The integrals in Gradshteyn and Ryzhik. Part9: Combinations of logarithms, rational and trigonometric functions

Abstract

The classical table of integrals by I. S. Gradshteyn and I. M. Ryzhik contains many definite integrals where the integrand is the product of a rational function times the logarithm of another rational function. We begin the systematic evaluation of these integrals.

Full text

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arXiv:0707.2124v1 [math.CA] 14 Jul 2007
THE INTEGRALS IN GRADSHTEYN AND RYZHIK.
PART 9: COMBINATIONS OF LOGARITHMS, RATIONAL AND
TRIGONOMETRIC FUNCTIONS.
TEWODROS AMDEBERHAN, VICTOR H. MOLL, JASON ROSENBERG,
ARMIN STRAUB, AND PAT WHITWORTH
Abstract. The table of Gradshteyn and Ryzhik contains many integrals with
integrands of the form R1(x) (lnR2(x))m, where R1and R2are rational func-
tions. In this paper we describe some examples where the logarithm appears
to a single power, that is m = 1, and the poles of R1 are either real or purely
imaginary.
1. Introduction
The table of integrals [3] contains many examples of the form
(1.1)
?b
a
R1(x) (lnR2(x))mdx,
where R1 and R2 are rational functions, a, b ∈ R+and m ∈ N. For example,
4.231.1 states that
?1
This result can be established by the elementary methods described here.
Other examples, such as 4.233.1
(1.2)
0
lnxdx
1 + x
= −π2
12.
(1.3)
?1
0
lnxdx
1 + x + x2=2
9
?2π2
3
− ψ′(1
3)
?
,
and 4.261.8
(1.4)
?1
0
ln2x1 − x
1 − x6dx =8√3π3+ 351ζ(3)
486
,
require more sophisticated special functions. Here ψ is the digamma function de-
fined by
(1.5)ψ(x) =Γ′(x)
Γ(x),
and ζ(s) is the classical Riemann zeta function. These results will be described in
a future publication.
2000 Mathematics Subject Classification. Primary 33.
Key words and phrases. Logarithms, rational and trigonometric functions.
The authors wish to the partial support of nsf-dms 0409968 and nsf-ccli 0633223.
1

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2 T. AMDEBERHAN, V. MOLL, J. ROSENBERG, A. STRAUB, AND P. WHITWORTH
The integrals discussed here can also be framed in the context of trigonometric
functions. For example, the change of variables x = tant yields the identity
(1.6)
?1
0
lnxdx
1 + x2=
?π/4
0
lntantdt = −G.
Here G is the Catalan’s constant defined by
(1.7)G :=
∞
?
k=1
(−1)k
(2k + 1)2.
In this paper we concentrate on integrals of the type (1.1) where the logarithm
appears to the first power and the poles of the rational function are either real of
purely imaginary. The method of partial fractions and scaling of the independent
variable, show that such integrals are linear combinations of
(1.8)hn,1(b) :=
?b
0
lntdt
(1 + t)n,
and
(1.9)hn,2(b) :=
?b
0
lntdt
(1 + t2)n.
The function hn,1 was evaluated in [4], where is was denoted simply by h. We
complete this evaluation in Section 4, by identifying a polynomial defined in [4]. The
closed-form of hn,1involves the Stirling numbers of the first kind. The evaluation
of hn,2is discussed in Section 6. The value of hn,2involves the tangent integral
?x
The case of integrals with more complicated pole structure will be described in
a future publication.
(1.10)Ti2(x) :=
0
tan−1t
t
.
2. Some elementary examples
We begin our discussion with an elementary example. Entry 4.291.1 states that
(2.1)
?1
0
ln(1 + x)
x
dx =π2
12.
To establish this value we consider first a useful series. The result is expressed in
terms of the Riemann zeta function
(2.2)ζ(s) :=
∞
?
k=1
1
ks.
Lemma 2.1. Let s > 1. Then
(2.3)
∞
?
k=1
(−1)k
ks
= −2s−1− 1
2s−1
ζ(s)
and
(2.4)
∞
?
k=1
1
(2k − 1)s=2s− 1
2s
ζ(s)

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LOGARITHMS, RATIONAL AND TRIGONOMETRIC3
Proof. The second sum is
(2.5)
∞
?
k=1
1
(2k − 1)s=
∞
?
k=1
1
ks−
∞
?
k=1
1
(2k)s= (1 − 2−s)ζ(s).
To evaluate the first sum, split it into even and odd values of the index k:
(2.6)
∞
?
k=1
(−1)k
ks
=
∞
?
k=1
1
(2k)s−
∞
?
k=1
1
(2k − 1)s
and use the value of the first sum.
?
To evaluate (2.1) we employ the expansion
(2.7)ln(1 + x) =
∞
?
k=1
(−1)k−1
k
xk
and integrate term by term we obtain
(2.8)
?1
0
ln(1 + x)
x
dx = −
∞
?
k=1
(−1)k
k2
.
The result now follows from the lemma and the classical value ζ(2) = π2/6.
A similar calculation yields 4.291.2:
(2.9)
?1
0
ln(1 − x)
x
dx = −π2
6.
The change of variables x = e−tproduces the evaluation of 4.223.1
(2.10)
?∞
0
ln(1 + e−t)dt =π2
12
and 4.223.2:
(2.11)
?∞
0
ln(1 − e−t)dt = −π2
6
3. More elementary examples
In [4] we analyze the case in which the rational function R1has a single multiple
pole and R2(x) = x. There are simple examples where the evaluation can be
obtained directly. For instance, formula 4.231.1 states that
(3.1)
?1
0
lnx
1 + xdx = −π2
12.
This can be established by simply expading the term 1/(1+x) as a geometric series
and integrate term by term. The same is true for 4.231.2
(3.2)
?1
0
lnx
1 − xdx = −π2
6.
The evaluation of 4.231.3
(3.3)
?1
0
x lnx
1 − xdx = 1 −π2
6,

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4 T. AMDEBERHAN, V. MOLL, J. ROSENBERG, A. STRAUB, AND P. WHITWORTH
and 4.231.4
(3.4)
?1
0
1 + x
1 − xlnxdx = 1 −π2
3,
follow directly from 3.2. Similar elementary algebraicmanipulations produce 4.231.19
?1
and 4.231.20
?1
The same is true for 4.231.14
?1
and 4.231.15
?1
via the change of variables t = x2.
(3.5)
0
x lnx
1 + xdx = −1 +π2
2,
(3.6)
0
(1 − x) lnx
1 + x
dx = 1 −π2
6.
(3.7)
0
x lnx
1 + x2dx = −π2
48,
(3.8)
0
x lnx
1 − x2dx = −π2
24,
The evaluation of 4.231.13:
(3.9)
?1
0
lnxdx
1 − x2= −π2
48
is a direct consequence of the partial fraction decomposition
1 − x2=1
and the identites (3.1) and (3.2).
(3.10)
1
2
1
1 − x+1
2
1
1 + x
It is often the case that a simple change of variables reduces an integral to one
that has previously been evaluated. For example, the change of variables t = 1−x2
produces
?1
The right-hand side is given in (3.2) and we obtain the value of 4.295.11:
?1
4. A single multiple pole
(3.11)
0
ln(1 − x2)
x
dx =1
2
?1
0
lntdt
1 − t.
(3.12)
0
ln(1 − x2)
x
dx = −π2
12.
The situation for a single multiple pole is more delicate. The pole pole may be
placed at x = −1 by scaling. The main result established in [4] is:
Theorem 4.1. Let n ≥ 2 and b > 0. Define
?b
and introduce the function
qn(b) = (1 + b)n−1hn,1(b).
(4.1)hn,1(b) =
0
lntdt
(1 + t)n
(4.2)

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LOGARITHMS, RATIONAL AND TRIGONOMETRIC5
Then
(4.3)qn(b) = Xn(b)lnb + Yn(b)ln(1 + b) + Zn(b),
where
(4.4)Xn(b) =(1 + b)n−1− 1
n − 1
,Yn(b) = −(1 + b)n−1
n − 1
.
Finally, the function
(4.5)Tn(b) := −(n − 1)!Zn(b)
b(1 + b)
,
satisfies T2(b) = 0 and for n ≥ 1 it satisfies the recurrence
(4.6)Tn+2(b) = n(1 + b)Tn+1(b) + (n − 1)!
?(1 + b)n− 1
b
?
.
It follows that Tn(b) is a polynomial in b of degree n − 3 with positive integer
coefficients.
Note 4.2. The case n = 1 is expressed in terms of the polylogarithm function
(4.7) PolyLog[n,z] :=
∞
?
k=1
zk
kn.
Indeed, we have
(4.8)
?b
0
lnx
1 + xdx = lnb ln(1 + b) + PolyLog[2,−b].
We now identify the polynomial Tn. The first few are given by
T3(b)
T4(b)
T5(b)
T6(b)
=1, (4.9)
=3b + 4,
11b2+ 27b + 18,
50b3+ 176b2+ 216b+ 96.
=
=
For n ≥ 2, define
(4.10)An(b) =
1
(n − 2)!Tn(b).
Then (4.6) becomes
(4.11)An+2(b) = (1 + b)An+1(b) +(1 + b)n− 1
bn
,
with initial condition A2(b) = 0.
The polynomial Anis written as
(4.12)An(b) =
n−3
?
j=0
an,jbj
The recursion (4.11) produces

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6 T. AMDEBERHAN, V. MOLL, J. ROSENBERG, A. STRAUB, AND P. WHITWORTH
Lemma 4.3. Let n ≥ 4. Then the coefficients an,jsatisfy
an,0
=an−1,0+ 1, (4.13)
an,j
=an−1,j+ an−1,j−1+
(n − 3)!
(j + 1)!(n − 3 − j)!, for 1 ≤ j ≤ n − 4,
an,n−3
=an−1,n−4+
1
n − 2.
The expressions an,0= n−2 and an,n−3= Hn−2, the harmonic number, are easy
to determine from (4.13). We now find closed-form expressions for the remaining
coefficients. These involve the Stirling numbers of the first kind s(n,j) defined by
the expansion
(4.14)(x)n=
n
?
j=1
s(n,j)xj,
where (x)n:= x(x+1)(x+2)···(x+n−1) is the Pochhammer symbol (also called
rising factorial). The numbers s(n,1) are given by
(4.15)s(n,1) = (−1)n−1(n − 1)!,
and the sequence s(n,j) satisfies the recurrence
(4.16)s(n + 1,j) = s(n,j − 1) − ns(n,j).
Theorem 4.4. The coefficients an,jare given by
(4.17)an,j=
(−1)j
(j + 1)!
?n − 2
j + 1
?
s(j + 2,2).
Proof. Define
(4.18)bn,j:= (−1)jan,j× (j + 1)!
?n − 2
j + 1
?−1
.
The recurrence (4.13) shows that bn,jis independent of n and satisfies
(4.19)bn,j+1= −jbn,j+ (−1)j+1j!
and this is (4.16) in the special case j = 2. Formula (4.17) has been established.
?
We now restate the value of hn,1. The index n is increased by 1 and the identity
|s(n,k)| = (−1)n+ks(n,k) is used in order to make the formula look cleaner.
Corollary 4.5. Assume b > 0 and n ∈ R. Then
?b
1
n(1 + b)n−1
0
lntdt
(1 + t)n+1
=
1
n
?1 − (1 + b)−n?lnb −1
n−1
?
nln(1 + b) (4.20)
−
j=1
1
j!
?n − 1
j
?
|s(j + 1,2)|bj.
The special case b = 1 provides the evaluation
?1
0
lntdt
(1 + t)n+1= −ln2
n
−
1
n2n−1
n−1
?
j=1
1
j!
?n − 1
j
?
|s(j + 1,2)|.

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LOGARITHMS, RATIONAL AND TRIGONOMETRIC7
Elementary changes of variables, starting with t = tan2ϕ, convert (4.20) into
?1
a
sn−1ln(1 − s)ds=
(1 − an)
n2
[nln(1 − a) − 1](4.21)
−
1
n
n−1
?
j=1
1
j!
?n − 1
j
?
|s(j + 1,2)|an+1−j(1 − a)j.
The special case a = 1/2 produces
?1
1/2
sn−1ln(1 − s)ds =
1
n2n−1
n−1
?
j=1
?n−1
j
?|s(j + 1,2)|
j!
−(2n− 1)
n22n
(nln2 + 1),
and a = 0 gives
(4.22)
?1
0
sn−1ln(1 − s)ds = −
?1
n2+|s(n,2)|
n!
?
.
5. Denominators with complex roots
In this section we consider the simplest example of the type (1.1), where the
rational function R1(x) has a complex (non-real) pole. This is
G := −
?1
0
lnx
1 + x2dx.
The constant G is called Catalan’s constant and is given by
(5.1)G =
∞
?
k=0
(−1)k
(2k + 1)2.
Entry 4.231.12 of [3] states
(5.2)
?1
0
lnx
1 + x2dx = −G.
To verify (5.2) simply expand the integrand to produce
?1
0
lnxdx
1 + x2
=−
∞
?
∞
?
∞
?
k=0
(−1)k
?1
?∞
0
x2klnxdx
=−
k=0
(−1)k
0
te−(2k+1)tdt
=−
k=0
(−1)k
(2k + 1)2
?∞
0
σe−σdσ
The integral is evaluated by integration by parts or recognizing its value as Γ(2) = 1.
Integration by parts gives the alternative form
(5.3)
?1
0
tan−1x
x
dx = G,
that appears as 4.531.1.

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8 T. AMDEBERHAN, V. MOLL, J. ROSENBERG, A. STRAUB, AND P. WHITWORTH
There are many definite integrals in [3] that are related to Catalan’s constant.
For example, the change of variables t = 1/x yields from (5.2), the value
?∞
This is the second part of 4.231.12. Adding these two expressions for G, we conclude
that
?∞
The change of variables x = at in (5.2) yields 4.231.11:
?a
We now introduce material that will provide a genaralization of (5.5) to the case
of a multiple pole at i. The integral is expressed in terms of the polygamma function
ψ(x) =Γ′(x)
(5.4)
1
lntdt
1 + t2= G.
(5.5)
0
lnxdx
1 + x2= 0.
(5.6)
0
ln dx
x2+ a2=π lna
4a
−G
a
(5.7)
Γ(a).
Lemma 5.1. Let a, b ∈ R. Then
(5.8)
?∞
0
lntdt
(1 + t2)b=Γ(1
2)ψ(1
2) − Γ(b −1
2Γ(b)
2)ψ(b −1
2)
.
Proof. Define
(5.9)f(a,b) :=
?∞
0
tadt
(1 + t2)b.
Then
(5.10)
d
daf(a,b) =
?∞
0
talnt
(1 + t2)bdt.
In particular,
(5.11)
d
daf(a,b)
???
a=0=
?∞
0
lntdt
(1 + t2)b.
The change of variables s = t2gives
(5.12)f(a,b) =1
2
?∞
0
s(a−1)/2ds
(1 + s)b.
Now use the integral representation
(5.13)B(u,v) =
?∞
0
su−1ds
(1 + s)u+v
(given in 8.380.3 in [3] and proved in [5]) with u = (a+1)/2 and v = b−(a+1)/2
to obtain
?a + 1
Therefore?∞
=
2Γ(b)(Γ′((a + 1)/2) − Γ′(b − (a + 1)/2))
(5.14)f(a,b) = B
2
,b −a + 1
2
?
=Γ((a + 1)/2)Γ(b − (a + 1)/2)
Γ(b)
.
0
lntdt
(1 + t2)b
=
d
daf(a,b)
1
???
a=0
(5.15)
???
a=0.

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LOGARITHMS, RATIONAL AND TRIGONOMETRIC9
Now use the relation Γ′(x) = ψ(x)Γ(x) to obtain the result.
?
The case b = n ∈ N requieres the value
(5.16) Γ(n +1
2) =
√π
22n
(2n)!
n!
,
and
(5.17)ψ(n +1
2) = −γ + 2ln2 − 2
n
?
k=1
1
2k − 1= −γ − 2ln2 + 2H2n− Hn,
that appears in 8.366.3. Here Hnis the n-th harmonic number. The reader will
find a proof of this evaluation in [2], page 212. A proof of (5.16) appears as Exercise
10.1.3 on page 190 of [2].
Corollary 5.2. Let n ∈ N. Then
(5.18)
?∞
0
lnxdx
(1 + x2)n+1= −
π
22n+1
?2n
n
?
n
?
k=1
1
2k − 1.
We now provide a proof of Entry 4.231.7 in [3]:
?∞
0
lnxdx
(a2+ b2x2)n=
Γ(n −1
4(n − 1)!a2n−1b
2)√π
?
2ln
?a
2b
?
− γ − ψ(n −1
2)
?
.
Taking the factor b2out of the parenthesis on the left and letting c = a/b yields
the equivalent form
(c2+ x2)n=Γ(n −1
4(n − 1)!
It is more convenient to replace n by n + 1 to obtain
(c2+ x2)n+1=Γ(n +1
4n!
?∞
0
lnxdx
2)√π
?
2ln
?c
2
?
− γ − ψ(n −1
2)
?
.
?∞
0
lnxdx
2)√π
?
2ln
?c
2
?
− γ − ψ(n +1
2)
?
.
Using (5.16) and (5.17) the requested evaluation amounts to
(5.19)
?∞
0
lnxdx
(c2+ x2)n+1=
π
(2c)2n+1
?2n
n
??
lnc −
n
?
k=1
1
2k − 1
?
.
This can be written as
(5.20)
?∞
0
lnxdx
(c2+ x2)n+1=
π
(2c)2n+1
?2n
n
?
(lnc − Hn+ 2H2n).
To establish this, make the change of variables x = ct to obtain
?∞
Using Wallis’ formula
?∞
the required evaluation now follows from Corollary 5.2.
The special case n = 0 yields 4.231.8:
?∞
0
lnxdx
(x2+ c2)n+1=
lnc
c2n+1
?∞
0
dx
(t2+ 1)n+1+
1
c2n+1
?∞
0
lntdt
(t2+ 1)n+1.
(5.21)
0
dt
(1 + t2)n+1=
π
22n+1
?2n
n
?
,
(5.22)
0
lnxdx
a2+ b2x2=
π
2abln
?a
b
?
.

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10 T. AMDEBERHAN, V. MOLL, J. ROSENBERG, A. STRAUB, AND P. WHITWORTH
This evaluation also appears as 4.231.9 in the form
(5.23)
?∞
0
lnpxdx
q2+ x2=π
2qlnpq.
6. The case of a single purely imaginary pole
In this section we evaluate the integral
(6.1)hn,2(a,b) :=
?b
0
lntdt
(t2+ a2)n+1,
for a, b > 0 and n ∈ N. This is the generalization of (4.1) to the case in which the
integrand has a multiple pole at t = ia. The change of variables t = ax yields
hn,2(a,b) = a−2n−1gn(b/a), (6.2)
where
(6.3)gn(x) :=
?x
0
lntdt
(t2+ 1)n+1.
We produce first a recurrence for the companion integral
(6.4)fn(x) :=
?x
0
dt
(t2+ 1)n+1.
Lemma 6.1. Let n ∈ N and x > 0. Then
(6.5)2nfn(x) = (2n − 1)fn−1(x) +
and f0(x) = tan−1x.
x
(x2+ 1)n,
Proof. Integrate by parts.
?
The recurrence (6.5) is now solved using the following result established in [1].
Lemma 6.2. Let n ∈ N and define λj = 22j?2j
j
?−1. Suppose the sequence zn
λkrk
2k
satisfy the recurrence 2nzn− (2n − 1)zn−1= rn, with rngiven. Then
1
λn
(6.6)zn=
?
z0+
n
?
k=1
?
.
We conclude with an explicit expression for fn(x).
Proposition 6.3. Let n ∈ N. Then
?x
(6.7)
0
dt
(1 + t2)n+1=
?2n
n
?
22n

tan−1x +
n
?
j=1
22j
2j?2j
j
?
x
(x2+ 1)j

.
Note 6.4. This expression for fnappears as 2.148.4 of [3] in the alternative form
?x
k=1
(2n − 1)!!
2nn!
0
dt
(1 + t2)n+1
=
x
2n + 1
n
?
(2n + 1)!!
(2n − 2k + 1)!!
(n − k)!
2kn!
1
(1 + x2)n+1−k
(6.8)
+
tan−1x.
We now produce a recurrence for the integral gn(x).

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LOGARITHMS, RATIONAL AND TRIGONOMETRIC11
Lemma 6.5. Let n ∈ N. Then the integrals gn(x) satisfy
(6.9)2ngn(x) − (2n − 1)gn−1(x) = 2nlnxfn(x) − [(2n − 1)lnx + 1]fn−1(x).
Proof. Integration by parts yields
(6.10)gn(x) = fn(x)lnx −
?x
0
fn(t)dt
t.
From the recurrence (6.5) we obtain
(6.11)2n
?x
0
fn(t)dt
t
− (2n − 1)
?x
0
fn−1(t)dt
t
= fn−1(x).
Now replace the integral term from (6.10) to obtain the result.
?
In order to produce a closed-form expression for gn(x) using Lemma 6.2, we need
the initial condition
?x
Lemma 6.6. The function gn(x) is given by
(6.12)g0(x) =
0
lntdt
1 + t2.
(6.13)gn(x) =
∞
?
k=0
(−1)k
?n + k
k
?x2k+1
2k + 1
?
lnx −
1
2k + 1
?
.
Proof. The binomial theorem gives
(6.14) (1 + t2)−n−1=
∞
?
k=0
(−1)k
?n + k
k
?
t2k.
The expression for gn(x) now follows by integrating term by term and the evaluation
?x
(6.15)
0
t2klntdt =
?
lnx −
1
2k + 1
?x2k+1
2k + 1.
?
In particular, the initial condition g0(x) of the recurrence (6.9) is given by
g0(x) = lnxtan−1x − L(x),
where
∞
?
The recurrence (6.9) is now solved using (6.2) to produce
(6.16)
(6.17)L(x) =
k=0
(−1)kx2k+1
(2k + 1)2
=
?x
0
tan−1t
t
dt.
gn(x)=2−2n
?2n
n
?
n
?
g0(x)
+
?2n
n
?
22n
j=1
22j
2j?2j
j
? [{2jfj(x) − (2j − 1)fj−1(x)}lnx − fj−1(x)].
Using the recurrence (6.5), this can be written as
(6.18)gn(x) =
?2n
n
?
22n

g0(x) +
n
?
j=1
22j
2j?2j
j
?
?
x lnx
(x2+ 1)j− fj−1(x)
?
.

Page 12

12T. AMDEBERHAN, V. MOLL, J. ROSENBERG, A. STRAUB, AND P. WHITWORTH
Solving the recurrence yields:
Theorem 6.7. Let n ∈ N. Define the rational function
(6.19)pj(x) =
j
?
k=1
22k
2k?2k
k
?
x
(1 + x2)k
Then the integral gn(x) is given by
?x
0
lntdt
(1 + t2)n+1=
?2n
n
?
22n
?
g0(x) + pn(x)lnx −
n
?
k=1
tan−1x + pk−1(x)
2k − 1
?
.
Note 6.8. The special case x = 1 in (6.7) produces
(6.20)
?1
0
lntdt
(t2+ 1)n+1= 2−2n
?2n
n
??
G −
n
?
k=1
π
4+ pk−1(1)
2k − 1
?
.
The values
(6.21)pn(1) =1
2
n
?
j=1
2j
j?2j
j
?
do not admit a closed-form, but they do satisfy the three term recurrence
(6.22)(2n + 1)pn+1(1) − (3n + 1)pn(1) + npn−1(1) = 0.
The reader is invited to verify the expansion
(6.23)
∞
?
k=1
xk
k?2k
k
? =2√x sin−1(√x/2)
∞
?
√4 − x
,
from which it follows that
(6.24)
k=1
2k
k?2k
k
? =π
2.
An alternative derivation. Integration by parts produces
(6.25)
?a
0
lnsds
s2+ b=
1
√blnatan−1a
√b−
1
√b
?a/√b
0
tan−1x
x
dx.
Differentiating n-times with respect to b and using
?j
and
?d
we obtain
?a
?d
db
1
√b
=(−1)j(2j)!
j!22jbj+1/2,
?d
db
?j
1
a2+ b=
(−1)jj!
(a2+ b)j+1,
db
?j
tan−1a
√b
= (−1)j(j − 1)!a
j−1
?
k=0
?2k
k
?
22k+1bk+1/2(a2+ b)j−k,
0
lnsds
(s2+ b)n+1
= lnatan−1
?a
√b
?
Fn(b) +alna
2
n
?
k=1
Fn−k(b)
k
k−1
?
j=0
Fj(b)
(a2+ b)k−j
−Fn(b)
?a
0
tan−1x
x
dx −1
2
n
?
k=1
Fn−k(b)
k
j−1
?
j=0
Fj(b)
?a
0
dt
(t2+ b)k−j

Page 13

LOGARITHMS, RATIONAL AND TRIGONOMETRIC13
with Fj(b) = 2−2jb−j−1/2?2j
the parameter b by b2to produce a cleaner formula.
j
?. The last integral in this expression can be evaluated
using Proposition 6.3 to produce a generalization of Theorem 6.7. We have replaced
Theorem 6.9. Let a, b ∈ R with a > 0 and n ∈ N. Introduce the notation
(6.26)Fj(b) =
?2j
j
?
22jb2j+1.
Then
?a
0
lnsds
(s2+ b2)n+1
=Fn(b)lnatan−1(a/b) +alna
2
n
?
k=1
Fn−k(b)
k
k−1
?
n
?
j=0
Fj(b)
(a2+ b2)k−1
−Fn(b)
?a/b
0
tan−1x
x
dx −1
2tan−1(a/b)


?
k=1
Fn−k(b)
k
k−1
?
j=0
Fj(b)Fk−j−1(b)


−
a
4√b
n
?
k=1
Fn−k(b)
k
k−1
?
j=0
Fj(b)Fk−j−1(b)
k−j−1
r=1
1
rFr(b)(a2+ b2)r.
7. Some trigonometric versions
In this section we provide trigonometric versions of some of the evaluations
provided in the previous sections. Many of these integrals correspond to special
values of the Lobachevsky function defined by
?x
This appears as entry 8.260 in [3] and also as 4.224.4. The change of variables
t = π/2 − x provides a proof of 4.224.1:
?x
The special value
(7.1)L(x) := −
0
lncostdt.
(7.2)
0
lnsintdt = L(π/2 − x) − L(π/2).
(7.3)L(π/2) =
?π/2
0
lnsinxdx = −π
2ln2,
appears as 4.224.3. The change of variables t =π
2− x yields 4.224.6:
(7.4)
?π/2
0
lncosxdx = −π
2ln2.
To establish these evaluations, observe that, by symmetry,
2
?π/2
0
lnsinxdx=
?π/2
?π/2
?π/2
0
lnsinxdx +
?π/2
0
lncosxdx
=
0
ln
?sin2x
ln(sin2x)dx −π
2
?
dx
=
0
2ln2.

Page 14

14 T. AMDEBERHAN, V. MOLL, J. ROSENBERG, A. STRAUB, AND P. WHITWORTH
Now let t = 2x in the last integral to obtain the result.
Combining (7.1) and (7.2) we obtain 4.227.1
?u
The identity (1.6) and the evaluation (5.2) yield the value of 4.227.2:
(7.5)
0
lntanxdx = L(u) + L(π/2 − u) +π
2ln2.
(7.6)
?π/4
0
lntantdt = −G.
Now observe that
(7.7)
?π/4
0
lntantdt =
?π/4
0
lnsintdt −
?π/4
0
lncostdt = −G
and
(7.8)
?π/2
0
lnsintdt =
?π/4
0
lnsintdt +
?π/4
0
lncostdt = −π
2ln2.
Solving this system of equations yields
(7.9)
?π/4
0
lnsintdt = −π
4ln2 −G
2
that appears as 4.224.2 in [3] and
(7.10)
?π/4
0
lncostdt = −π
4ln2 +G
2
that appears as 4.224.5.
We now make use of the identity
(7.11) cosx − sinx =
√2cos(x + π/4)
to obtain
?π/4
0
ln(cosx − sinx)dx=
π
8+
?π/2
0
lncos(x + π/4)
=
π
8ln2 +
?π/2
0
lncostdt −
?π/4
0
lncostdt.
The first integral is given in (7.4) as −π
(7.10). We conclude with the evaluation of 4.225.1
2ln2 and the second one as −π
4ln2 +G
2in
(7.12)
?π/4
0
ln(cosx − sinx)dx = −π
8ln2 −G
2.
A similar analysis produces 4.225.2:
(7.13)
?π/4
0
ln(cosx + sinx)dx = −π
8ln2 +G
2.
These evaluations can be used to produce 4.227.9:
(7.14)
?π/4
0
ln(1 + tanx)dx =π
8ln2,

Page 15

LOGARITHMS, RATIONAL AND TRIGONOMETRIC15
and 4.227.11:
(7.15)
?π/4
0
ln(1 − tanx)dx =π
8ln2 − G.
To prove these formulas, simply write
(7.16)ln(1 ± tanx) = ln(cos±sinx) − lncosx.
The same type of calculations provide verification of 4.227.13
?π/4
and 4.227.14
?π/4
The next example of this type is 4.227.15:
?π/4
This is evaluated by writing the integral as
?π/4
using (7.9) and (7.10).
The evaluation of 4.227.10
?π/2
follows from the integrals evaluated here. Indeed,
?π/2
(7.17)
0
ln(1 + cotx)dx =π
8ln2 + G
(7.18)
0
ln(cotx − 1)dx =π
8ln2.
(7.19)
0
ln(tanx + cotx)dx =π
2ln2,
(7.20)−
0
ln(sinx)dx −
?π/2
0
ln(cosx)dx =π
2ln2,
(7.21)
0
ln(1 + tanx)dx =π
4ln2 + G,
0
ln(1 + tanx)dx=
?π/2
?π/4
2
0
ln(sinx + cosx)dx −
?π/2
?π/2
0
ln(cosx)dx
=2
0
−π
ln(sinx + cosx)dx −
8ln2 +G
2
0
ln(cosx)dx
=
?
?
+π
2ln2,
where we have used (7.4) and (7.13).
The identity (5.5) yields
(7.22)
?π/2
0
lntantdt = 0.
The apparent generalization
(7.23)
?π/2
0
ln(atant)dt =π
2lna,
with a > 0, appears as 4.227.3.
The evaluations (7.9) and (7.10) can be brought back into rational form. The
change of variables t = tan−1u produces from (7.10):
(7.24)−π
4ln2 +G
2= −1
2
?1
0
ln(1 + u2)
1 + u2
du.