Clausen Function, Logsine Integrals and Central Binomial Series

In today´s post we will review the Clausen function and prove some of its properties. Than we will show it´s connection with logsine integrals and Binomial series.

The Clausen function is defined as

$\mathrm{Cl}_{n}(x) \equiv \begin{cases}S_{n}(x)=\sum_{k=1}^{\infty} \frac{\sin (k x)}{k^{n}} & n \text { even } \\ C_{n}(x)=\sum_{k=1}^{\infty} \frac{\cos (k x)}{k^{n}} & n \text { odd },\end{cases}$

Recall the Fourier expansion (see here)

$\ln\left(2 \sin\left( \frac{x}{2}\right) \right)=-\sum_{n=1}^\infty \frac{\cos\left( n x\right)}{n}$ (1)

Integrating both sides from 0 to x we obtain

$\begin{aligned} \int_0^x\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds&=-\sum_{n=1}^\infty \frac{\sin\left( n x\right)}{n^2}\\ &=-\operatorname{Cl}_2(x) \end{aligned}$ (2)

Claim 2:

$\int_0^x s\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds=-x\operatorname{Cl}_2(x)-\operatorname{Cl}_3(x)+\zeta(3)$ (3)

Proof:

$\begin{aligned} \int_0^x\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds&=-\sum_{n=1}^\infty \frac{1}{n}\int_0^x \cos\left( n s\right)\,ds\\ &=-\sum_{n=1}^\infty \frac{\sin\left( n s\right)}{n^2}\\ &=-\operatorname{Cl}_2(x) \end{aligned}$

$\begin{aligned} \int_0^x\int_0^t\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds\,dt&=-\sum_{n=1}^\infty \frac{1}{n^2}\int_0^x \sin\left( n s\right)\,ds\\ &=\sum_{n=1}^\infty \frac{\cos\left( n s\right)}{n^3}-\zeta(3)\\ &=\operatorname{Cl}_3(x)-\zeta(3) \end{aligned}$

On the other hand

$\begin{aligned} \int_0^x\int_0^t\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds\,dt&=\int_0^x \ln\left(2 \sin\left( \frac{s}{2}\right) \right) \left(\int_s^x\,dt\right)\,ds\\ &=\int_0^x \left(x-s \right)\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds\\ &=x \int_0^x\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds-\int_0^x s\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds\\ &=-x\operatorname{Cl}_2(x)-\int_0^x s\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds\\ \end{aligned}$

Hence

$\int_0^x s\ln\left(2 \sin\left( \frac{s}{2}\right) \right)\,ds=-x\operatorname{Cl}_2(x)-\operatorname{Cl}_3(x)+\zeta(3)$

Sometimes it´s useful to integrate the Clausen function:

$\int_0^x\operatorname{Cl}_{2m}(t)\,dt=\zeta(2m+1)-\operatorname{Cl}_{2m+1}(x)$ (4)

Proof:

$\begin{aligned} \int_0^x\operatorname{Cl}_{2m}(t)\,dt&=\int_0^x \sum_{n=1}^\infty\frac{\sin(nt)}{n^{2m}}\,dt\\ &=\sum_{n=1}^\infty\frac{1}{n^{2m}}\int_0^x \sin(nt)\,dt\\ &=\sum_{n=1}^\infty\frac{1}{n^{2m}}\left( -\frac{\cos(nt)}{n}\Bigg|_0^x\right)\\ &=\sum_{n=1}^\infty\frac{1}{n^{2m+1}}-\sum_{n=1}^\infty\frac{\cos(nx)}{n^{2m+1}}\\ &=\zeta(2m+1)-\operatorname{Cl}_{2m+1}(x) \qquad \blacksquare \end{aligned}$

$\int_0^x\operatorname{Cl}_{2m+1}(t)\,dt=\operatorname{Cl}_{2m+2}(x)$ (5)

Proof:

$\begin{aligned} \int_0^x\operatorname{Cl}_{2m+1}(t)\,dt&=\int_0^x \sum_{n=1}^\infty\frac{\cos(nt)}{n^{2m+1}}\,dt\\ &=\sum_{n=1}^\infty\frac{1}{n^{2m+1}}\int_0^x \cos(nt)\,dt\\ &=\sum_{n=1}^\infty\frac{1}{n^{2m+1}}\left( \frac{\sin(nt)}{n}\Bigg|_0^x\right)\\ &=\sum_{n=1}^\infty\frac{\sin(nx)}{n^{2m+2}}\\ &=\operatorname{Cl}_{2m+2}(x) \qquad \blacksquare \end{aligned}$

From equation (2) we can derive a duplication formula for $\operatorname{Cl}_2(x)$

$\operatorname{Cl}_2(2x)=2\operatorname{Cl}_2(x)-2\operatorname{Cl}_2(\pi-x)$ (6)

Proof:

$\begin{aligned} \operatorname{Cl}_2(2x)&=-\int_0^{2x}\ln\left(2 \sin\left( \frac{t}{2}\right) \right)\,dt\\ &=-2\int_0^{x}\ln\left(2 \sin\left( t\right) \right)\,dt\\ &=-2\int_0^{x}\ln\left(4 \sin\left( \frac{t}{2}\right)\cos\left( \frac{t}{2}\right) \right)\,dt\\ &=-2\int_0^x\ln\left(2 \sin\left( \frac{t}{2}\right) \right)\,dt-2\int_0^x\ln\left(2 \cos\left( \frac{t}{2}\right) \right)\,dt\\ &=2\operatorname{Cl}_2(x)-4\int_0^{x/2}\ln\left(2 \cos\left(t\right) \right)\,dt\\ &=2\operatorname{Cl}_2(x)-4\int_{\frac{\pi-x}{2}}^{\pi/2}\ln\left(2 \sin\left(t\right) \right)\,dt\\ &=2\operatorname{Cl}_2(x)-4\int_{0}^{\pi/2}\ln\left(2 \sin\left(t\right) \right)\,dt+4\int_{0}^{\frac{\pi-x}{2}}\ln\left(2 \sin\left(t\right) \right)\,dt\\ &=2\operatorname{Cl}_2(x)-2\int_{0}^{\pi}\ln\left(2 \sin\left(\frac{t}{2}\right) \right)\,dt+2\int_{0}^{\pi-x}\ln\left(2 \sin\left(\frac{t}{2}\right) \right)\,dt\\ &=2\operatorname{Cl}_2(x)+2\operatorname{Cl}_2(\pi)-2\operatorname{Cl}_2(\pi-x)\\ &=2\operatorname{Cl}_2(x)-2\operatorname{Cl}_2(\pi-x) \qquad \blacksquare\\ \end{aligned}$

the same procedure we can obtain a duplication formula for $\operatorname{Cl}_3(x)$

$\operatorname{Cl}_3(2x)=4\operatorname{Cl}_3(x)+4\operatorname{Cl}_3(\pi-x)$ (7)

Proof:

Integrating both sides of (6) form 0 to x we have:

$\int_0^x\operatorname{Cl}_2(2t)\,dt=2\int_0^x\operatorname{Cl}_2(t)\,dt-2\int_0^x\operatorname{Cl}_2(\pi-t)\,dt$

Lets evaluate each of these integrals separately

$\begin{aligned} \int_0^x \operatorname{Cl}_2(2t)\,dt&=\frac12\int_{0}^{2x} \operatorname{Cl}_2(t)\,dt\\ &=\frac12\left(\zeta(3)- \operatorname{Cl}_3(2x)\right) \qquad \blacksquare \end{aligned}$

$\begin{aligned} \int_0^x \operatorname{Cl}_2(t)\,dt&=\zeta(3)-\operatorname{Cl}_3(x) \qquad \blacksquare \end{aligned}$

$\begin{aligned} \int_0^x \operatorname{Cl}_2(\pi-t)\,dt&=\int_{\pi-x}^{\pi} \operatorname{Cl}_2(t)\,dt\\ &=\int_0^\pi \operatorname{Cl}_2(t)\,dt-\int_0^{\pi-x} \operatorname{Cl}_2(t)\,dt\\ &=\zeta(3)-\operatorname{Cl}_3(\pi)-\left(\zeta(3)-\operatorname{Cl}_3(\pi-x) \right)\\ &=-\sum_{n=1}^\infty\frac{\cos(n\pi)}{n^3}+\operatorname{Cl}_3(\pi-x) \right)\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^3}+\operatorname{Cl}_3(\pi-x) \right)\\ &=\eta(3)+\operatorname{Cl}_3(\pi-x) \right)\\ &=\frac34\zeta(3)+\operatorname{Cl}_3(\pi-x) \right) \qquad \blacksquare\\ \end{aligned}$

Putting all together

$\begin{aligned} &\frac12\left(\zeta(3)- \operatorname{Cl}_3(2x)\right)=2\left( \zeta(3)-\operatorname{Cl}_3(x)\right)-2\left( \frac34\zeta(3)+\operatorname{Cl}_3(\pi-x) \right)\right)\\ &\operatorname{Cl}_3(2x)=4\operatorname{Cl}_3(x)+4\operatorname{Cl}_3(\pi-x) \qquad \blacksquare \end{aligned}$

The general formula is given by:

$\mathrm{Cl}_{m+1}(2 \theta)=2^{m}\left[\mathrm{Cl}_{m+1}(\theta)+(-1)^{m} \mathrm{Cl}_{m+1}(\pi-\theta)\right]$ (8)

First note that (very easy to prove just by expanding the R.H.S.):

$2\sum_{k=1}^\infty a_{2k}=\sum_{k=1}^\infty a_{k}+\sum_{k=1}^\infty (-1)^ka_{k}$ (9)

Then

$\begin{aligned} 2\sum_{k=1}^\infty \frac{\cos\left(2x k \right)}{(2k)^{2n+1}}&=\frac{1}{2^{2n}}\sum_{k=1}^\infty \frac{\cos\left(2x k \right)}{k^{2n+1}}\\ &=\frac{1}{2^{2n}}\mathrm{Cl}_{2n+1}(2 x) \end{aligned}$ (10)

On the other hand

$\begin{aligned} \sum_{k=1}^\infty \frac{\cos\left(x k \right)}{k^{2n+1}}+\sum_{k=1}^\infty \frac{\cos\left( k(\pi-x) \right)}{k^{2n+1}} &=\sum_{k=1}^\infty \frac{\cos\left(x k \right)}{k^{2n+1}}+\sum_{k=1}^\infty \frac{(-1)^k\cos\left( k x \right)}{k^{2n+1}}\\ &=\mathrm{Cl}_{2n+1}( x)+\mathrm{Cl}_{2n+1}\left( \pi-x\right) \end{aligned}$ (11)

By (9) we can equate (10) and (11)

$\frac{1}{2^{2n}}\mathrm{Cl}_{2n+1}(2 x)=\mathrm{Cl}_{2n+1}( x)+\mathrm{Cl}_{2n+1}\left( \pi-x\right)$ (12)

Proving the duplication formula for odd indices.

For even indices we differentiate the above equation w.r. to x to obtain

$\begin{aligned} &\frac{d}{dx}\left(\frac{1}{2^{2n}}\mathrm{Cl}_{2n+1}(2 x)=\mathrm{Cl}_{2n+1}( x)+\mathrm{Cl}_{2n+1}\left( \pi-x\right)\right)\\ &\frac{1}{2^{2n}} \sum_{k=1}^\infty \frac{-2k\sin(2kx)}{k^{2n+1}}=\sum_{k=1}^\infty \frac{-k\sin(kx)}{k^{2n+1}}+\sum_{k=1}^\infty \frac{-k\sin(k(\pi-x))}{k^{2n+1}}\\ &\mathrm{Cl}_{2n}(2 x)=2^{2n-1}\left(\mathrm{Cl}_{2n}( x)-\mathrm{Cl}_{2n}\left( \pi-x\right)\right) \qquad \blacksquare \end{aligned}$

A useful formula

Recall the triplication formula for the Gamma function (see here)

$\Gamma(3 x)=\frac{3^{3 x-1 / 2}}{2 \pi} \Gamma(x) \Gamma\left(x+\frac{1}{3}\right) \Gamma\left(x+\frac{2}{3}\right)$ (13)

Taking Logs on both sides of (13) we obtain

$\ln \Gamma(3 x)=\left(3 x-\frac{1}{2}\right) \ln 3-\ln 2 \pi+\ln \Gamma(x)+\ln \Gamma\left(x+\frac{1}{3}\right)+\ln \Gamma\left(x+\frac{2}{3}\right)$

Differentiating w.r. to x

$3 \psi(3 x)=3 \ln 3+\psi(x)+\psi\left(x+\frac{1}{3}\right)+\psi\left(x+\frac{2}{3}\right)$

If we keep that process we may obtain

$3^{2n+1} \psi^{(2n)}(3 x)=\psi^{(2n)}(x)+\psi^{(2n)}\left(x+\frac{1}{3}\right)+\psi^{(2n)}\left(x+\frac{2}{3}\right)$ (14)

Letting $x=\frac13$ in (14) we obtain

$(3^{2n+1} -1)\psi^{(2n)}(1)=\psi^{(2n)}\left(\frac13\right)+\psi^{(2n)}\left(\frac{2}{3}\right)$

$\psi^{(2n)}\left(\frac13\right)+\psi^{(2n)}\left(\frac{2}{3}\right)=(2n!)(1-3^{2n+1} )\zeta(2n+1)$ (15)

Now let´s focus on the Clauset function

Claim:

$\operatorname{Cl}_{2n+1}\left(\frac{2 \pi}{3}\right)=-\frac12\left(1-3^{-2n} \right)\zeta(2n+1)$ (16)

Proof:

$\begin{aligned} \operatorname{Cl}_{2n+1}\left(\frac{2 \pi}{3}\right)&=\sum_{k=1}^\infty\frac{\cos\left(\frac{2 k \pi}{3}\right)}{k^{2n+1}}\\ &=\sum_{k=1}^\infty \frac{1}{(3k)^{2n+1}}-\frac12\left(\sum_{k=0}^\infty \frac{1}{(3n+1)^{2n+1}}+\sum_{k=0}^\infty \frac{1}{(3n+2)^{2n+1}} \right)\\ &=\frac{1}{3^{2n+1}}\zeta(2n+1)-\frac12 \cdot \frac{1}{3^{2n+1}}\left(\sum_{k=0}^\infty \frac{1}{\left(n+\frac13\right)^{2n+1}}+\sum_{k=0}^\infty \frac{1}{\left(n+\frac23\right)^{2n+1}} \right)\\ &=\frac{1}{3^{2n+1}}\left(\zeta(2n+1)-\frac12\cdot\frac{1}{(2n!)(-1)^{2n+1}}\left( \psi^{(2n)}\left(\frac13\right)+\psi^{(2n)}\left(\frac{2}{3}\right)\right)\\ &=\frac{1}{3^{2n+1}}\left(\zeta(2n+1)-\frac12\cdot\frac{1}{(2n!)(-1)^{2n+1}}\left((2n!)(1-3^{2n+1} )\zeta(2n+1) \right) &(\text{by eq.(3) above})\\ &=\frac{1}{3^{2n+1}}\left(\zeta(2n+1)+\frac12\left((1-3^{2n+1} )\zeta(2n+1) \right)\\ &=-\frac12\left(1-3^{-2n} \right)\zeta(2n+1) \qquad \blacksquare \end{aligned}$

From the duplication formula (8)

$\operatorname{Cl}_{2n+1}\left(2x\right)=2^{2n}\left( \operatorname{Cl}_{2n+1}\left(x\right)\right)+(-1)^{2n}\operatorname{Cl}_{2n+1}\left(\pi-x\right)\right)$ (17)

Letting $x=\frac{\pi}{3}$ in (17) we obtain

$\operatorname{Cl}_{2n+1}\left(\frac{\pi}{3}\right)=-(1-2^{-2n})\operatorname{Cl}_{2n+1}\left(\frac{2\pi}{3}\right)$ (18)

Plugging (16) in (18) we obtain

$\operatorname{Cl}_{2n+1}\left(\frac{ \pi}{3}\right)=\frac12(1-2^{-2n})\left(1-3^{-2n} \right)\zeta(2n+1)$ (19)

Now recall the identity (proved here)

$\frac{\arcsin\left(\frac{x}{2} \right)}{\sqrt{1-\left( \frac{x}{2}\right)^2}}=\sum_{n=1}^\infty\frac{x^{2n-1}}{n\binom{2n}{n}}$ (20)

Let $x \to \sqrt{a}x$ , to obtain:

$\frac{\sqrt{a}\arcsin\left(\frac{\sqrt{a}x}{2} \right)}{\sqrt{1-\left( \frac{\sqrt{a}x}{2}\right)^2}}=\sum_{n=1}^\infty\frac{a^n x^{2n-1}}{n\binom{2n}{n}}$ (21)

Now recall the following integral representation (see here a proof)

$\frac{1}{n^k}=\frac{1}{\Gamma(k)}\int_0^1 \left(-\ln(x) \right)^{k-1}x^{n-1}\,dx$ (22)

I want to show that

$\sum_{n=1}^\infty\frac{ a^n}{\binom{2n}{n}n^k}=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{2\operatorname{arcsin}\left(\frac{\sqrt{a}}{2} \right)}x\ln^{k-2}\left(\frac{2}{\sqrt{a}}\sin\left( \frac{x}{2}\right) \right)\,dx$ (23)

Proof:

$\begin{aligned} \sum_{n=1}^\infty\frac{ a^n}{\binom{2n}{n}n^k}&=\frac{(-1)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ a^n}{\binom{2n}{n}}\int_0^1 \ln^{k-1}(x) x^{n-1}\,dx\\ &=\frac{2(-1)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ a^n}{\binom{2n}{n}}\int_0^1 \ln^{k-1}\left(x^2\right) x^{2n-1}\,dx & \left(x \to x^2\right)\\ &=\frac{2(-2)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ a^n}{\binom{2n}{n}}\int_0^1 \ln^{k-1}\left(x\right) x^{2n-1}\,dx \\ &=\frac{2(-2)^{k-1}}{(k-1)!}\sum_{n=1}^\infty\frac{ a^n}{\binom{2n}{n}}\left(\frac{x^{2n}\ln^{k-1}(x)}{2n}\Bigg|_0^1-\frac{(k-1)}{2n}\int_0^1 \ln^{k-2}\left(x\right) x^{2n-1}\,dx \right)\\ &=-\frac{(-2)^{k-1}}{(k-2)!}\sum_{n=1}^\infty\frac{ a^n}{\binom{2n}{n}n}\int_0^1 \ln^{k-2}\left(x\right) x^{2n-1}\,dx \\ &=-\frac{(-2)^{k-1}}{(k-2)!}\int_0^1 \ln^{k-2}\left(x\right)\left(\sum_{n=1}^\infty\frac{a^n x^{2n-1}}{\binom{2n}{n}n} \right) \,dx \\ &=-\frac{(-2)^{k-1}}{(k-2)!}\int_0^1 \ln^{k-2}\left(x\right)\left(\frac{\sqrt{a} \arcsin\left(\frac{\sqrt{a}x}{2} \right)}{\sqrt{1-\left( \frac{\sqrt{a}x}{2}\right)^2}} \right) \,dx & \left( \text{by eq. (2)}\right)\\ &=\frac{(-2)^{k-2}\,a}{(k-2)!}\int_0^{\frac{2}{\sqrt{a}}\operatorname{arcsin}\left(\frac{\sqrt{a}}{2} \right)} \frac{x\ln^{k-2}\left(\frac{2}{\sqrt{a}}\sin\left(\frac{\sqrt{a}x}{2}\right)\right) \cos\left(\frac{\sqrt{a}x}{2} \right)}{\sqrt{1-\sin^2\left( \frac{\sqrt{a}x}{2}\right)}} \,dx & \left( \frac{\sqrt{a}x}{2} \to \sin\left(\frac{\sqrt{a}x}{2} \right)\right)\\ &=\frac{(-2)^{k-2}}{(k-2)!}\int_0^{2\operatorname{arcsin}\left(\frac{\sqrt{a}}{2} \right)}x\ln^{k-2}\left(\frac{2}{\sqrt{a}}\sin\left( \frac{x}{2}\right) \right)\,dx & (\sqrt{a}x \to x) \end{aligned}$

Letting a=2 and k=3 in (23), and taking the fact that $\operatorname{arcsin}\left(\frac{\sqrt{2}}{2} \right)=\frac{\pi}{4}$ We obtain:

$\sum_{n=1}^\infty\frac{ 2^n}{\binom{2n}{n}n^3}=\frac{\pi^2\ln(2)}{8}+\pi \beta(2)-\frac{35}{16}\zeta(3)$ (24)

Proof:

$\begin{aligned} \sum_{n=1}^\infty\frac{ 2^n}{\binom{2n}{n}n^3}&=-2\int_0^{\pi/2}x\ln\left(\frac{2}{\sqrt{2}}\sin\left( \frac{x}{2}\right) \right)\,dx\\ &=\ln(2)\int_0^{\pi/2}x\,dx-2\int_0^{\pi/2}x\ln\left(2\sin\left( \frac{x}{2}\right) \right)\,dx\\ &=\frac{\pi^2\ln(2)}{8}-2\left(-\frac{\pi}{2}\operatorname{Cl}_2\left(\frac{\pi}{2}\right)-\operatorname{Cl}_3\left(\frac{\pi}{2}\right)+\zeta(3) \right) & (\text{by eq.(3) })\\ &=\frac{\pi^2\ln(2)}{8}+\pi \beta(2)+2\sum_{n=1}^\infty\frac{\cos\left(\frac{ \pi n}{2} \right)}{n^3}-2\zeta(3) \\ &=\frac{\pi^2\ln(2)}{8}+\pi \beta(2)-2\zeta(3) +2\sum_{n=1}^\infty\frac{(-1)^n}{(2n)^3}\\ &=\frac{\pi^2\ln(2)}{8}+\pi \beta(2)-2\zeta(3) +\frac14 \eta(3)\\ &=\frac{\pi^2\ln(2)}{8}+\pi \beta(2)-2\zeta(3)-\frac{3}{16}\zeta(3) \\ &=\frac{\pi^2\ln(2)}{8}+\pi \beta(2)-\frac{35}{16}\zeta(3) \qquad \blacksquare\\ \end{aligned}$

Letting a=1 and k=3 in (23), and taking the fact that $\operatorname{arcsin}\left(\frac{1}{2} \right)=\frac{\pi}{6}$

$\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = \frac{ \pi \sqrt{3}}{18} \Bigg[ \psi^{\prime} \left(\frac{1}{3} \right) - \psi^{\prime} \left(\frac{2}{3} \right) \Bigg]- \frac{4}{3} \zeta(3)$

Proof:

$\begin{aligned} \sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} &= -2\int_0^{\pi/3}x\ln\left(2\sin\left( \frac{x}{2}\right) \right)\,dx\\ &=\frac{2 \pi}{3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)+2\operatorname{Cl}_2\left(\frac{\pi}{3}\right)-2 \zeta(3) & (\text{by eq.(3)})\\ &= \frac{2 \pi}{3}\left(\frac{1}{4\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right) \right) + 2 \left(\frac{\zeta(3)}{3} \right) - 2 \zeta(3)\\ &=\frac{ \pi \sqrt{3}}{18} \Bigg[ \psi^{\prime} \left(\frac{1}{3} \right) - \psi^{\prime} \left(\frac{2}{3} \right) \Bigg]- \frac{4}{3} \zeta(3) \qquad \blacksquare \end{aligned}$

Where we used that

$\operatorname{Cl}_2\left(\frac\pi3\right)=\frac{1}{4\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)$

Proof:

$\begin{aligned} \operatorname{Cl}_2\left(\frac{2 \pi}{3} \right)&=\sum_{n=1}^\infty\frac{\sin\left(\frac{2 n \pi}{3} \right)}{n^2}\\ &=\frac{\sin\left(\frac{2 \pi}{3} \right)}{1^2}+\frac{\sin\left(\frac{4 \pi}{3} \right)}{2^2}+\frac{\sin\left(\frac{6 \pi}{3} \right)}{3^2}+\frac{\sin\left(\frac{8 \pi}{3} \right)}{4^2}+\frac{\sin\left(\frac{10\pi}{3} \right)}{5^2}+\frac{\sin\left(\frac{12 \pi}{3} \right)}{6^2}+\cdots\\ &=\frac{\left(\frac{\sqrt{3}}{2}\right)}{1}+\frac{\left(-\frac{\sqrt{3}}{2}\right)}{4}+\frac{0}{9}+\frac{\left(\frac{\sqrt{3}}{2}\right)}{16}+\frac{\left(-\frac{\sqrt{3}}{2}\right)}{25}+\frac{0}{36}+\cdots\\ &=\frac{\sqrt{3}}{2}\sum_{k=0}^\infty \frac{1}{(3k+1)^2}-\frac{\sqrt{3}}{2}\sum_{k=0}^\infty \frac{1}{(3k+2)^2}\\ &=\frac{\sqrt{3}}{18}\sum_{k=0}^\infty \frac{1}{\left(k+\frac13\right)^2}-\frac{\sqrt{3}}{18}\sum_{k=0}^\infty \frac{1}{\left(k+\frac{2}{3}\right)^2}\\ &=\frac{1}{6\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)\\ \end{aligned}$

setting $x=\frac\pi3$ in (6) we obtain

$\operatorname{Cl}_2\left(\frac\pi3\right)=\frac32\operatorname{Cl}_2\left(\frac{2\pi}3\right)$

Since

$\operatorname{Cl}_2\left(\frac{2 \pi}{3} \right)=\frac{1}{6\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)$

We conclude that

$\operatorname{Cl}_2\left(\frac\pi3\right)=\frac{1}{4\sqrt{3}}\left(\psi^\prime\left(\frac13 \right)-\psi^\prime\left(\frac23 \right)\right)$

And that

$\operatorname{Cl}_3\left(\frac{\pi}{3}\right)=\frac{1}{2}\left(1-2^{-2}\right)\left(1-3^{-2}\right)\zeta(3)=\frac{1}{3}\zeta(3)$

Letting n=1 in equation (19).

Search This Blog

MY MATEX NOTES