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Clausen function

Inmathematics, the Clausen function, introduced by Thomas Clausen (1832), 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:

${\displaystyle \operatorname {Cl} _{2}(\varphi )=-\int _{0}^{\varphi }\log \left|2\sin {\frac {x}{2}}\right|\,dx:}$

In the range ${\displaystyle 0<\varphi <2\pi \,}$ the sine function inside the absolute value sign remains strictly positive, so the absolute value signs may be omitted. The Clausen function also has the Fourier series representation:

${\displaystyle \operatorname {Cl} _{2}(\varphi )=\sum _{k=1}^{\infty }{\frac {\sin k\varphi }{k^{2}}}=\sin \varphi +{\frac {\sin 2\varphi }{2^{2}}}+{\frac {\sin 3\varphi }{3^{2}}}+{\frac {\sin 4\varphi }{4^{2}}}+\cdots }$

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.

Basic properties[edit]

The Clausen function (of order 2) has simple zeros at all (integer) multiples of ${\displaystyle \pi ,\,}$ since if ${\displaystyle k\in \mathbb {Z} \,}$ is an integer, then ${\displaystyle \sin k\pi =0}$

${\displaystyle \operatorname {Cl} _{2}(m\pi )=0,\quad m=0,\,\pm 1,\,\pm 2,\,\pm 3,\,\cdots }$

It has maxima at ${\displaystyle \theta ={\frac {\pi }{3}}+2m\pi \quad [m\in \mathbb {Z} ]}$

${\displaystyle \operatorname {Cl} _{2}\left({\frac {\pi }{3}}+2m\pi \right)=1.01494160\ldots }$

and minima at ${\displaystyle \theta =-{\frac {\pi }{3}}+2m\pi \quad [m\in \mathbb {Z} ]}$

${\displaystyle \operatorname {Cl} _{2}\left(-{\frac {\pi }{3}}+2m\pi \right)=-1.01494160\ldots }$

The following properties are immediate consequences of the series definition:

${\displaystyle \operatorname {Cl} _{2}(\theta +2m\pi )=\operatorname {Cl} _{2}(\theta )}$

${\displaystyle \operatorname {Cl} _{2}(-\theta )=-\operatorname {Cl} _{2}(\theta )}$

See Lu & Perez (1992).

General definition[edit]

Standard Clausen functions

More generally, one defines the two generalized Clausen functions:

${\displaystyle \operatorname {S} _{z}(\theta )=\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{z}}}}$

${\displaystyle \operatorname {C} _{z}(\theta )=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{z}}}}$

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:

${\displaystyle \operatorname {Cl} _{2m+2}(\theta )=\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{2m+2}}}}$

${\displaystyle \operatorname {Cl} _{2m+1}(\theta )=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{2m+1}}}}$

${\displaystyle \operatorname {Sl} _{2m+2}(\theta )=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{2m+2}}}}$

${\displaystyle \operatorname {Sl} _{2m+1}(\theta )=\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{2m+1}}}}$

N.B. The SL-type Clausen functions have the alternative notation ${\displaystyle \operatorname {Gl} _{m}(\theta )\,}$ and are sometimes referred to as the Glaisher–Clausen functions (after James Whitbread Lee Glaisher, hence the GL-notation).

Relation to the Bernoulli polynomials[edit]

The SL-type Clausen function are polynomials in ${\displaystyle \,\theta \,}$, and are closely related to the Bernoulli polynomials. This connection is apparent from the Fourier series representations of the Bernoulli polynomials:

${\displaystyle B_{2n-1}($x)={\frac {2(-1)^{n}(2n-1)!}{(2\pi )^{2n-1}}}\,\sum _{k=1}^{\infty }{\frac {\sin 2\pi kx}{k^{2n-1}}}.}

${\displaystyle B_{2n}($x)={\frac {2(-1)^{n-1}(2n)!}{(2\pi )^{2n}}}\,\sum _{k=1}^{\infty }{\frac {\cos 2\pi kx}{k^{2n}}}.}

Setting ${\displaystyle \,x=\theta /2\pi \,}$ in the above, and then rearranging the terms gives the following closed form (polynomial) expressions:

${\displaystyle \operatorname {Sl} _{2m}(\theta )={\frac {(-1)^{m-1}(2\pi )^{2m}}{2(2m)!}}B_{2m}\left({\frac {\theta }{2\pi }}\right),}$

${\displaystyle \operatorname {Sl} _{2m-1}(\theta )={\frac {(-1)^{m}(2\pi )^{2m-1}}{2(2m-1)!}}B_{2m-1}\left({\frac {\theta }{2\pi }}\right),}$

where the Bernoulli polynomials ${\displaystyle \,B_{n}($x)\,} are defined in terms of the Bernoulli numbers ${\displaystyle \,B_{n}\equiv B_{n}(0)\,}$ by the relation:

${\displaystyle B_{n}($x)=\sum _{j=0}^{n}{\binom {n}{j}}B_{j}x^{n-j}.}

Explicit evaluations derived from the above include:

${\displaystyle \operatorname {Sl} _{1}(\theta )={\frac {\pi }{2}}-{\frac {\theta }{2}},}$

${\displaystyle \operatorname {Sl} _{2}(\theta )={\frac {\pi ^{2}}{6}}-{\frac {\pi \theta }{2}}+{\frac {\theta ^{2}}{4}},}$

${\displaystyle \operatorname {Sl} _{3}(\theta )={\frac {\pi ^{2}\theta }{6}}-{\frac {\pi \theta ^{2}}{4}}+{\frac {\theta ^{3}}{12}},}$

${\displaystyle \operatorname {Sl} _{4}(\theta )={\frac {\pi ^{4}}{90}}-{\frac {\pi ^{2}\theta ^{2}}{12}}+{\frac {\pi \theta ^{3}}{12}}-{\frac {\theta ^{4}}{48}}.}$

Duplication formula[edit]

For ${\displaystyle 0<\theta <\pi }$, the duplication formula can be proven directly from the integral definition (see also Lu & Perez (1992) for the result – although no proof is given):

${\displaystyle \operatorname {Cl} _{2}(2\theta )=2\operatorname {Cl} _{2}(\theta )-2\operatorname {Cl} _{2}(\pi -\theta )}$

Denoting Catalan's constantby${\displaystyle K=\operatorname {Cl} _{2}\left({\frac {\pi }{2}}\right)}$, immediate consequences of the duplication formula include the relations:

${\displaystyle \operatorname {Cl} _{2}\left({\frac {\pi }{4}}\right)-\operatorname {Cl} _{2}\left({\frac {3\pi }{4}}\right)={\frac {K}{2}}}$

${\displaystyle 2\operatorname {Cl} _{2}\left({\frac {\pi }{3}}\right)=3\operatorname {Cl} _{2}\left({\frac {2\pi }{3}}\right)}$

For higher order Clausen functions, duplication formulae can be obtained from the one given above; simply replace ${\displaystyle \,\theta \,}$ with the dummy variable ${\displaystyle x}$, and integrate over the interval ${\displaystyle \,[0,\theta ].\,}$ Applying the same process repeatedly yields:

${\displaystyle \operatorname {Cl} _{3}(2\theta )=4\operatorname {Cl} _{3}(\theta )+4\operatorname {Cl} _{3}(\pi -\theta )}$

${\displaystyle \operatorname {Cl} _{4}(2\theta )=8\operatorname {Cl} _{4}(\theta )-8\operatorname {Cl} _{4}(\pi -\theta )}$

${\displaystyle \operatorname {Cl} _{5}(2\theta )=16\operatorname {Cl} _{5}(\theta )+16\operatorname {Cl} _{5}(\pi -\theta )}$

${\displaystyle \operatorname {Cl} _{6}(2\theta )=32\operatorname {Cl} _{6}(\theta )-32\operatorname {Cl} _{6}(\pi -\theta )}$

And more generally, upon induction on ${\displaystyle \,m,\;m\geq 1}$

${\displaystyle \operatorname {Cl} _{m+1}(2\theta )=2^{m}\left[\operatorname {Cl} _{m+1}(\theta )+(-1)^{m}\operatorname {Cl} _{m+1}(\pi -\theta )\right]}$

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 ${\displaystyle \,m\in \mathbb {Z} \geq 1\,}$

${\displaystyle \operatorname {Cl} _{2m}\left({\frac {\pi }{2}}\right)=2^{2m-1}\left[\operatorname {Cl} _{2m}\left({\frac {\pi }{4}}\right)-\operatorname {Cl} _{2m}\left({\frac {3\pi }{4}}\right)\right]=\beta (2m)}$

Where ${\displaystyle \,\beta ($x)\,} is the Dirichlet beta function.

Proof of the duplication formula[edit]

From the integral definition,

${\displaystyle \operatorname {Cl} _{2}(2\theta )=-\int _{0}^{2\theta }\log \left|2\sin {\frac {x}{2}}\right|\,dx}$

Apply the duplication formula for the sine function, ${\displaystyle \sin x=2\sin {\frac {x}{2}}\cos {\frac {x}{2}}}$ to obtain

${\displaystyle {\begin{aligned}&-\int _{0}^{2\theta }\log \left|\left(2\sin {\frac {x}{4}}\right)\left(2\cos {\frac {x}{4}}\right)\right|\,dx\\={}&-\int _{0}^{2\theta }\log \left|2\sin {\frac {x}{4}}\right|\,dx-\int _{0}^{2\theta }\log \left|2\cos {\frac {x}{4}}\right|\,dx\end{aligned}}}$

Apply the substitution ${\displaystyle x=2y,dx=2\,dy}$ on both integrals:

${\displaystyle {\begin{aligned}&-2\int _{0}^{\theta }\log \left|2\sin {\frac {x}{2}}\right|\,dx-2\int _{0}^{\theta }\log \left|2\cos {\frac {x}{2}}\right|\,dx\\={}&2\,\operatorname {Cl} _{2}(\theta )-2\int _{0}^{\theta }\log \left|2\cos {\frac {x}{2}}\right|\,dx\end{aligned}}}$

On that last integral, set ${\displaystyle y=\pi -x,\,x=\pi -y,\,dx=-dy}$, and use the trigonometric identity ${\displaystyle \cos(x-y)=\cos x\cos y-\sin x\sin y}$ to show that:

${\displaystyle {\begin{aligned}&\cos \left({\frac {\pi -y}{2}}\right)=\sin {\frac {y}{2}}\\\Longrightarrow \qquad &\operatorname {Cl} _{2}(2\theta )=2\,\operatorname {Cl} _{2}(\theta )-2\int _{0}^{\theta }\log \left|2\cos {\frac {x}{2}}\right|\,dx\\={}&2\,\operatorname {Cl} _{2}(\theta )+2\int _{\pi }^{\pi -\theta }\log \left|2\sin {\frac {y}{2}}\right|\,dy\\={}&2\,\operatorname {Cl} _{2}(\theta )-2\,\operatorname {Cl} _{2}(\pi -\theta )+2\,\operatorname {Cl} _{2}(\pi )\end{aligned}}}$

${\displaystyle \operatorname {Cl} _{2}(\pi )=0\,}$

Therefore,

${\displaystyle \operatorname {Cl} _{2}(2\theta )=2\,\operatorname {Cl} _{2}(\theta )-2\,\operatorname {Cl} _{2}(\pi -\theta )\,.\,\Box }$

Derivatives of general-order Clausen functions[edit]

Direct differentiation of the Fourier series expansions for the Clausen functions give:

${\displaystyle {\frac {d}{d\theta }}\operatorname {Cl} _{2m+2}(\theta )={\frac {d}{d\theta }}\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{2m+2}}}=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{2m+1}}}=\operatorname {Cl} _{2m+1}(\theta )}$

${\displaystyle {\frac {d}{d\theta }}\operatorname {Cl} _{2m+1}(\theta )={\frac {d}{d\theta }}\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{2m+1}}}=-\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{2m}}}=-\operatorname {Cl} _{2m}(\theta )}$

${\displaystyle {\frac {d}{d\theta }}\operatorname {Sl} _{2m+2}(\theta )={\frac {d}{d\theta }}\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{2m+2}}}=-\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{2m+1}}}=-\operatorname {Sl} _{2m+1}(\theta )}$

${\displaystyle {\frac {d}{d\theta }}\operatorname {Sl} _{2m+1}(\theta )={\frac {d}{d\theta }}\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{2m+1}}}=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{2m}}}=\operatorname {Sl} _{2m}(\theta )}$

By appealing to the First Fundamental Theorem Of Calculus, we also have:

${\displaystyle {\frac {d}{d\theta }}\operatorname {Cl} _{2}(\theta )={\frac {d}{d\theta }}\left[-\int _{0}^{\theta }\log \left|2\sin {\frac {x}{2}}\right|\,dx\,\right]=-\log \left|2\sin {\frac {\theta }{2}}\right|=\operatorname {Cl} _{1}(\theta )}$

Relation to the inverse tangent integral[edit]

The inverse tangent integral is defined on the interval ${\displaystyle 0<z<1}$by

${\displaystyle \operatorname {Ti} _{2}($z)=\int _{0}^{z}{\frac {\tan ^{-1}x}{x}}\,dx=\sum _{k=0}^{\infty }(-1)^{k}{\frac {z^{2k+1}}{(2k+1)^{2}}}}

It has the following closed form in terms of the Clausen function:

${\displaystyle \operatorname {Ti} _{2}(\tan \theta )=\theta \log(\tan \theta )+{\frac {1}{2}}\operatorname {Cl} _{2}(2\theta )+{\frac {1}{2}}\operatorname {Cl} _{2}(\pi -2\theta )}$

Proof of the inverse tangent integral relation[edit]

From the integral definition of the inverse tangent integral, we have

${\displaystyle \operatorname {Ti} _{2}(\tan \theta )=\int _{0}^{\tan \theta }{\frac {\tan ^{-1}x}{x}}\,dx}$

Performing an integration by parts

${\displaystyle \int _{0}^{\tan \theta }{\frac {\tan ^{-1}x}{x}}\,dx=\tan ^{-1}x\log x\,{\Bigg |}_{0}^{\tan \theta }-\int _{0}^{\tan \theta }{\frac {\log x}{1+x^{2}}}\,dx=}$

${\displaystyle \theta \log \tan \theta -\int _{0}^{\tan \theta }{\frac {\log x}{1+x^{2}}}\,dx}$

Apply the substitution ${\displaystyle x=\tan y,\,y=\tan ^{-1}x,\,dy={\frac {dx}{1+x^{2}}}\,}$ to obtain

${\displaystyle \theta \log \tan \theta -\int _{0}^{\theta }\log(\tan y)\,dy}$

For that last integral, apply the transform :${\displaystyle y=x/2,\,dy=dx/2\,}$ to get

${\displaystyle {\begin{aligned}&\theta \log \tan \theta -{\frac {1}{2}}\int _{0}^{2\theta }\log \left(\tan {\frac {x}{2}}\right)\,dx\\[6pt]={}&\theta \log \tan \theta -{\frac {1}{2}}\int _{0}^{2\theta }\log \left({\frac {\sin(x/2)}{\cos(x/2)}}\right)\,dx\\[6pt]={}&\theta \log \tan \theta -{\frac {1}{2}}\int _{0}^{2\theta }\log \left({\frac {2\sin(x/2)}{2\cos(x/2)}}\right)\,dx\\[6pt]={}&\theta \log \tan \theta -{\frac {1}{2}}\int _{0}^{2\theta }\log \left(2\sin {\frac {x}{2}}\right)\,dx+{\frac {1}{2}}\int _{0}^{2\theta }\log \left(2\cos {\frac {x}{2}}\right)\,dx\\[6pt]={}&\theta \log \tan \theta +{\frac {1}{2}}\operatorname {Cl} _{2}(2\theta )+{\frac {1}{2}}\int _{0}^{2\theta }\log \left(2\cos {\frac {x}{2}}\right)\,dx.\end{aligned}}}$

Finally, as with the proof of the Duplication formula, the substitution ${\displaystyle x=(\pi -y)\,}$ reduces that last integral to

${\displaystyle \int _{0}^{2\theta }\log \left(2\cos {\frac {x}{2}}\right)\,dx=\operatorname {Cl} _{2}(\pi -2\theta )-\operatorname {Cl} _{2}(\pi )=\operatorname {Cl} _{2}(\pi -2\theta )}$

Thus

${\displaystyle \operatorname {Ti} _{2}(\tan \theta )=\theta \log \tan \theta +{\frac {1}{2}}\operatorname {Cl} _{2}(2\theta )+{\frac {1}{2}}\operatorname {Cl} _{2}(\pi -2\theta )\,.\,\Box }$

Relation to the Barnes' G-function[edit]

For real ${\displaystyle 0<z<1}$, the Clausen function of second order can be expressed in terms of the Barnes G-function and (Euler) Gamma function:

${\displaystyle \operatorname {Cl} _{2}(2\pi z)=2\pi \log \left({\frac {G(1-z)}{G(1+z)}}\right)+2\pi z\log \left({\frac {\pi }{\sin \pi z}}\right)}$

Or equivalently

${\displaystyle \operatorname {Cl} _{2}(2\pi z)=2\pi \log \left({\frac {G(1-z)}{G($z)}}\right)-2\pi \log \Gamma (z)+2\pi z\log \left({\frac {\pi }{\sin \pi z}}\right)}

See Adamchik (2003).

Relation to the polylogarithm[edit]

The Clausen functions represent the real and imaginary parts of the polylogarithm, on the unit circle:

${\displaystyle \operatorname {Cl} _{2m}(\theta )=\Im (\operatorname {Li} _{2m}(e^{i\theta })),\quad m\in \mathbb {Z} \geq 1}$

${\displaystyle \operatorname {Cl} _{2m+1}(\theta )=\Re (\operatorname {Li} _{2m+1}(e^{i\theta })),\quad m\in \mathbb {Z} \geq 0}$

This is easily seen by appealing to the series definition of the polylogarithm.

${\displaystyle \operatorname {Li} _{n}($z)=\sum _{k=1}^{\infty }{\frac {z^{k}}{k^{n}}}\quad \Longrightarrow \operatorname {Li} _{n}\left(e^{i\theta }\right)=\sum _{k=1}^{\infty }{\frac {\left(e^{i\theta }\right)^{k}}{k^{n}}}=\sum _{k=1}^{\infty }{\frac {e^{ik\theta }}{k^{n}}}}

By Euler's theorem,

${\displaystyle e^{i\theta }=\cos \theta +i\sin \theta }$

and by de Moivre's Theorem (De Moivre's formula)

${\displaystyle (\cos \theta +i\sin \theta )^{k}=\cos k\theta +i\sin k\theta \quad \Rightarrow \operatorname {Li} _{n}\left(e^{i\theta }\right)=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{n}}}+i\,\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{n}}}}$

Hence

${\displaystyle \operatorname {Li} _{2m}\left(e^{i\theta }\right)=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{2m}}}+i\,\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{2m}}}=\operatorname {Sl} _{2m}(\theta )+i\operatorname {Cl} _{2m}(\theta )}$

${\displaystyle \operatorname {Li} _{2m+1}\left(e^{i\theta }\right)=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{2m+1}}}+i\,\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{2m+1}}}=\operatorname {Cl} _{2m+1}(\theta )+i\operatorname {Sl} _{2m+1}(\theta )}$

Relation to the polygamma function[edit]

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:

${\displaystyle \operatorname {Cl} _{2m}\left({\frac {q\pi }{p}}\right)={\frac {1}{(2p)^{2m}(2m-1)!}}\,\sum _{j=1}^{p}\sin \left({\tfrac {qj\pi }{p}}\right)\,\left[\psi _{2m-1}\left({\tfrac {j}{2p}}\right)+(-1)^{q}\psi _{2m-1}\left({\tfrac {j+p}{2p}}\right)\right].}$