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Clausen function
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<h2 id="basic-properties">Basic properties</h2> <p>The Clausen function (of order 2) has simple zeros at all (integer) multiples of π , {\displaystyle \pi ,\,} since if k ∈ Z {\displaystyle k\in \mathbb {Z} \,} is an integer, then sin k π = 0 {\displaystyle \sin k\pi =0} </p> Cl 2 ( m π ) = 0 , m = 0 , ± 1 , ± 2 , ± 3 , ⋯ {\displaystyle \operatorname {Cl} _{2}(m\pi )=0,\quad m=0,\,\pm 1,\,\pm 2,\,\pm 3,\,\cdots } <p>It has maxima at θ = π 3 + 2 m π [ m ∈ Z ] {\displaystyle \theta ={\frac {\pi }{3}}+2m\pi \quad [m\in \mathbb {Z} ]} </p> Cl 2 ( π 3 + 2 m π ) = 1.01494160 … {\displaystyle \operatorname {Cl} _{2}\left({\frac {\pi }{3}}+2m\pi \right)=1.01494160\ldots } <p>and minima at θ = − π 3 + 2 m π [ m ∈ Z ] {\displaystyle \theta =-{\frac {\pi }{3}}+2m\pi \quad [m\in \mathbb {Z} ]} </p> Cl 2 ( − π 3 + 2 m π ) = − 1.01494160 … {\displaystyle \operatorname {Cl} _{2}\left(-{\frac {\pi }{3}}+2m\pi \right)=-1.01494160\ldots } <p>The following properties are immediate consequences of the series definition: </p> Cl 2 ( θ + 2 m π ) = Cl 2 ( θ ) {\displaystyle \operatorname {Cl} _{2}(\theta +2m\pi )=\operatorname {Cl} _{2}(\theta )} Cl 2 ( − θ ) = − Cl 2 ( θ ) {\displaystyle \operatorname {Cl} _{2}(-\theta )=-\operatorname {Cl} _{2}(\theta )} <p>See Lu & Perez (1992). </p> <h2 id="general-definition">General definition</h2> <p>More generally, one defines the two generalized Clausen functions: </p> S z ( θ ) = ∑ k = 1 ∞ sin k θ k z {\displaystyle \operatorname {S} _{z}(\theta )=\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{z}}}} C z ( θ ) = ∑ k = 1 ∞ cos k θ k z {\displaystyle \operatorname {C} _{z}(\theta )=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{z}}}} <p>which are valid for complex <i>z</i> with Re <i>z</i> >1. The definition may be extended to all of the complex plane through <a href="/facts/Analytic_continuation/aBCuEgvt">analytic continuation</a>. </p><p>When <i>z</i> is replaced with a non-negative integer, the standard Clausen functions are defined by the following <a href="/facts/Fourier_series/s3fsxPAT">Fourier series</a>: </p> Cl 2 m + 2 ( θ ) = ∑ k = 1 ∞ sin k θ k 2 m + 2 {\displaystyle \operatorname {Cl} _{2m+2}(\theta )=\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{2m+2}}}} Cl 2 m + 1 ( θ ) = ∑ k = 1 ∞ cos k θ k 2 m + 1 {\displaystyle \operatorname {Cl} _{2m+1}(\theta )=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{2m+1}}}} Sl 2 m + 2 ( θ ) = ∑ k = 1 ∞ cos k θ k 2 m + 2 {\displaystyle \operatorname {Sl} _{2m+2}(\theta )=\sum _{k=1}^{\infty }{\frac {\cos k\theta }{k^{2m+2}}}} Sl 2 m + 1 ( θ ) = ∑ k = 1 ∞ sin k θ k 2 m + 1 {\displaystyle \operatorname {Sl} _{2m+1}(\theta )=\sum _{k=1}^{\infty }{\frac {\sin k\theta }{k^{2m+1}}}} <p>N.B. The SL-type Clausen functions have the alternative notation Gl m ( θ ) {\displaystyle \operatorname {Gl} _{m}(\theta )\,} and are sometimes referred to as the Glaisher–Clausen functions (after <a href="/facts/James_Whitbread_Lee_Glaisher/VUhEg8Jk">James Whitbread Lee Glaisher</a>, hence the GL-notation). </p> <h2 id="relation-to-the-bernoulli-polynomials">Relation to the Bernoulli polynomials</h2> <p>The SL-type Clausen function are polynomials in θ {\displaystyle \,\theta \,} , and are closely related to the <a href="/facts/Bernoulli_polynomials/o3JBxUTN">Bernoulli polynomials</a>. This connection is apparent from the <a href="/facts/Fourier_series/s3fsxPAT">Fourier series</a> representations of the Bernoulli polynomials: </p> B 2 n − 1 ( x ) = 2 ( − 1 ) n ( 2 n − 1 ) ! ( 2 π ) 2 n − 1 ∑ k = 1 ∞ sin 2 π k x k 2 n − 1 . {\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}}}.} B 2 n ( x ) = 2 ( − 1 ) n − 1 ( 2 n ) ! ( 2 π ) 2 n ∑ k = 1 ∞ cos 2 π k x k 2 n . {\displaystyle B_{2n}(x)={\frac {2(-1)^{n-1}(2n)!}{(2\pi )^{2n}}}\,\sum _{k=1}^{\infty }{\frac {\cos 2\pi kx}{k^{2n}}}.} <p>Setting x = θ / 2 π {\displaystyle \,x=\theta /2\pi \,} in the above, and then rearranging the terms gives the following closed form (polynomial) expressions: </p> Sl 2 m ( θ ) = ( − 1 ) m − 1 ( 2 π ) 2 m 2 ( 2 m ) ! B 2 m ( θ 2 π ) , {\displaystyle \operatorname {Sl} _{2m}(\theta )={\frac {(-1)^{m-1}(2\pi )^{2m}}{2(2m)!}}B_{2m}\left({\frac {\theta }{2\pi }}\right),} Sl 2 m − 1 ( θ ) = ( − 1 ) m ( 2 π ) 2 m − 1 2 ( 2 m − 1 ) ! B 2 m − 1 ( θ 2 π ) , {\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),} <p>where the <a href="/facts/Bernoulli_polynomials/o3JBxUTN">Bernoulli polynomials</a> B n ( x ) {\displaystyle \,B_{n}(x)\,} are defined in terms of the <a href="/facts/Bernoulli_numbers/RH0mblhs">Bernoulli numbers</a> B n ≡ B n ( 0 ) {\displaystyle \,B_{n}\equiv B_{n}(0)\,} by the relation: </p> B n ( x ) = ∑ j = 0 n ( n j ) B j x n − j . {\displaystyle B_{n}(x)=\sum _{j=0}^{n}{\binom {n}{j}}B_{j}x^{n-j}.} <p>Explicit evaluations derived from the above include: </p> Sl 1 ( θ ) = π 2 − θ 2 , {\displaystyle \operatorname {Sl} _{1}(\theta )={\frac {\pi }{2}}-{\frac {\theta }{2}},} Sl 2 ( θ ) = π 2 6 − π θ 2 + θ 2 4 , {\displaystyle \operatorname {Sl} _{2}(\theta )={\frac {\pi ^{2}}{6}}-{\frac {\pi \theta }{2}}+{\frac {\theta ^{2}}{4}},} Sl 3 ( θ ) = π 2 θ 6 − π θ 2 4 + θ 3 12 , {\displaystyle \operatorname {Sl} _{3}(\theta )={\frac {\pi ^{2}\theta }{6}}-{\frac {\pi \theta ^{2}}{4}}+{\frac {\theta ^{3}}{12}},} Sl 4 ( θ ) = π 4 90 − π 2 θ 2 12 + π θ 3 12 − θ 4 48 . {\displaystyle \operatorname {Sl} _{4}(\theta )={\frac {\pi ^{4}}{90}}-{\frac {\pi ^{2}\theta ^{2}}{12}}+{\frac {\pi \theta ^{3}}{12}}-{\frac {\theta ^{4}}{48}}.} <h2 id="duplication-formula">Duplication formula</h2> <p>For 0 < θ < π {\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): </p> Cl 2 ( 2 θ ) = 2 Cl 2 ( θ ) − 2 Cl 2 ( π − θ ) {\displaystyle \operatorname {Cl} _{2}(2\theta )=2\operatorname {Cl} _{2}(\theta )-2\operatorname {Cl} _{2}(\pi -\theta )} <p>Denoting <a href="/facts/Catalan%2527s_constant/c9fNqG8j">Catalan's constant</a> by K = Cl 2 ( π 2 ) {\displaystyle K=\operatorname {Cl} _{2}\left({\frac {\pi }{2}}\right)} , immediate consequences of the duplication formula include the relations: </p> Cl 2 ( π 4 ) − Cl 2 ( 3 π 4 ) = K 2 {\displaystyle \operatorname {Cl} _{2}\left({\frac {\pi }{4}}\right)-\operatorname {Cl} _{2}\left({\frac {3\pi }{4}}\right)={\frac {K}{2}}} 2 Cl 2 ( π 3 ) = 3 Cl 2 ( 2 π 3 ) {\displaystyle 2\operatorname {Cl} _{2}\left({\frac {\pi }{3}}\right)=3\operatorname {Cl} _{2}\left({\frac {2\pi }{3}}\right)} <p>For higher order Clausen functions, duplication formulae can be obtained from the one given above; simply replace θ {\displaystyle \,\theta \,} with the <a href="/facts/Bound_variable/ndWnXXYs">dummy variable</a> x {\displaystyle x} , and integrate over the interval [ 0 , θ ] . {\displaystyle \,[0,\theta ].\,} Applying the same process repeatedly yields: </p> Cl 3 ( 2 θ ) = 4 Cl 3 ( θ ) + 4 Cl 3 ( π − θ ) {\displaystyle \operatorname {Cl} _{3}(2\theta )=4\operatorname {Cl} _{3}(\theta )+4\operatorname {Cl} _{3}(\pi -\theta )} Cl 4 ( 2 θ ) = 8 Cl 4 ( θ ) − 8 Cl 4 ( π − θ ) {\displaystyle \operatorname {Cl} _{4}(2\theta )=8\operatorname {Cl} _{4}(\theta )-8\operatorname {Cl} _{4}(\pi -\theta )} Cl 5 ( 2 θ ) = 16 Cl 5 ( θ ) + 16 Cl 5 ( π − θ ) {\displaystyle \operatorname {Cl} _{5}(2\theta )=16\operatorname {Cl} _{5}(\theta )+16\operatorname {Cl} _{5}(\pi -\theta )} Cl 6 ( 2 θ ) = 32 Cl 6 ( θ ) − 32 Cl 6 ( π − θ ) {\displaystyle \operatorname {Cl} _{6}(2\theta )=32\operatorname {Cl} _{6}(\theta )-32\operatorname {Cl} _{6}(\pi -\theta )} <p>And more generally, upon induction on m , m ≥ 1 {\displaystyle \,m,\;m\geq 1} </p> Cl m + 1 ( 2 θ ) = 2 m [ Cl m + 1 ( θ ) + ( − 1 ) m Cl m + 1 ( π − θ ) ] {\displaystyle \operatorname {Cl} _{m+1}(2\theta )=2^{m}\left[\operatorname {Cl} _{m+1}(\theta )+(-1)^{m}\operatorname {Cl} _{m+1}(\pi -\theta )\right]} <p>Use of the generalized duplication formula allows for an extension of the result for the Clausen function of order 2, involving <a href="/facts/Catalan%2527s_constant/c9fNqG8j">Catalan's constant</a>. For m ∈ Z ≥ 1 {\displaystyle \,m\in \mathbb {Z} \geq 1\,} </p> Cl 2 m ( π 2 ) = 2 2 m − 1 [ Cl 2 m ( π 4 ) − Cl 2 m ( 3 π 4 ) ] = β ( 2 m ) {\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)} <p>Where β ( x ) {\displaystyle \,\beta (x)\,} is the <a href="/facts/Dirichlet_beta_function/FKwkVguL">Dirichlet beta function</a>. </p> <h3>Proof of the duplication formula</h3> <p>From the integral definition, </p> Cl 2 ( 2 θ ) = − ∫ 0 2 θ log | 2 sin x 2 | d x {\displaystyle \operatorname {Cl} _{2}(2\theta )=-\int _{0}^{2\theta }\log \left|2\sin {\frac {x}{2}}\right|\,dx} <p>Apply the duplication formula for the <a href="/facts/Sine_function/gQ6LXK8z">sine function</a>, sin x = 2 sin x 2 cos x 2 {\displaystyle \sin x=2\sin {\frac {x}{2}}\cos {\frac {x}{2}}} to obtain </p> − ∫ 0 2 θ log | ( 2 sin x 4 ) ( 2 cos x 4 ) | d x = − ∫ 0 2 θ log | 2 sin x 4 | d x − ∫ 0 2 θ log | 2 cos x 4 | d x {\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}}} <p>Apply the substitution x = 2 y , d x = 2 d y {\displaystyle x=2y,dx=2\,dy} on both integrals: </p> − 2 ∫ 0 θ log | 2 sin x 2 | d x − 2 ∫ 0 θ log | 2 cos x 2 | d x = 2 Cl 2 ( θ ) − 2 ∫ 0 θ log | 2 cos x 2 | d x {\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}}} <p>On that last integral, set y = π − x , x = π − y , d x = − d y {\displaystyle y=\pi -x,\,x=\pi -y,\,dx=-dy} , and use the trigonometric identity cos ( x − y ) = cos x cos y − sin x sin y {\displaystyle \cos(x-y)=\cos x\cos y-\sin x\sin y} to show that: </p> cos ( π − y 2 ) = sin y 2 ⟹ Cl 2 ( 2 θ ) = 2 Cl 2 ( θ ) − 2 ∫ 0 θ log | 2 cos x 2 | d x = 2 Cl 2 ( θ ) + 2 ∫ π π − θ log | 2 sin y 2 | d y = 2 Cl 2 ( θ ) − 2 Cl 2 ( π − θ ) + 2 Cl 2 ( π ) {\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}}} Cl 2 ( π ) = 0 {\displaystyle \operatorname {Cl} _{2}(\pi )=0\,} <p>Therefore, </p> Cl 2 ( 2 θ ) = 2 Cl 2 ( θ ) − 2 Cl 2 ( π − θ ) . ◻ {\displaystyle \operatorname {Cl} _{2}(2\theta )=2\,\operatorname {Cl} _{2}(\theta )-2\,\operatorname {Cl} _{2}(\pi -\theta )\,.\,\Box } <h2 id="derivatives-of-general-order-clausen-functions">Derivatives of general-order Clausen functions</h2> <p>Direct differentiation of the <a href="/facts/Fourier_series/s3fsxPAT">Fourier series</a> expansions for the Clausen functions give: </p> d d θ Cl 2 m + 2 ( θ ) = d d θ ∑ k = 1 ∞ sin k θ k 2 m + 2 = ∑ k = 1 ∞ cos k θ k 2 m + 1 = Cl 2 m + 1 ( θ ) {\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 )} d d θ Cl 2 m + 1 ( θ ) = d d θ ∑ k = 1 ∞ cos k θ k 2 m + 1 = − ∑ k = 1 ∞ sin k θ k 2 m = − Cl 2 m ( θ ) {\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 )} d d θ Sl 2 m + 2 ( θ ) = d d θ ∑ k = 1 ∞ cos k θ k 2 m + 2 = − ∑ k = 1 ∞ sin k θ k 2 m + 1 = − Sl 2 m + 1 ( θ ) {\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 )} d d θ Sl 2 m + 1 ( θ ) = d d θ ∑ k = 1 ∞ sin k θ k 2 m + 1 = ∑ k = 1 ∞ cos k θ k 2 m = Sl 2 m ( θ ) {\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 )} <p>By appealing to the <a href="/facts/First_Fundamental_Theorem_Of_Calculus/pHeSGtUx">First Fundamental Theorem Of Calculus</a>, we also have: </p> d d θ Cl 2 ( θ ) = d d θ [ − ∫ 0 θ log | 2 sin x 2 | d x ] = − log | 2 sin θ 2 | = Cl 1 ( θ ) {\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 )} <h2 id="relation-to-the-inverse-tangent-integral">Relation to the inverse tangent integral</h2> <p>The <a href="/facts/Inverse_tangent_integral/CKrkJMKn">inverse tangent integral</a> is defined on the interval 0 < z < 1 {\displaystyle 0<z<1} by </p> Ti 2 ( z ) = ∫ 0 z tan − 1 x x d x = ∑ k = 0 ∞ ( − 1 ) k z 2 k + 1 ( 2 k + 1 ) 2 {\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}}}} <p>It has the following closed form in terms of the Clausen function: </p> Ti 2 ( tan θ ) = θ log ( tan θ ) + 1 2 Cl 2 ( 2 θ ) + 1 2 Cl 2 ( π − 2 θ ) {\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 )} <h3>Proof of the inverse tangent integral relation</h3> <p>From the integral definition of the <a href="/facts/Inverse_tangent_integral/CKrkJMKn">inverse tangent integral</a>, we have </p> Ti 2 ( tan θ ) = ∫ 0 tan θ tan − 1 x x d x {\displaystyle \operatorname {Ti} _{2}(\tan \theta )=\int _{0}^{\tan \theta }{\frac {\tan ^{-1}x}{x}}\,dx} <p>Performing an integration by parts </p> ∫ 0 tan θ tan − 1 x x d x = tan − 1 x log x | 0 tan θ − ∫ 0 tan θ log x 1 + x 2 d x = {\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=} θ log tan θ − ∫ 0 tan θ log x 1 + x 2 d x {\displaystyle \theta \log \tan \theta -\int _{0}^{\tan \theta }{\frac {\log x}{1+x^{2}}}\,dx} <p>Apply the substitution x = tan y , y = tan − 1 x , d y = d x 1 + x 2 {\displaystyle x=\tan y,\,y=\tan ^{-1}x,\,dy={\frac {dx}{1+x^{2}}}\,} to obtain </p> θ log tan θ − ∫ 0 θ log ( tan y ) d y {\displaystyle \theta \log \tan \theta -\int _{0}^{\theta }\log(\tan y)\,dy} <p>For that last integral, apply the transform : y = x / 2 , d y = d x / 2 {\displaystyle y=x/2,\,dy=dx/2\,} to get </p> θ log tan θ − 1 2 ∫ 0 2 θ log ( tan x 2 ) d x = θ log tan θ − 1 2 ∫ 0 2 θ log ( sin ( x / 2 ) cos ( x / 2 ) ) d x = θ log tan θ − 1 2 ∫ 0 2 θ log ( 2 sin ( x / 2 ) 2 cos ( x / 2 ) ) d x = θ log tan θ − 1 2 ∫ 0 2 θ log ( 2 sin x 2 ) d x + 1 2 ∫ 0 2 θ log ( 2 cos x 2 ) d x = θ log tan θ + 1 2 Cl 2 ( 2 θ ) + 1 2 ∫ 0 2 θ log ( 2 cos x 2 ) d x . {\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}}} <p>Finally, as with the proof of the Duplication formula, the substitution x = ( π − y ) {\displaystyle x=(\pi -y)\,} reduces that last integral to </p> ∫ 0 2 θ log ( 2 cos x 2 ) d x = Cl 2 ( π − 2 θ ) − Cl 2 ( π ) = Cl 2 ( π − 2 θ ) {\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 )} <p>Thus </p> Ti 2 ( tan θ ) = θ log tan θ + 1 2 Cl 2 ( 2 θ ) + 1 2 Cl 2 ( π − 2 θ ) . ◻ {\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 } <h2 id="relation-to-the-barnes-g-function">Relation to the Barnes' G-function</h2> <p>For real 0 < z < 1 {\displaystyle 0<z<1} , the Clausen function of second order can be expressed in terms of the <a href="/facts/Barnes_G-function/4bmYxJrE">Barnes G-function</a> and (Euler) <a href="/facts/Gamma_function/SjGQcnyw">Gamma function</a>: </p> Cl 2 ( 2 π z ) = 2 π log ( G ( 1 − z ) G ( 1 + z ) ) + 2 π z log ( π sin π z ) {\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)} <p>Or equivalently </p> Cl 2 ( 2 π z ) = 2 π log ( G ( 1 − z ) G ( z ) ) − 2 π log Γ ( z ) + 2 π z log ( π sin π z ) {\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)} <p>See Adamchik (2003). </p> <h2 id="relation-to-the-polylogarithm">Relation to the polylogarithm</h2> <p>The Clausen functions represent the real and imaginary parts of the polylogarithm, on the <a href="/facts/Unit_circle/VsTffjPK">unit circle</a>: </p> Cl 2 m ( θ ) = ℑ ( Li 2 m ( e i θ ) ) , m ∈ Z ≥ 1 {\displaystyle \operatorname {Cl} _{2m}(\theta )=\Im (\operatorname {Li} _{2m}(e^{i\theta })),\quad m\in \mathbb {Z} \geq 1} Cl 2 m + 1 ( θ ) = ℜ ( Li 2 m + 1 ( e i θ ) ) , m ∈ Z ≥ 0 {\displaystyle \operatorname {Cl} _{2m+1}(\theta )=\Re (\operatorname {Li} _{2m+1}(e^{i\theta })),\quad m\in \mathbb {Z} \geq 0} <p>This is easily seen by appealing to the series definition of the <a href="/facts/Polylogarithm/CKrkJMKn">polylogarithm</a>. </p> Li n ( z ) = ∑ k = 1 ∞ z k k n ⟹ Li n ( e i θ ) = ∑ k = 1 ∞ ( e i θ ) k k n = ∑ k = 1 ∞ e i k θ k n {\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}}}} <p>By Euler's theorem, </p> e i θ = cos θ + i sin θ {\displaystyle e^{i\theta }=\cos \theta +i\sin \theta } <p>and by de Moivre's Theorem (<a href="/facts/De_Moivre%2527s_formula/uyG4k2uC">De Moivre's formula</a>) </p> ( cos θ + i sin θ ) k = cos k θ + i sin k θ ⇒ Li n ( e i θ ) = ∑ k = 1 ∞ cos k θ k n + i ∑ k = 1 ∞ sin k θ k n {\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}}}} <p>Hence </p> Li 2 m ( e i θ ) = ∑ k = 1 ∞ cos k θ k 2 m + i ∑ k = 1 ∞ sin k θ k 2 m = Sl 2 m ( θ ) + i Cl 2 m ( θ ) {\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 )} Li 2 m + 1 ( e i θ ) = ∑ k = 1 ∞ cos k θ k 2 m + 1 + i ∑ k = 1 ∞ sin k θ k 2 m + 1 = Cl 2 m + 1 ( θ ) + i Sl 2 m + 1 ( θ ) {\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 )} <h2 id="relation-to-the-polygamma-function">Relation to the polygamma function</h2> <p>The Clausen functions are intimately connected to the <a href="/facts/Polygamma_function/IF6W68aL">polygamma function</a>. 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: </p> Cl 2 m ( q π p ) = 1 ( 2 p ) 2 m ( 2 m − 1 ) ! ∑ j = 1 p sin ( q j π p ) [ ψ 2 m − 1 ( j 2 p ) + ( − 1 ) q ψ 2 m − 1 ( j + p 2 p ) ] . {\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].} <p>An immediate corollary is this equivalent formula in terms of the Hurwitz zeta function: </p> Cl 2 m ( q π p ) = 1 ( 2 p ) 2 m ∑ j = 1 p sin ( q j π p ) [ ζ ( 2 m , j 2 p ) + ( − 1 ) q ζ ( 2 m , j + p 2 p ) ] . {\displaystyle \operatorname {Cl} _{2m}\left({\frac {q\pi }{p}}\right)={\frac {1}{(2p)^{2m}}}\,\sum _{j=1}^{p}\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].} <table><tbody><tr><th>Proof of the formula</th></tr><tr><td><p>Let p {\displaystyle \,p\,} and q {\displaystyle \,q\,} be positive integers, such that q / p {\displaystyle \,q/p\,} is a rational number 0 < q / p < 1 {\displaystyle \,0<q/p<1\,} , then, by the series definition for the higher order Clausen function (of even index):</p> Cl 2 m ( q π p ) = ∑ k = 1 ∞ sin ( k q π / p ) k 2 m {\displaystyle \operatorname {Cl} _{2m}\left({\frac {q\pi }{p}}\right)=\sum _{k=1}^{\infty }{\frac {\sin(kq\pi /p)}{k^{2m}}}} <p>We split this sum into exactly p-parts, so that the first series contains all, and only, those terms congruent to k p + 1 , {\displaystyle \,kp+1,\,} the second series contains all terms congruent to k p + 2 , {\displaystyle \,kp+2,\,} etc., up to the final p-th part, that contain all terms congruent to k p + p {\displaystyle \,kp+p\,} </p> Cl 2 m ( q π p ) = ∑ k = 0 ∞ sin [ ( k p + 1 ) q π p ] ( k p + 1 ) 2 m + ∑ k = 0 ∞ sin [ ( k p + 2 ) q π p ] ( k p + 2 ) 2 m + ∑ k = 0 ∞ sin [ ( k p + 3 ) q π p ] ( k p + 3 ) 2 m + ⋯ ⋯ + ∑ k = 0 ∞ sin [ ( k p + p − 2 ) q π p ] ( k p + p − 2 ) 2 m + ∑ k = 0 ∞ sin [ ( k p + p − 1 ) q π p ] ( k p + p − 1 ) 2 m + ∑ k = 0 ∞ sin [ ( k p + p ) q π p ] ( k p + p ) 2 m {\displaystyle {\begin{aligned}&\operatorname {Cl} _{2m}\left({\frac {q\pi }{p}}\right)\\={}&\sum _{k=0}^{\infty }{\frac {\sin \left[(kp+1){\frac {q\pi }{p}}\right]}{(kp+1)^{2m}}}+\sum _{k=0}^{\infty }{\frac {\sin \left[(kp+2){\frac {q\pi }{p}}\right]}{(kp+2)^{2m}}}+\sum _{k=0}^{\infty }{\frac {\sin \left[(kp+3){\frac {q\pi }{p}}\right]}{(kp+3)^{2m}}}+\cdots \\&\cdots +\sum _{k=0}^{\infty }{\frac {\sin \left[(kp+p-2){\frac {q\pi }{p}}\right]}{(kp+p-2)^{2m}}}+\sum _{k=0}^{\infty }{\frac {\sin \left[(kp+p-1){\frac {q\pi }{p}}\right]}{(kp+p-1)^{2m}}}+\sum _{k=0}^{\infty }{\frac {\sin \left[(kp+p){\frac {q\pi }{p}}\right]}{(kp+p)^{2m}}}\end{aligned}}} <p>We can index these sums to form a double sum:</p> Cl 2 m ( q π p ) = ∑ j = 1 p { ∑ k = 0 ∞ sin [ ( k p + j ) q π p ] ( k p + j ) 2 m } = ∑ j = 1 p 1 p 2 m { ∑ k = 0 ∞ sin [ ( k p + j ) q π p ] ( k + ( j / p ) ) 2 m } {\displaystyle {\begin{aligned}&\operatorname {Cl} _{2m}\left({\frac {q\pi }{p}}\right)=\sum _{j=1}^{p}\left\{\sum _{k=0}^{\infty }{\frac {\sin \left[(kp+j){\frac {q\pi }{p}}\right]}{(kp+j)^{2m}}}\right\}\\={}&\sum _{j=1}^{p}{\frac {1}{p^{2m}}}\left\{\sum _{k=0}^{\infty }{\frac {\sin \left[(kp+j){\frac {q\pi }{p}}\right]}{(k+(j/p))^{2m}}}\right\}\end{aligned}}} <p>Applying the addition formula for the <a href="/facts/Sine_function/gQ6LXK8z">sine function</a>, sin ( x + y ) = sin x cos y + cos x sin y , {\displaystyle \,\sin(x+y)=\sin x\cos y+\cos x\sin y,\,} the sine term in the numerator becomes:</p> sin [ ( k p + j ) q π p ] = sin ( k q π + q j π p ) = sin k q π cos q j π p + cos k q π sin q j π p {\displaystyle \sin \left[(kp+j){\frac {q\pi }{p}}\right]=\sin \left(kq\pi +{\frac {qj\pi }{p}}\right)=\sin kq\pi \cos {\frac {qj\pi }{p}}+\cos kq\pi \sin {\frac {qj\pi }{p}}} sin m π ≡ 0 , cos m π ≡ ( − 1 ) m ⟺ m = 0 , ± 1 , ± 2 , ± 3 , … {\displaystyle \sin m\pi \equiv 0,\quad \,\cos m\pi \equiv (-1)^{m}\quad \Longleftrightarrow m=0,\,\pm 1,\,\pm 2,\,\pm 3,\,\ldots } sin [ ( k p + j ) q π p ] = ( − 1 ) k q sin q j π p {\displaystyle \sin \left[(kp+j){\frac {q\pi }{p}}\right]=(-1)^{kq}\sin {\frac {qj\pi }{p}}} <p>Consequently,</p> Cl 2 m ( q π p ) = ∑ j = 1 p 1 p 2 m sin ( q j π p ) { ∑ k = 0 ∞ ( − 1 ) k q ( k + ( j / p ) ) 2 m } {\displaystyle \operatorname {Cl} _{2m}\left({\frac {q\pi }{p}}\right)=\sum _{j=1}^{p}{\frac {1}{p^{2m}}}\sin \left({\frac {qj\pi }{p}}\right)\,\left\{\sum _{k=0}^{\infty }{\frac {(-1)^{kq}}{(k+(j/p))^{2m}}}\right\}} <p>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:</p> ∑ k = 0 ∞ ( − 1 ) k q ( k + ( j / p ) ) 2 m = ∑ k = 0 ∞ ( − 1 ) ( 2 k ) q ( ( 2 k ) + ( j / p ) ) 2 m + ∑ k = 0 ∞ ( − 1 ) ( 2 k + 1 ) q ( ( 2 k + 1 ) + ( j / p ) ) 2 m = ∑ k = 0 ∞ 1 ( 2 k + ( j / p ) ) 2 m + ( − 1 ) q ∑ k = 0 ∞ 1 ( 2 k + 1 + ( j / p ) ) 2 m = 1 2 p [ ∑ k = 0 ∞ 1 ( k + ( j / 2 p ) ) 2 m + ( − 1 ) q ∑ k = 0 ∞ 1 ( k + ( j + p 2 p ) ) 2 m ] {\displaystyle {\begin{aligned}&\sum _{k=0}^{\infty }{\frac {(-1)^{kq}}{(k+(j/p))^{2m}}}=\sum _{k=0}^{\infty }{\frac {(-1)^{(2k)q}}{((2k)+(j/p))^{2m}}}+\sum _{k=0}^{\infty }{\frac {(-1)^{(2k+1)q}}{((2k+1)+(j/p))^{2m}}}\\={}&\sum _{k=0}^{\infty }{\frac {1}{(2k+(j/p))^{2m}}}+(-1)^{q}\,\sum _{k=0}^{\infty }{\frac {1}{(2k+1+(j/p))^{2m}}}\\={}&{\frac {1}{2^{p}}}\left[\sum _{k=0}^{\infty }{\frac {1}{(k+(j/2p))^{2m}}}+(-1)^{q}\,\sum _{k=0}^{\infty }{\frac {1}{(k+\left({\frac {j+p}{2p}}\right))^{2m}}}\right]\end{aligned}}} <p>For m ∈ Z ≥ 1 {\displaystyle \,m\in \mathbb {Z} \geq 1\,} , the <a href="/facts/Polygamma_function/IF6W68aL">polygamma function</a> has the series representation</p> ψ m ( z ) = ( − 1 ) m + 1 m ! ∑ k = 0 ∞ 1 ( k + z ) m + 1 {\displaystyle \psi _{m}(z)=(-1)^{m+1}m!\sum _{k=0}^{\infty }{\frac {1}{(k+z)^{m+1}}}} <p>So, in terms of the polygamma function, the previous inner sum becomes:</p> 1 2 2 m ( 2 m − 1 ) ! [ ψ 2 m − 1 ( j 2 p ) + ( − 1 ) q ψ 2 m − 1 ( j + p 2 p ) ] {\displaystyle {\frac {1}{2^{2m}(2m-1)!}}\left[\psi _{2m-1}\left({\tfrac {j}{2p}}\right)+(-1)^{q}\psi _{2m-1}\left({\tfrac {j+p}{2p}}\right)\right]} <p>Plugging this back into the double sum gives the desired result:</p> Cl 2 m ( q π p ) = 1 ( 2 p ) 2 m ( 2 m − 1 ) ! ∑ j = 1 p sin ( q j π p ) [ ψ 2 m − 1 ( j 2 p ) + ( − 1 ) q ψ 2 m − 1 ( j + p 2 p ) ] {\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]} </td></tr></tbody></table> <h2 id="relation-to-the-generalized-logsine-integral">Relation to the generalized logsine integral</h2> <p>The generalized logsine integral is defined by: </p> L s n m ( θ ) = − ∫ 0 θ x m log n − m − 1 | 2 sin x 2 | d x {\displaystyle {\mathcal {L}}s_{n}^{m}(\theta )=-\int _{0}^{\theta }x^{m}\log ^{n-m-1}\left|2\sin {\frac {x}{2}}\right|\,dx} <p>In this generalized notation, the Clausen function can be expressed in the form: </p> Cl 2 ( θ ) = L s 2 0 ( θ ) {\displaystyle \operatorname {Cl} _{2}(\theta )={\mathcal {L}}s_{2}^{0}(\theta )} <h2 id="kummers-relation">Kummer's relation</h2> <p><a href="/facts/Ernst_Kummer/dGnaTQyG">Ernst Kummer</a> and Rogers give the relation </p> Li 2 ( e i θ ) = ζ ( 2 ) − θ ( 2 π − θ ) / 4 + i Cl 2 ( θ ) {\displaystyle \operatorname {Li} _{2}(e^{i\theta })=\zeta (2)-\theta (2\pi -\theta )/4+i\operatorname {Cl} _{2}(\theta )} <p>valid for 0 ≤ θ ≤ 2 π {\displaystyle 0\leq \theta \leq 2\pi } . </p> <h2 id="relation-to-the-lobachevsky-function">Relation to the Lobachevsky function</h2> <p>The Lobachevsky function Λ or Л is essentially the same function with a change of variable: </p> Λ ( θ ) = − ∫ 0 θ log | 2 sin ( t ) | d t = Cl 2 ( 2 θ ) / 2 {\displaystyle \Lambda (\theta )=-\int _{0}^{\theta }\log |2\sin(t)|\,dt=\operatorname {Cl} _{2}(2\theta )/2} <p>though the name "Lobachevsky function" is not quite historically accurate, as Lobachevsky's formulas for hyperbolic volume used the slightly different function </p> ∫ 0 θ log | sec ( t ) | d t = Λ ( θ + π / 2 ) + θ log 2. {\displaystyle \int _{0}^{\theta }\log |\sec(t)|\,dt=\Lambda (\theta +\pi /2)+\theta \log 2.} <h2 id="relation-to-dirichlet-l-functions">Relation to Dirichlet L-functions</h2> <p>For rational values of θ / π {\displaystyle \theta /\pi } (that is, for θ / π = p / q {\displaystyle \theta /\pi =p/q} for some integers <i>p</i> and <i>q</i>), the function sin ( n θ ) {\displaystyle \sin(n\theta )} can be understood to represent a periodic orbit of an element in the <a href="/facts/Cyclic_group/JL5zfFuL">cyclic group</a>, and thus Cl s ( θ ) {\displaystyle \operatorname {Cl} _{s}(\theta )} can be expressed as a simple sum involving the <a href="/facts/Hurwitz_zeta_function/4EH7sKXB">Hurwitz zeta function</a>. This allows relations between certain <a href="/facts/Dirichlet_L-function/my8hc6Qy">Dirichlet L-functions</a> to be easily computed. </p> <h2 id="series-acceleration">Series acceleration</h2> <p>A <a href="/facts/Series_acceleration/XmWMXVFh">series acceleration</a> for the Clausen function is given by </p> Cl 2 ( θ ) θ = 1 − log | θ | + ∑ n = 1 ∞ ζ ( 2 n ) n ( 2 n + 1 ) ( θ 2 π ) 2 n {\displaystyle {\frac {\operatorname {Cl} _{2}(\theta )}{\theta }}=1-\log |\theta |+\sum _{n=1}^{\infty }{\frac {\zeta (2n)}{n(2n+1)}}\left({\frac {\theta }{2\pi }}\right)^{2n}} <p>which holds for | θ | < 2 π {\displaystyle |\theta |<2\pi } . Here, ζ ( s ) {\displaystyle \zeta (s)} is the <a href="/facts/Riemann_zeta_function/MnfRu7lY">Riemann zeta function</a>. A more rapidly convergent form is given by </p> Cl 2 ( θ ) θ = 3 − log [ | θ | ( 1 − θ 2 4 π 2 ) ] − 2 π θ log ( 2 π + θ 2 π − θ ) + ∑ n = 1 ∞ ζ ( 2 n ) − 1 n ( 2 n + 1 ) ( θ 2 π ) 2 n . {\displaystyle {\frac {\operatorname {Cl} _{2}(\theta )}{\theta }}=3-\log \left[|\theta |\left(1-{\frac {\theta ^{2}}{4\pi ^{2}}}\right)\right]-{\frac {2\pi }{\theta }}\log \left({\frac {2\pi +\theta }{2\pi -\theta }}\right)+\sum _{n=1}^{\infty }{\frac {\zeta (2n)-1}{n(2n+1)}}\left({\frac {\theta }{2\pi }}\right)^{2n}.} <p>Convergence is aided by the fact that ζ ( n ) − 1 {\displaystyle \zeta (n)-1} approaches zero rapidly for large values of <i>n</i>. Both forms are obtainable through the types of resummation techniques used to obtain <a href="/facts/Rational_zeta_series/dKEtUnLV">rational zeta series</a> (Borwein et al. 2000). </p> <h2 id="special-values">Special values</h2> <p>Recall the <a href="/facts/Barnes_G-function/4bmYxJrE">Barnes G-function</a>, the <a href="/facts/Catalan_constant/c9fNqG8j">Catalan's constant</a> <i>K</i> and the <a href="/facts/Gieseking_manifold/KTI5jr7D">Gieseking constant</a> <i>V</i>. Some special values include </p> Cl 2 ( π 2 ) = K {\displaystyle \operatorname {Cl} _{2}\left({\frac {\pi }{2}}\right)=K} Cl 2 ( π 3 ) = V {\displaystyle \operatorname {Cl} _{2}\left({\frac {\pi }{3}}\right)=V} Cl 2 ( π 3 ) = 3 π log ( G ( 2 3 ) G ( 1 3 ) ) − 3 π log Γ ( 1 3 ) + π log ( 2 π 3 ) {\displaystyle \operatorname {Cl} _{2}\left({\frac {\pi }{3}}\right)=3\pi \log \left({\frac {G\left({\frac {2}{3}}\right)}{G\left({\frac {1}{3}}\right)}}\right)-3\pi \log \Gamma \left({\frac {1}{3}}\right)+\pi \log \left({\frac {2\pi }{\sqrt {3}}}\right)} Cl 2 ( 2 π 3 ) = 2 π log ( G ( 2 3 ) G ( 1 3 ) ) − 2 π log Γ ( 1 3 ) + 2 π 3 log ( 2 π 3 ) {\displaystyle \operatorname {Cl} _{2}\left({\frac {2\pi }{3}}\right)=2\pi \log \left({\frac {G\left({\frac {2}{3}}\right)}{G\left({\frac {1}{3}}\right)}}\right)-2\pi \log \Gamma \left({\frac {1}{3}}\right)+{\frac {2\pi }{3}}\log \left({\frac {2\pi }{\sqrt {3}}}\right)} Cl 2 ( π 4 ) = 2 π log ( G ( 7 8 ) G ( 1 8 ) ) − 2 π log Γ ( 1 8 ) + π 4 log ( 2 π 2 − 2 ) {\displaystyle \operatorname {Cl} _{2}\left({\frac {\pi }{4}}\right)=2\pi \log \left({\frac {G\left({\frac {7}{8}}\right)}{G\left({\frac {1}{8}}\right)}}\right)-2\pi \log \Gamma \left({\frac {1}{8}}\right)+{\frac {\pi }{4}}\log \left({\frac {2\pi }{\sqrt {2-{\sqrt {2}}}}}\right)} Cl 2 ( 3 π 4 ) = 2 π log ( G ( 5 8 ) G ( 3 8 ) ) − 2 π log Γ ( 3 8 ) + 3 π 4 log ( 2 π 2 + 2 ) {\displaystyle \operatorname {Cl} _{2}\left({\frac {3\pi }{4}}\right)=2\pi \log \left({\frac {G\left({\frac {5}{8}}\right)}{G\left({\frac {3}{8}}\right)}}\right)-2\pi \log \Gamma \left({\frac {3}{8}}\right)+{\frac {3\pi }{4}}\log \left({\frac {2\pi }{\sqrt {2+{\sqrt {2}}}}}\right)} Cl 2 ( π 6 ) = 2 π log ( G ( 11 12 ) G ( 1 12 ) ) − 2 π log Γ ( 1 12 ) + π 6 log ( 2 π 2 3 − 1 ) {\displaystyle \operatorname {Cl} _{2}\left({\frac {\pi }{6}}\right)=2\pi \log \left({\frac {G\left({\frac {11}{12}}\right)}{G\left({\frac {1}{12}}\right)}}\right)-2\pi \log \Gamma \left({\frac {1}{12}}\right)+{\frac {\pi }{6}}\log \left({\frac {2\pi {\sqrt {2}}}{{\sqrt {3}}-1}}\right)} Cl 2 ( 5 π 6 ) = 2 π log ( G ( 7 12 ) G ( 5 12 ) ) − 2 π log Γ ( 5 12 ) + 5 π 6 log ( 2 π 2 3 + 1 ) {\displaystyle \operatorname {Cl} _{2}\left({\frac {5\pi }{6}}\right)=2\pi \log \left({\frac {G\left({\frac {7}{12}}\right)}{G\left({\frac {5}{12}}\right)}}\right)-2\pi \log \Gamma \left({\frac {5}{12}}\right)+{\frac {5\pi }{6}}\log \left({\frac {2\pi {\sqrt {2}}}{{\sqrt {3}}+1}}\right)} <p>In general, from the <a href="/facts/Barnes_G-function/4bmYxJrE">Barnes G-function reflection formula</a>, </p> Cl 2 ( 2 π z ) = 2 π log ( G ( 1 − z ) G ( z ) ) − 2 π log Γ ( z ) + 2 π z log ( π sin π z ) {\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)} <p>Equivalently, using Euler's <a href="/facts/Reflection_formula/aKUhTbH0">reflection formula</a> for the gamma function, then, </p> Cl 2 ( 2 π z ) = 2 π log ( G ( 1 − z ) G ( z ) ) − 2 π log Γ ( z ) + 2 π z log ( Γ ( z ) Γ ( 1 − z ) ) {\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 {\big (}\Gamma (z)\Gamma (1-z){\big )}} <h2 id="generalized-special-values">Generalized special values</h2> <p>Some special values for higher order Clausen functions include </p> Cl 2 m ( 0 ) = Cl 2 m ( π ) = Cl 2 m ( 2 π ) = 0 {\displaystyle \operatorname {Cl} _{2m}(0)=\operatorname {Cl} _{2m}(\pi )=\operatorname {Cl} _{2m}(2\pi )=0} Cl 2 m ( π 2 ) = β ( 2 m ) {\displaystyle \operatorname {Cl} _{2m}\left({\frac {\pi }{2}}\right)=\beta (2m)} Cl 2 m + 1 ( 0 ) = Cl 2 m + 1 ( 2 π ) = ζ ( 2 m + 1 ) {\displaystyle \operatorname {Cl} _{2m+1}(0)=\operatorname {Cl} _{2m+1}(2\pi )=\zeta (2m+1)} Cl 2 m + 1 ( π ) = − η ( 2 m + 1 ) = − ( 2 2 m − 1 2 2 m ) ζ ( 2 m + 1 ) {\displaystyle \operatorname {Cl} _{2m+1}(\pi )=-\eta (2m+1)=-\left({\frac {2^{2m}-1}{2^{2m}}}\right)\zeta (2m+1)} Cl 2 m + 1 ( π 2 ) = − 1 2 2 m + 1 η ( 2 m + 1 ) = − ( 2 2 m − 1 2 4 m + 1 ) ζ ( 2 m + 1 ) {\displaystyle \operatorname {Cl} _{2m+1}\left({\frac {\pi }{2}}\right)=-{\frac {1}{2^{2m+1}}}\eta (2m+1)=-\left({\frac {2^{2m}-1}{2^{4m+1}}}\right)\zeta (2m+1)} <p>where β ( x ) {\displaystyle \beta (x)} is the <a href="/facts/Dirichlet_beta_function/FKwkVguL">Dirichlet beta function</a>, η ( x ) {\displaystyle \eta (x)} is the <a href="/facts/Dirichlet_eta_function/32tUb6a7">Dirichlet eta function</a> (also called the alternating zeta function), and ζ ( x ) {\displaystyle \zeta (x)} is the <a href="/facts/Riemann_zeta_function/MnfRu7lY">Riemann zeta function</a>. </p> <h2 id="integrals-of-the-direct-function">Integrals of the direct function</h2> <p>The following integrals are easily proven from the series representations of the Clausen function: </p> ∫ 0 θ Cl 2 m ( x ) d x = ζ ( 2 m + 1 ) − Cl 2 m + 1 ( θ ) {\displaystyle \int _{0}^{\theta }\operatorname {Cl} _{2m}(x)\,dx=\zeta (2m+1)-\operatorname {Cl} _{2m+1}(\theta )} ∫ 0 θ Cl 2 m + 1 ( x ) d x = Cl 2 m + 2 ( θ ) {\displaystyle \int _{0}^{\theta }\operatorname {Cl} _{2m+1}(x)\,dx=\operatorname {Cl} _{2m+2}(\theta )} ∫ 0 θ Sl 2 m ( x ) d x = Sl 2 m + 1 ( θ ) {\displaystyle \int _{0}^{\theta }\operatorname {Sl} _{2m}(x)\,dx=\operatorname {Sl} _{2m+1}(\theta )} ∫ 0 θ Sl 2 m + 1 ( x ) d x = ζ ( 2 m + 2 ) − Cl 2 m + 2 ( θ ) {\displaystyle \int _{0}^{\theta }\operatorname {Sl} _{2m+1}(x)\,dx=\zeta (2m+2)-\operatorname {Cl} _{2m+2}(\theta )} <p>Fourier-analytic methods can be used to find the first moments of the square of the function Cl 2 ( x ) {\displaystyle \operatorname {Cl} _{2}(x)} on the interval [ 0 , π ] {\displaystyle [0,\pi ]} :<a class="footnote-ref" id="fnref:1" href="#fn:1"><sup>1</sup></a> </p> ∫ 0 π Cl 2 2 ( x ) d x = ζ ( 4 ) , {\displaystyle \int _{0}^{\pi }\operatorname {Cl} _{2}^{2}(x)\,dx=\zeta (4),} ∫ 0 π t Cl 2 2 ( x ) d x = 221 90720 π 6 − 4 ζ ( 5 ¯ , 1 ) − 2 ζ ( 4 ¯ , 2 ) , {\displaystyle \int _{0}^{\pi }t\operatorname {Cl} _{2}^{2}(x)\,dx={\frac {221}{90720}}\pi ^{6}-4\zeta ({\overline {5}},1)-2\zeta ({\overline {4}},2),} ∫ 0 π t 2 Cl 2 2 ( x ) d x = − 2 3 π [ 12 ζ ( 5 ¯ , 1 ) + 6 ζ ( 4 ¯ , 2 ) − 23 10080 π 6 ] . {\displaystyle \int _{0}^{\pi }t^{2}\operatorname {Cl} _{2}^{2}(x)\,dx=-{\frac {2}{3}}\pi \left[12\zeta ({\overline {5}},1)+6\zeta ({\overline {4}},2)-{\frac {23}{10080}}\pi ^{6}\right].} <p>Here ζ {\displaystyle \zeta } denotes the <a href="/facts/Multiple_zeta_function/tBycfN4h">multiple zeta function</a>. </p> <h2 id="integral-evaluations-involving-the-direct-function">Integral evaluations involving the direct function</h2> <p>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 {\displaystyle \,K\,} (<a href="/facts/Catalan%2527s_constant/c9fNqG8j">Catalan's constant</a>), log 2 {\displaystyle \,\log 2\,} , and the special cases of the <a href="/facts/Zeta_function/lf43angF">zeta function</a>, ζ ( 2 ) {\displaystyle \,\zeta (2)\,} and ζ ( 3 ) {\displaystyle \,\zeta (3)\,} . </p><p>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 <a href="/facts/Fourier_series/s3fsxPAT">Fourier series</a> definitions of the Clausen functions. </p> ∫ 0 θ log ( sin x ) d x = − 1 2 Cl 2 ( 2 θ ) − θ log 2 {\displaystyle \int _{0}^{\theta }\log(\sin x)\,dx=-{\tfrac {1}{2}}\operatorname {Cl} _{2}(2\theta )-\theta \log 2} ∫ 0 θ log ( cos x ) d x = 1 2 Cl 2 ( π − 2 θ ) − θ log 2 {\displaystyle \int _{0}^{\theta }\log(\cos x)\,dx={\tfrac {1}{2}}\operatorname {Cl} _{2}(\pi -2\theta )-\theta \log 2} ∫ 0 θ log ( tan x ) d x = − 1 2 Cl 2 ( 2 θ ) − 1 2 Cl 2 ( π − 2 θ ) {\displaystyle \int _{0}^{\theta }\log(\tan x)\,dx=-{\tfrac {1}{2}}\operatorname {Cl} _{2}(2\theta )-{\tfrac {1}{2}}\operatorname {Cl} _{2}(\pi -2\theta )} ∫ 0 θ log ( 1 + cos x ) d x = 2 Cl 2 ( π − θ ) − θ log 2 {\displaystyle \int _{0}^{\theta }\log(1+\cos x)\,dx=2\operatorname {Cl} _{2}(\pi -\theta )-\theta \log 2} ∫ 0 θ log ( 1 − cos x ) d x = − 2 Cl 2 ( θ ) − θ log 2 {\displaystyle \int _{0}^{\theta }\log(1-\cos x)\,dx=-2\operatorname {Cl} _{2}(\theta )-\theta \log 2} ∫ 0 θ log ( 1 + sin x ) d x = 2 K − 2 Cl 2 ( π 2 + θ ) − θ log 2 {\displaystyle \int _{0}^{\theta }\log(1+\sin x)\,dx=2K-2\operatorname {Cl} _{2}\left({\frac {\pi }{2}}+\theta \right)-\theta \log 2} ∫ 0 θ log ( 1 − sin x ) d x = − 2 K + 2 Cl 2 ( π 2 − θ ) − θ log 2 {\displaystyle \int _{0}^{\theta }\log(1-\sin x)\,dx=-2K+2\operatorname {Cl} _{2}\left({\frac {\pi }{2}}-\theta \right)-\theta \log 2} <ul><li><a href="/facts/Milton_Abramowitz/6yRyez7P">Abramowitz, Milton</a>; <a href="/facts/Irene_Stegun/FjSS1LDR">Stegun, Irene Ann</a>, eds. (1983) [June 1964]. <a href="http://www.math.ubc.ca/~cbm/aands/page_1005.htm">"Chapter 27.8"</a>. <a href="/facts/Abramowitz_and_Stegun/CNGSV3Ed"><i>Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables</i></a>. Applied Mathematics Series. Vol. 55 (Ninth reprint with additional corrections of tenth original printing with corrections (December 1972); first ed.). 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"A C99 implementation of the Clausen sums". <a href="/facts/ArXiv_(identifier)/H6EtgnBe">arXiv</a>:<a href="https://arxiv.org/abs/1309.7504">1309.7504</a> [<a href="https://arxiv.org/archive/math.NA">math.NA</a>].</li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>István, Mező (2020). "Log-sine integrals and alternating Euler sums". Acta Mathematica Hungarica (160): 45–57. doi:10.1007/s10474-019-00975-w. <a href="/wiki/Acta_Mathematica_Hungarica" target="_blank">/wiki/Acta_Mathematica_Hungarica</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> </ol>