Cn−1
l=0
i=0
n
ln 2
759
Integrals of fractional parts
n
Qn,m (N ) =
Cni
i=0
+
n−1
m
j(−1)m+i+j−1 n+j−m
−
N
i+j−m
j=1,j=m−i
min{m−1,n}
Cni (m − i)N n−i ln N
i=0
n
i
(−1)l Cn−1
(N + 1)n−2−l + m (−1)l Cnl (N + 1)n−l ln(N + 1)
l=0
n−1
−n
l=0
i
(−1)l Cn−1
(N + 1)n−1−l ln(N + 1).
l=0
Noticing the identities
n
(−1)i i
n!
Cn =
, for α = 0, −1, ..., −n,
α
+
i
α(α
+
1)...(α
+
n)
i=0
n
(−1)j jCnj aj−1 bn−j = −n(b − a)n − 1,
j=1
n
(−1)i
i=1
i
n
(−1)i
i=2
i−1
Cni ai
=
n
(1 − a)i
i=1
i
− Hn ,
n−1
(1 − a)i (1 − a)n − 1 − na
= nHn−1 − n
−
.
i
a
i=1
Cni ai−1
one gets
Pn,m
n−1
j(−1)m+n+j m(−1)n
(−1)i
−
+ (n − m) ln 2 +
=
n+j−m
n
i
i=1
j=1,j=m−n
m
+
⎛
⎜
⎜
⎜
Qn,m (N ) = ⎜
⎜
⎝
(−1)n
− δ n−1,0
2
n
i=0
Cni
m
(4.14)
j(−1)m+i+j−1
N n+j−m
i+j−m
i=0
+ mN ln N − nN
n−1
ln N ⎟
⎟
⎟
⎟.
1
i
n−i
n−1
Cn (m − i)N
ln N − nN
ln(1 + N )
⎟
⎠
1
1
n−1
n
+ mN ln(1 + N ).
+ N +1 N
j=1,j=m−i
min{m−1,n}
−
⎞
n
(4.15)
By the proof of Theorem 1 and noticing
k−1
1
(−1)i−1 k−i (−1)k−1
1
N
+ O( ),
+
N ln(1 + ) =
N
i
k
N
i=1
k
(−1)i
1
1
N n−1 = N n−2
+ O( ),
i
N +1
N
N
i=i
n−2
760
Huizeng Qin and Youmin Lu
and R(k) = 0 for k ≥ 0, we have
⎛
n
Cni
m−i−1
m−i−l
l+n+j−m j(−1)n+l+i−1
Cn−i
R(−l)
i+j−m
⎞
n−1
+ (−1) γ
⎜
i=0
l=1 j=1,j=m−i
⎜
⎜ n
m−n−1
j(−1)m+i+j−1
n
⎜ − Ci
R(n − m + j) + n−1
(−1)n−1 H(n − 1)
⎜
n
i+j−m
⎜ i=0
j=1,j=m−i
⎜
min{m−1,n}
⎜
L(n, m) = ⎜ +
Cni (m − i)J(n − i) + (−1)n−2 H(n − 1) + m
(−1)n−1
n
⎜
i=0
⎜
min{m−1,n}
n−i
⎜
i
l
⎜
−
C
(m
−
i)
(−1)l Cn−i
J(n − i − l)
n
⎜
i=0
l=0
⎜
n
n−1
⎝
l
(−1)l Cn−1
J(n − 1 − l) + Pn,m
+m (−1)l Cnl J(n − l) − n
l=1
l=1
(4.16)
When n = m, (4.16) becomes (4.5). The proof is completed.
when n = m = 1, 2, 3,or 4, by (4.5) we obtain
0
0
0
1
1
1
1
1
}dx = 2γ − 1,
{ }{
x 1−x
1
1 2
{ }2 {
} dx = 4ln(2π) − 4γ − 5,
x
1−x
1
1 3
18ς (2)
} dx = 6γ + 2 − ς(2) − 3 ln(2π) −
{ }3 {
x
1−x
π2
and
1
2
24ς (2) 24ς(3)
1
1 3
2
11
} dx = ς(2) − ς(3) +
.
{ }3 {
+
+
4ln(2π)
−
8γ
−
x
1−x
3
3
π2
π2
3
0
References
[1] F. W. J. Olver, Asymptotics and Special Functions, Academic Press, New
York and London 1974
[2] Havil, J. Gamma, Exploring Euler ’s Constants, Princeton University
Press, pp. 109-111, 2003
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Integrals of fractional parts
761
[3] H. Qin, Solutions of Problem 150, Missouri Journal of Mathematical Sciences, 2005(3)
[4] N. N. Lebedev, Special Functions and Their Applications, Dover Publications, Inc. New York 1972
[5] Ovidiu Furdui, Campia Turzii, Cluj, Romania, Crux, Problem 3366, Oct.
2009, pp. 403-404
Received: July, 2010