In mathematics, the Clausen function, introduced by, is a transcendental, special function of a single variable. It can variously be expressed in the form of a definite integral, a trigonometric series, and various other forms. It is intimately connected with the polylogarithm, inverse tangent integral, polygamma function, Riemann zeta function, Dirichlet eta function, and Dirichlet beta function.
The Clausen function of order 2 – often referred to as the Clausen function, despite being but one of a class of many – is given by the integral:
\operatorname{Cl}2(\varphi)=-\int
\varphi | ||
log\left|2\sin | ||
0 |
x | |
2 |
\right|dx:
In the range
0<\varphi<2\pi
\operatorname{Cl}2(\varphi)=\sum
infty | |
k=1 |
\sink\varphi | |
k2 |
=\sin\varphi+
\sin2\varphi | + | |
22 |
\sin3\varphi | + | |
32 |
\sin4\varphi | |
42 |
+ …
The Clausen functions, as a class of functions, feature extensively in many areas of modern mathematical research, particularly in relation to the evaluation of many classes of logarithmic and polylogarithmic integrals, both definite and indefinite. They also have numerous applications with regard to the summation of hypergeometric series, summations involving the inverse of the central binomial coefficient, sums of the polygamma function, and Dirichlet L-series.
The Clausen function (of order 2) has simple zeros at all (integer) multiples of
\pi,
k\inZ
\sink\pi=0
\operatorname{Cl}2(m\pi)=0, m=0,\pm1,\pm2,\pm3, …
It has maxima at
\theta=
\pi | |
3 |
+2m\pi [m\inZ]
\operatorname{Cl} | ||||
|
+2m\pi\right)=1.01494160\ldots
and minima at
\theta=-
\pi | |
3 |
+2m\pi [m\inZ]
\operatorname{Cl} | ||||
|
+2m\pi\right)=-1.01494160\ldots
The following properties are immediate consequences of the series definition:
\operatorname{Cl}2(\theta+2m\pi)=\operatorname{Cl}2(\theta)
\operatorname{Cl}2(-\theta)=-\operatorname{Cl}2(\theta)
See .
More generally, one defines the two generalized Clausen functions:
\operatorname{S}z(\theta)=
infty | |
\sum | |
k=1 |
\sink\theta | |
kz |
\operatorname{C}z(\theta)=
infty | |
\sum | |
k=1 |
\cosk\theta | |
kz |
which are valid for complex z with Re z >1. The definition may be extended to all of the complex plane through analytic continuation.
When z is replaced with a non-negative integer, the standard Clausen functions are defined by the following Fourier series:
\operatorname{Cl}2m+2(\theta)=
infty | |
\sum | |
k=1 |
\sink\theta | |
k2m+2 |
\operatorname{Cl}2m+1(\theta)=
infty | |
\sum | |
k=1 |
\cosk\theta | |
k2m+1 |
\operatorname{Sl}2m+2(\theta)=
infty | |
\sum | |
k=1 |
\cosk\theta | |
k2m+2 |
\operatorname{Sl}2m+1(\theta)=
infty | |
\sum | |
k=1 |
\sink\theta | |
k2m+1 |
N.B. The SL-type Clausen functions have the alternative notation
\operatorname{Gl}m(\theta)
The SL-type Clausen function are polynomials in
\theta
B2n-1(x)=
2(-1)n(2n-1)! | |
(2\pi)2n-1 |
infty | |
\sum | |
k=1 |
\sin2\pikx | |
k2n-1 |
.
B2n(x)=
2(-1)n-1(2n)! | |
(2\pi)2n |
infty | |
\sum | |
k=1 |
\cos2\pikx | |
k2n |
.
Setting
x=\theta/2\pi
\operatorname{Sl}2m(\theta)=
(-1)m-1(2\pi)2m | |
2(2m)! |
B2m\left(
\theta | |
2\pi |
\right),
\operatorname{Sl}2m-1(\theta)=
(-1)m(2\pi)2m-1 | |
2(2m-1)! |
B2m-1\left(
\theta | |
2\pi |
\right),
Bn(x)
Bn\equivBn(0)
Bn(x)=\sum
n\binom{n}{j} | |
j=0 |
n-j | |
B | |
jx |
.
Explicit evaluations derived from the above include:
\operatorname{Sl}1(\theta)=
\pi | - | |
2 |
\theta | |
2, |
\operatorname{Sl}2(\theta)=
\pi2 | - | |
6 |
\pi\theta | + | |
2 |
\theta2 | |
4 |
,
\operatorname{Sl}3(\theta)=
\pi2\theta | - | |
6 |
\pi\theta2 | + | |
4 |
\theta3 | |
12 |
,
\operatorname{Sl}4(\theta)=
\pi4 | - | |
90 |
\pi2\theta2 | + | |
12 |
\pi\theta3 | - | |
12 |
\theta4 | |
48 |
.
For
0<\theta<\pi
\operatorname{Cl}2(2\theta)=2\operatorname{Cl}2(\theta)-2\operatorname{Cl}2(\pi-\theta)
Denoting Catalan's constant by
K=\operatorname{Cl} | ||||
|
\right)
\operatorname{Cl} | ||||
|
\right)-\operatorname{Cl}2\left(
3\pi | |||
|
2\operatorname{Cl} | ||||
|
\right)=3\operatorname{Cl}2\left(
2\pi | |
3\right) |
For higher order Clausen functions, duplication formulae can be obtained from the one given above; simply replace
\theta
x
[0,\theta].
\operatorname{Cl}3(2\theta)=4\operatorname{Cl}3(\theta)+4\operatorname{Cl}3(\pi-\theta)
\operatorname{Cl}4(2\theta)=8\operatorname{Cl}4(\theta)-8\operatorname{Cl}4(\pi-\theta)
\operatorname{Cl}5(2\theta)=16\operatorname{Cl}5(\theta)+16\operatorname{Cl}5(\pi-\theta)
\operatorname{Cl}6(2\theta)=32\operatorname{Cl}6(\theta)-32\operatorname{Cl}6(\pi-\theta)
And more generally, upon induction on
m, m\ge1
\operatorname{Cl}m+1(2\theta)=
m\left[\operatorname{Cl} | |
2 | |
m+1 |
(\theta)+(-1)m\operatorname{Cl}m+1(\pi-\theta)\right]
Use of the generalized duplication formula allows for an extension of the result for the Clausen function of order 2, involving Catalan's constant. For
m\inZ\ge1
\operatorname{Cl}2m\left(
\pi | |
2 |
\right)=22m-1\left[\operatorname{Cl}2m\left(
\pi | |
4 |
\right)-\operatorname{Cl}2m\left(
3\pi | |
4 |
\right)\right]=\beta(2m)
Where
\beta(x)
From the integral definition,
\operatorname{Cl}2(2\theta)=-\int
2\theta | |
0 |
log\left|2\sin
x | |
2 |
\right|dx
Apply the duplication formula for the sine function,
\sinx=2\sin
x | \cos | |
2 |
x | |
2 |
\begin{align} &
2\theta | |
-\int | |
0 |
log\left|\left(2\sin
x | |
4 |
\right)\left(2\cos
x | |
4 |
\right)\right|dx\\ ={}&
2\theta | |
-\int | |
0 |
log\left|2\sin
x | |
4 |
\right|dx
2\theta | |
-\int | |
0 |
log\left|2\cos
x | |
4 |
\right|dx \end{align}
Apply the substitution
x=2y,dx=2dy
\begin{align} &
\theta | |
-2\int | |
0 |
log\left|2\sin
x | |
2 |
\right|dx
\theta | |
-2\int | |
0 |
log\left|2\cos
x | |
2 |
\right|dx\\ ={}&2\operatorname{Cl}2(\theta)
\theta | |
-2\int | |
0 |
log\left|2\cos
x | |
2 |
\right|dx \end{align}
On that last integral, set
y=\pi-x,x=\pi-y,dx=-dy
\cos(x-y)=\cosx\cosy-\sinx\siny
\begin{align} &\cos\left(
\pi-y | |
2 |
\right)=\sin
y | |
2 |
\\ \Longrightarrow &\operatorname{Cl}2(2\theta)=2\operatorname{Cl}2(\theta)
\theta | |
-2\int | |
0 |
log\left|2\cos
x | |
2 |
\right|dx\\ ={}&2\operatorname{Cl}2(\theta)
\pi-\theta | |
+2\int | |
\pi |
log\left|2\sin
y | |
2 |
\right|dy\\ ={}&2\operatorname{Cl}2(\theta)-2\operatorname{Cl}2(\pi-\theta)+2\operatorname{Cl}2(\pi) \end{align}
\operatorname{Cl}2(\pi)=0
Therefore,
\operatorname{Cl}2(2\theta)=2\operatorname{Cl}2(\theta)-2\operatorname{Cl}2(\pi-\theta).\Box
Direct differentiation of the Fourier series expansions for the Clausen functions give:
d | |
d\theta |
\operatorname{Cl}2m+2(\theta)=
d | |
d\theta |
infty | |
\sum | |
k=1 |
\sink\theta | |
k2m+2 |
infty | |
=\sum | |
k=1 |
\cosk\theta | |
k2m+1 |
=\operatorname{Cl}2m+1(\theta)
d | |
d\theta |
\operatorname{Cl}2m+1(\theta)=
d | |
d\theta |
infty | |
\sum | |
k=1 |
\cosk\theta | |
k2m+1 |
infty | |
=-\sum | |
k=1 |
\sink\theta | |
k2m |
=-\operatorname{Cl}2m(\theta)
d | |
d\theta |
\operatorname{Sl}2m+2(\theta)=
d | |
d\theta |
infty | |
\sum | |
k=1 |
\cosk\theta | |
k2m+2 |
=
infty | |
-\sum | |
k=1 |
\sink\theta | |
k2m+1 |
=-\operatorname{Sl}2m+1(\theta)
d | |
d\theta |
\operatorname{Sl}2m+1(\theta)=
d | |
d\theta |
infty | |
\sum | |
k=1 |
\sink\theta | |
k2m+1 |
infty | |
=\sum | |
k=1 |
\cosk\theta | |
k2m |
=\operatorname{Sl}2m(\theta)
By appealing to the First Fundamental Theorem Of Calculus, we also have:
d | |
d\theta |
\operatorname{Cl}2(\theta)=
d | |
d\theta |
\left[
\theta | |
-\int | |
0 |
log\left|2\sin
x | |
2 |
\right|dx\right]=-log\left|2\sin
\theta | |
2 |
\right|=\operatorname{Cl}1(\theta)
The inverse tangent integral is defined on the interval
0<z<1
\operatorname{Ti}2(z)=\int
z | |
0 |
\tan-1x | |
x |
dx=
infty | |
\sum | |
k=0 |
(-1)k
z2k+1 | |
(2k+1)2 |
It has the following closed form in terms of the Clausen function:
\operatorname{Ti}2(\tan\theta)=\thetalog(\tan\theta)+
1 | |
2 |
\operatorname{Cl}2(2\theta)+
1 | |
2 |
\operatorname{Cl}2(\pi-2\theta)
From the integral definition of the inverse tangent integral, we have
\operatorname{Ti}2(\tan\theta)=
\tan\theta | |
\int | |
0 |
\tan-1x | |
x |
dx
Performing an integration by parts
\tan\theta | |
\int | |
0 |
\tan-1x | |
x |
dx=\tan-1xlogx
\tan\theta | |
| | |
0 |
-
\tan\theta | |
\int | |
0 |
logx | |
1+x2 |
dx=
\thetalog\tan\theta-
\tan\theta | |
\int | |
0 |
logx | |
1+x2 |
dx
Apply the substitution
x=\tany,y=\tan-1x,dy=
dx | |
1+x2 |
\thetalog\tan\theta-
\theta | |
\int | |
0 |
log(\tany)dy
For that last integral, apply the transform :
y=x/2,dy=dx/2
\begin{align} &\thetalog\tan\theta-
1 | |
2 |
2\theta | |
\int | |
0 |
log\left(\tan
x | |
2 |
\right)dx\\[6pt] ={}&\thetalog\tan\theta-
1 | |
2 |
2\theta | ||
\int | log\left( | |
0 |
\sin(x/2) | |
\cos(x/2) |
\right)dx\\[6pt] ={}&\thetalog\tan\theta-
1 | |
2 |
2\theta | ||
\int | log\left( | |
0 |
2\sin(x/2) | |
2\cos(x/2) |
\right)dx\\[6pt] ={}&\thetalog\tan\theta-
1 | |
2 |
2\theta | |
\int | |
0 |
log\left(2\sin
x | |
2 |
\right)dx+
1 | |
2 |
2\theta | |
\int | |
0 |
log\left(2\cos
x | |
2 |
\right)dx\\[6pt] ={}&\thetalog\tan\theta+
1 | |
2 |
\operatorname{Cl}2(2\theta)+
1 | |
2 |
2\theta | |
\int | |
0 |
log\left(2\cos
x | |
2 |
\right)dx. \end{align}
Finally, as with the proof of the Duplication formula, the substitution
x=(\pi-y)
2\theta | |
\int | |
0 |
log\left(2\cos
x | |
2 |
\right)dx=\operatorname{Cl}2(\pi-2\theta)-\operatorname{Cl}2(\pi)=\operatorname{Cl}2(\pi-2\theta)
Thus
\operatorname{Ti}2(\tan\theta)=\thetalog\tan\theta+
1 | |
2 |
\operatorname{Cl}2(2\theta)+
1 | |
2 |
\operatorname{Cl}2(\pi-2\theta).\Box
For real
0<z<1
\operatorname{Cl}2(2\piz)=2\pilog\left(
G(1-z) | |
G(1+z) |
\right)+2\pizlog\left(
\pi | |
\sin\piz |
\right)
Or equivalently
\operatorname{Cl}2(2\piz)=2\pilog\left(
G(1-z) | |
G(z) |
\right)-2\pilog\Gamma(z)+2\pizlog\left(
\pi | |
\sin\piz |
\right)
See .
The Clausen functions represent the real and imaginary parts of the polylogarithm, on the unit circle:
\operatorname{Cl}2m(\theta)=\Im(\operatorname{Li}2m(ei)), m\inZ\ge1
\operatorname{Cl}2m+1(\theta)=\Re(\operatorname{Li}2m+1(ei)), m\inZ\ge0
This is easily seen by appealing to the series definition of the polylogarithm.
\operatorname{Li}n(z)=\sum
infty | |
k=1 |
zk | |
kn |
\Longrightarrow
i\theta | |
\operatorname{Li} | |
n\left(e |
infty | |
\right)=\sum | |
k=1 |
\left(ei\theta\right)k | |
kn |
=
infty | |
\sum | |
k=1 |
eik\theta | |
kn |
By Euler's theorem,
ei\theta=\cos\theta+i\sin\theta
and by de Moivre's Theorem (De Moivre's formula)
(\cos\theta+i\sin\theta)k=\cosk\theta+i\sink\theta ⇒
i\theta | |
\operatorname{Li} | |
n\left(e |
infty | |
\right)=\sum | |
k=1 |
\cosk\theta | |
kn |
+i
infty | |
\sum | |
k=1 |
\sink\theta | |
kn |
Hence
\operatorname{Li}2m\left(ei\theta
infty | |
\right)=\sum | |
k=1 |
\cosk\theta | |
k2m |
+i
infty | |
\sum | |
k=1 |
\sink\theta | |
k2m |
=\operatorname{Sl}2m(\theta)+i\operatorname{Cl}2m(\theta)
\operatorname{Li}2m+1\left(ei\theta
infty | |
\right)=\sum | |
k=1 |
\cosk\theta | |
k2m+1 |
+i
infty | |
\sum | |
k=1 |
\sink\theta | |
k2m+1 |
=\operatorname{Cl}2m+1(\theta)+i\operatorname{Sl}2m+1(\theta)
The Clausen functions are intimately connected to the polygamma function. Indeed, it is possible to express Clausen functions as linear combinations of sine functions and polygamma functions. One such relation is shown here, and proven below:
\operatorname{Cl}2m\left(
q\pi | |
p |
\right)=
1 | |
(2p)2m(2m-1)! |
p | |
\sum | |
j=1 |
\sin\left(\tfrac{qj\pi}{p}\right)\left[\psi2m-1
q\psi | |
\left(\tfrac{j}{2p}\right)+(-1) | |
2m-1 |
\left(\tfrac{j+p}{2p}\right)\right].
An immediate corollary is this equivalent formula in terms of the Hurwitz zeta function:
\operatorname{Cl}2m\left(
q\pi | |
p |
\right)=
1 | |
(2p)2m |
p | |
\sum | |
j=1 |
\sin\left(\tfrac{qj\pi}{p}\right)\left[\zeta\left(2m,\tfrac{j}{2p}\right)+(-1)q\zeta\left(2m,\tfrac{j+p}{2p}\right)\right].
Let
p
q
q/p
0<q/p<1
\operatorname{Cl}2m\left(
q\pi | |
p |
\right)=
infty | |
\sum | |
k=1 |
\sin(kq\pi/p) | |
k2m |
We split this sum into exactly p-parts, so that the first series contains all, and only, those terms congruent to
kp+1,
kp+2,
kp+p
\begin{align} &\operatorname{Cl}2m\left(
q\pi | |
p |
\right)\\ ={}&
infty | |
\sum | |
k=0 |
| ||||||
(kp+1)2m |
+
infty | |
\sum | |
k=0 |
| ||||||
(kp+2)2m |
infty | |
+ \sum | |
k=0 |
| ||||||
(kp+3)2m |
+ … \\ & … +
infty | |
\sum | |
k=0 |
| ||||||
(kp+p-2)2m |
+
infty | |
\sum | |
k=0 |
| ||||||
(kp+p-1)2m |
infty | |
+ \sum | |
k=0 |
| ||||||
(kp+p)2m |
\end{align}
We can index these sums to form a double sum:
\begin{align} &\operatorname{Cl}2m\left(
q\pi | |
p |
\right)=
p | |
\sum | |
j=1 |
\left\{
infty | |
\sum | |
k=0 |
| ||||||
(kp+j)2m |
\right\}\\ ={}&
p | |
\sum | |
j=1 |
1 | |
p2m |
\left\{
infty | |
\sum | |
k=0 |
| ||||||
(k+(j/p))2m |
\right\} \end{align}
Applying the addition formula for the sine function,
\sin(x+y)=\sinx\cosy+\cosx\siny,
\sin\left[(kp+j)
q\pi | \right]=\sin\left(kq\pi+ | |
p |
qj\pi | |
p |
\right)=\sinkq\pi\cos
qj\pi | |
p |
+\coskq\pi\sin
qj\pi | |
p |
\sinm\pi\equiv0, \cosm\pi\equiv(-1)m \Longleftrightarrowm=0,\pm1,\pm2,\pm3,\ldots
\sin\left[(kp+j)
q\pi | |
p |
\right]=(-1)kq\sin
qj\pi | |
p |
Consequently,
\operatorname{Cl}2m\left(
q\pi | |
p |
\right)=
p | |
\sum | |
j=1 |
1 | \sin\left( | |
p2m |
qj\pi | |
p |
\right)\left\{
infty | |
\sum | |
k=0 |
(-1)kq | |
(k+(j/p))2m |
\right\}
To convert the inner sum in the double sum into a non-alternating sum, split in two in parts in exactly the same way as the earlier sum was split into p-parts:
\begin{align} &
infty | |
\sum | |
k=0 |
(-1)kq | |
(k+(j/p))2m |
infty | |
=\sum | |
k=0 |
(-1)(2k)q | |
((2k)+(j/p))2m |
+
infty | |
\sum | |
k=0 |
(-1)(2k+1)q | |
((2k+1)+(j/p))2m |
\\ ={}&
infty | |
\sum | |
k=0 |
1 | |
(2k+(j/p))2m |
+(-1)q
infty | |
\sum | |
k=0 |
1 | |
(2k+1+(j/p))2m |
\\ ={}&
1 | |
2p |
\left[
infty | |
\sum | |
k=0 |
1 | |
(k+(j/2p))2m |
+(-1)q
infty | |
\sum | |
k=0 |
1 | ||||
|
\right] \end{align}
For
m\inZ\ge1
m+1 | |
\psi | |
m(z)=(-1) |
m!
infty | |
\sum | |
k=0 |
1 | |
(k+z)m+1 |
So, in terms of the polygamma function, the previous inner sum becomes:
1 | |
22m(2m-1)! |
\left[\psi2m-1
q\psi | |
\left(\tfrac{j}{2p}\right)+(-1) | |
2m-1 |
\left(\tfrac{j+p}{2p}\right)\right]
Plugging this back into the double sum gives the desired result:
\operatorname{Cl}2m\left(
q\pi | |
p |
\right)=
1 | |
(2p)2m(2m-1)! |
p | |
\sum | |
j=1 |
\sin\left(\tfrac{qj\pi}{p}\right)\left[\psi2m-1
q\psi | |
\left(\tfrac{j}{2p}\right)+(-1) | |
2m-1 |
\left(\tfrac{j+p}{2p}\right)\right]
The generalized logsine integral is defined by:
m | |
l{L}s | |
n |
(\theta)=
\theta | |
-\int | |
0 |
xmlogn-m-1\left|2\sin
x | |
2 |
\right|dx
In this generalized notation, the Clausen function can be expressed in the form:
\operatorname{Cl}2(\theta)=
0 | |
l{L}s | |
2 |
(\theta)
Ernst Kummer and Rogers give the relation
i\theta | |
\operatorname{Li} | |
2(e |
)=\zeta(2)-\theta(2\pi-\theta)/4+i\operatorname{Cl}2(\theta)
valid for
0\leq\theta\leq2\pi
The Lobachevsky function Λ or Л is essentially the same function with a change of variable:
Λ(\theta)=-
\theta | |
\int | |
0 |
log|2\sin(t)|dt=\operatorname{Cl}2(2\theta)/2
though the name "Lobachevsky function" is not quite historically accurate, as Lobachevsky's formulas for hyperbolic volume used the slightly different function
\theta | |
\int | |
0 |
log|\sec(t)|dt=Λ(\theta+\pi/2)+\thetalog2.
For rational values of
\theta/\pi
\theta/\pi=p/q
\sin(n\theta)
\operatorname{Cl}s(\theta)
A series acceleration for the Clausen function is given by
\operatorname{Cl | |
2(\theta)} |
\theta=1-log|\theta|+
infty | |
\sum | |
n=1 |
\zeta(2n) | \left( | |
n(2n+1) |
\theta | |
2\pi |
\right)2n
which holds for
|\theta|<2\pi
\zeta(s)
\operatorname{Cl | |
2(\theta)}{\theta} |
=3-log\left[|\theta|\left(1-
\theta2 | \right)\right] - | |
4\pi2 |
2\pi | |
\theta |
log\left(
2\pi+\theta | |
2\pi-\theta |
\right)
infty | |
+\sum | |
n=1 |
\zeta(2n)-1 | \left( | |
n(2n+1) |
\theta | |
2\pi |
\right)2n.
Convergence is aided by the fact that
\zeta(n)-1
Recall the Barnes G-function, the Catalan's constant K and the Gieseking constant V. Some special values include
\operatorname{Cl} | ||||
|
\right)=K
\operatorname{Cl} | ||||
|
\right)=V
\operatorname{Cl} | ||||
|
\right)=3\pilog\left(
| |||||
|
\right)-3\pilog\Gamma\left(
1 | |
3 |
\right)+\pilog\left(
2\pi | |
\sqrt{3 |
\operatorname{Cl} | ||||
|
\right)=2\pilog\left(
| |||||
|
\right)-2\pilog\Gamma\left(
1 | |
3 |
\right)+
2\pi | |
3 |
log\left(
2\pi | |
\sqrt{3 |
\operatorname{Cl} | ||||
|
\right)= 2\pilog\left(
| |||||
|
\right)-2\pilog\Gamma\left(
1 | \right)+ | |
8 |
\pi | |
4 |
log\left(
2\pi | |
\sqrt{2-\sqrt{2 |
\operatorname{Cl} | ||||
|
\right)= 2\pilog\left(
| |||||
|
\right)-2\pilog\Gamma\left(
3 | \right)+ | |
8 |
3\pi | |
4 |
log\left(
2\pi | |
\sqrt{2+\sqrt{2 |
\operatorname{Cl} | ||||
|
\right)= 2\pilog\left(
| |||||
|
\right)-2\pilog\Gamma\left(
1 | \right)+ | |
12 |
\pi | |
6 |
log\left(
2\pi\sqrt{2 | |
}{\sqrt{3}-1} |
\right)
\operatorname{Cl} | ||||
|
\right)= 2\pilog\left(
| |||||
|
\right)-2\pilog\Gamma\left(
5 | \right)+ | |
12 |
5\pi | |
6 |
log\left(
2\pi\sqrt{2 | |
}{\sqrt{3}+1} |
\right)
In general, from the Barnes G-function reflection formula,
\operatorname{Cl}2(2\piz)=2\pilog\left(
G(1-z) | |
G(z) |
\right)-2\pilog\Gamma(z)+2\pizlog\left(
\pi | |
\sin\piz |
\right)
Equivalently, using Euler's reflection formula for the gamma function, then,
\operatorname{Cl}2(2\piz)=2\pilog\left(
G(1-z) | |
G(z) |
\right)-2\pilog\Gamma(z)+2\pizlog(\Gamma(z)\Gamma(1-z))
Some special values for higher order Clausen functions include
\operatorname{Cl}2m(0)=\operatorname{Cl}2m(\pi)=\operatorname{Cl}2m(2\pi)=0
\operatorname{Cl}2m\left(
\pi | |
2 |
\right)=\beta(2m)
\operatorname{Cl}2m+1(0)=\operatorname{Cl}2m+1(2\pi)=\zeta(2m+1)
\operatorname{Cl}2m+1(\pi)=-η(2m+1)=-\left(
22m-1 | |
22m |
\right)\zeta(2m+1)
\operatorname{Cl}2m+1\left(
\pi | \right)=- | |
2 |
1 | η(2m+1)=-\left( | |
22m+1 |
22m-1 | |
24m+1 |
\right)\zeta(2m+1)
where
\beta(x)
η(x)
\zeta(x)
The following integrals are easily proven from the series representations of the Clausen function:
\theta | |
\int | |
0 |
\operatorname{Cl}2m(x)dx=\zeta(2m+1)-\operatorname{Cl}2m+1(\theta)
\theta | |
\int | |
0 |
\operatorname{Cl}2m+1(x)dx=\operatorname{Cl}2m+2(\theta)
\theta | |
\int | |
0 |
\operatorname{Sl}2m(x)dx=\operatorname{Sl}2m+1(\theta)
\theta | |
\int | |
0 |
\operatorname{Sl}2m+1(x)dx=\zeta(2m+2)-\operatorname{Cl}2m+2(\theta)
Fourier-analytic methods can be used to find the first moments of the square of the function
\operatorname{Cl}2(x)
[0,\pi]
\pi | |
\int | |
0 |
2(x)dx=\zeta(4), | |
\operatorname{Cl} | |
2 |
\pi | |
\int | |
0 |
| ||||
t\operatorname{Cl} | ||||
2 |
\pi6-4\zeta(\overline{5},1)-2\zeta(\overline{4},2),
\pi | |
\int | |
0 |
| ||||
t | ||||
2 |
\pi\left[12\zeta(\overline{5},1)+6\zeta(\overline{4},2)-
23 | |
10080 |
\pi6\right].
Here
\zeta
A large number of trigonometric and logarithmo-trigonometric integrals can be evaluated in terms of the Clausen function, and various common mathematical constants like
K
log2
\zeta(2)
\zeta(3)
The examples listed below follow directly from the integral representation of the Clausen function, and the proofs require little more than basic trigonometry, integration by parts, and occasional term-by-term integration of the Fourier series definitions of the Clausen functions.
\theta | |
\int | |
0 |
log(\sinx)dx=-\tfrac{1}{2}\operatorname{Cl}2(2\theta)-\thetalog2
\theta | |
\int | |
0 |
log(\cosx)dx=\tfrac{1}{2}\operatorname{Cl}2(\pi-2\theta)-\thetalog2
\theta | |
\int | |
0 |
log(\tanx)dx=-\tfrac{1}{2}\operatorname{Cl}2(2\theta)-\tfrac{1}{2}\operatorname{Cl}2(\pi-2\theta)
\theta | |
\int | |
0 |
log(1+\cosx)dx=2\operatorname{Cl}2(\pi-\theta)-\thetalog2
\theta | |
\int | |
0 |
log(1-\cosx)dx=-2\operatorname{Cl}2(\theta)-\thetalog2
\theta | |
\int | |
0 |
log(1+\sin
x)dx=2K-2\operatorname{Cl} | ||||
|
+\theta\right)-\thetalog2
\theta | |
\int | |
0 |
log(1-\sin
x)dx=-2K+2\operatorname{Cl} | ||||
|
-\theta\right)-\thetalog2