# A sum involving the central binomial coefficients

#### Random Variable

##### Well-known member
MHB Math Helper
Wolfram MathWorld states that $$\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = \frac{ \pi \sqrt{3}}{18} \Big[ \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{2}{3} \right) \Big]- \frac{4}{3} \zeta(3)$$

where $\psi_{1}(x)$ is the trigamma function.

But I can't get my answer in that form.

Using the Taylor expansion $\displaystyle \arcsin^{2}(x) = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2} \binom{2n}{n}} (2x)^{2n}$,

$$\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = 4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \ dx$$

Then integrating by parts

$$4 \int_{0}^{\frac{1}{2}} \frac{\arcsin^{2}(x)}{x} \ dx = \frac{\pi^{2}}{9} \ln \left(\frac{1}{2} \right) - 8 \int_{0}^{\frac{1}{2}} \frac{\arcsin (x) \ln (x)}{\sqrt{1-x^{2}}} \ dx$$

$$= - \frac{\pi^{2}}{9} \ln 2 - 8 \int_{0}^{\frac{\pi}{6}} u \ln (\sin u ) \ du = - 8 \ln 2 \int_{0}^{\frac{\pi}{6}} u \ du - 8\int_{0}^{\frac{\pi}{6}} u \ln (\sin u ) \ du$$

$$= - 8 \int_{0}^{\frac{\pi}{6}} u \ln ( 2 \sin u ) \ du = -8 \ \text{Re} \int_{0}^{\frac{\pi}{6}} u \ln (1-e^{2iu}) \ du = 8 \ \text{Re} \int_{0}^{\frac{\pi}{6}} u \sum_{n=1}^{\infty} \frac{e^{2in u}}{n} \ du$$

$$= 8 \ \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\frac{\pi}{6}} u \cos (2nu) \ du = \frac{2 \pi}{3} \sum_{n=1}^{\infty} \frac{\sin (\frac{n \pi}{3})}{n^{2}} + 2 \sum_{n=1}^{\infty} \frac{\cos (\frac{n \pi}{3})}{n^{3}} - 2 \zeta(3)$$

$$= \frac{2 \pi}{3} \Bigg( \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+1)^{2}} + \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+2)^{2}} - \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+4)^{2}} - \frac{\sqrt{3}}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+5)^{2}} \Bigg)$$

$$+ \ 2 \Bigg( \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+1)^{3}} - \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+2)^{3}} - \sum_{n=0}^{\infty} \frac{1}{(6n+3)^{3}} - \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+4)^{3}}$$

$$+ \ \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{(6n+5)^{3}} + \sum_{n=1}^{\infty} \frac{1}{(6n)^{3}} \Bigg) - 2 \zeta(3)$$

$$= \frac{\pi \sqrt{3}}{108} \Bigg( \psi_{1} \left(\frac{1}{6} \right) + \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{2}{3} \right) - \psi_{1}\left(\frac{5}{6} \right) \Bigg) + \frac{1}{432} \Bigg( - \psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) -28 \zeta(3)$$

$$+ \psi_{2} \left(\frac{2}{3} \right) - \psi_{2} \left(\frac{5}{6} \right) + 4 \zeta(3)\Bigg) - 2 \zeta (3)$$

EDIT:

Using the duplication formula for the trigamma function,

$$\psi_{1} \left(\frac{1}{6} \right) + \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{2}{3} \right) - \psi_{1}\left(\frac{5}{6} \right) = 4 \psi_{1} \left(\frac{1}{3} \right) -\psi_{1} \left(\frac{2}{3} \right) + \psi_{1} \left(\frac{1}{3} \right) -\psi_{1} \left(\frac{2}{3} \right) - 4 \psi_{1} \left(\frac{2}{3} \right) + \psi_{1} \left(\frac{1}{3} \right)$$

$$= 6 \psi_{1} \left(\frac{1}{3} \right) - 6 \psi_{1} \left(\frac{2}{3} \right)$$

Then

$$\sum_{n=1}^{\infty} \frac{1}{n^{3} \binom{2n}{n}} = \frac{\sqrt{3} \pi}{18} \Bigg( \psi_{1} \left(\frac{1}{3} \right) - \psi_{1} \left(\frac{2}{3} \right) \Bigg) + \frac{1}{432} \Bigg( - \psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) - \psi_{2} \left(\frac{5}{6} \right) -24 \zeta(3)\Bigg)$$

$$- 2 \zeta (3)$$

So what I need to show is that

$$-\psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) - \psi_{2} \left(\frac{5}{6} \right) = 312 \zeta(3)$$

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#### DreamWeaver

##### Well-known member
Hi RV!

When you get to the log-trig integral form, I'd recommend converting it into the form:

$$\displaystyle \int_0^{\pi/6}y\log(2\sin y)\,dy=\frac{1}{4}\int_0^{\pi/3}x\log\left(2\sin\frac{x}{2}\right)\,dx$$

$$\displaystyle -\log\left(2\sin\frac{x}{2}\right)=\frac{d}{dx} \text{Cl}_2(x)$$

And, in terms of Clausen Functions,

$$\displaystyle \int_0^{\theta}x\log\left(2\sin\frac{x}{2}\right)\,dx=\zeta(3)-\theta\text{Cl}_2(\theta)-\text{Cl}_3(\theta)$$

$$\displaystyle \Rightarrow$$

$$\displaystyle \int_0^{\pi/3}x\log\left(2\sin\frac{x}{2}\right)\,dx=\zeta(3)-\frac{\pi}{3}\text{Cl}_2\left(\frac{\pi}{3}\right)-\text{Cl}_3\left(\frac{\pi}{3}\right)$$

Now the polygamma part:

$$\displaystyle \psi_1\left(\frac{1}{3}\right)=\frac{2\pi^2}{3}+3 \sqrt{3}\text{Cl}_2\left(\frac{2\pi}{3}\right)$$

$$\displaystyle \psi_1\left(\frac{2}{3}\right)=\frac{2\pi^2}{3}-3 \sqrt{3}\text{Cl}_2\left(\frac{2\pi}{3}\right)$$

So

$$\displaystyle \psi_1\left(\frac{1}{3}\right)- \psi_1\left(\frac{2}{3}\right)=6 \sqrt{3}\text{Cl}_2\left(\frac{2\pi}{3}\right)$$

Now apply the duplication formula for the Clausen function (post #3 here --> Clausen functions (and related series, functions, integrals) : Tutorials ):

$$\displaystyle \text{Cl}_2(2\theta)=2\, \text{Cl}_2(\theta)-2\, \text{Cl}_2(\pi-\theta)$$

Set $$\displaystyle \theta = \pi/3\,$$ to obtain

$$\displaystyle \text{Cl}_2\left(\frac{2\pi}{3}\right)=\frac{2}{3}\text{Cl}_2\left(\frac{\pi}{3}\right)$$

So

$$\displaystyle \psi_1\left(\frac{1}{3}\right)- \psi_1\left(\frac{2}{3}\right)=6 \sqrt{3}\text{Cl}_2\left(\frac{2\pi}{3}\right)=4 \sqrt{3}\text{Cl}_2\left(\frac{\pi}{3}\right)$$

Plugging this back into the log-trig integral gives

$$\displaystyle \int_0^{\pi/3}x\log\left(2\sin\frac{x}{2}\right)\,dx=\zeta(3)-\frac{\pi}{3}\text{Cl}_2\left(\frac{\pi}{3}\right)-\text{Cl}_3\left(\frac{\pi}{3}\right)=$$

$$\displaystyle \zeta(3)-\frac{\pi}{12\sqrt{3}}\left[\psi_1\left(\frac{1}{3}\right)- \psi_1\left(\frac{2}{3}\right)\right]-\text{Cl}_3\left(\frac{\pi}{3}\right)$$

The remaining, third-order Clausen function can treated similarly.

Writing in a bit of a rush, so hope I haven't made any silly errors here...

The full proof is long and involved, but some time, when I get times, I'll post details about how to evaluate all higher order polygamma functions - with rational arguments of 1/2, 1/3, 2/3, 1/4, 3/4, 1/6, and 5/6 - in terms of Clausen functions.

Gotta dash...

#### Random Variable

##### Well-known member
MHB Math Helper
By using the duplication formula for $\psi_{2}(x)$ multiple times I imagine it's possible to show that

$$-\psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) - \psi_{2} \left(\frac{5}{6} \right) = 312 \zeta(3)$$

Then my evaluation is complete without having used properties of the Clausen functions.

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#### DreamWeaver

##### Well-known member
Hi RV!

I'll need to double check this at some point, just to make sure I haven't made any silly mistakes, but even if so, you'll see that the methodology is sound...

Note: with the two main relations below, at the end, you can just as easily eliminate the polygamma functions with arguments 1/3 and 2/3, and thereby obtain precise formulae for arguments 1/6 and 5/6...

----------------------------------------------

To start with, let's consider the following particular Clausen function of (arbitrary) odd order:

$$\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos (\pi k/3)}{k^{2m+1}}$$

We want to split this into six sums, where the first sum contains the first of every six terms, the second contains the second of every six terms, and so on. We also change summation index so our new series start at k=0, rather than k=1 above.

$$\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos (\pi k/3)}{k^{2m+1}}=$$

$$\sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+1)}{(6k+1)^{2m+1}}+ \sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+2)}{(6k+2)^{2m+1}}+ \sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+3)}{(6k+3)^{2m+1}}+$$

$$\sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+4)}{(6k+4)^{2m+1}}+ \sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+5)}{(6k+5)^{2m+1}}+ \sum_{k=0}^{\infty}\frac{\cos\frac{\pi}{3}(6k+6)}{(6k+6)^{2m+1}}=$$

Simplify the trig term in each series:

$$\cos \frac{\pi}{3}(6k+n)=\cos\left(2\pi k+\frac{\pi n}{3}\right)=$$

$$\cos 2\pi k\cos\frac{\pi n}{3}-\sin 2\pi k\sin\frac{\pi n}{3}\equiv \cos\frac{\pi n}{3}$$

Our new sextet of series is thus

$$\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)=$$

$$\cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+1)^{2m+1}}+ \cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^{2m+1}}+$$

$$\cos\left(\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+3)^{2m+1}}+ \cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^{2m+1}}+$$

$$\cos\left(\frac{5\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+5)^{2m+1}}+ \cos\left(2\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+6)^{2m+1}}$$

Multiply both sides by $$6^{2m+1}\,$$, and then subtract the third and sixth series on the RHS from the Clausen term on the LHS (using $$\cos\pi = -1\,$$ and $$\cos 2\pi=1\,$$ ) to obtain:

$$6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\sum_{k=0}^{\infty}\frac{1}{(k+1/2)^{2m+1}}-\sum_{k=0}^{\infty}\frac{1}{(k+1)^{2m+1}}=$$

$$\cos\left(\frac{\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+1/6)^{2m+1}}+ \cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+1/3)^{2m+1}}+$$

$$\cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+2/3)^{2m+1}}+\cos\left(\frac{5\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(k+5/6)^{2m+1}}$$

Express the cosine terms on the RHS in real/rational form to make the RHS

$$\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+1/6)^{2m+1}}- \frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+1/3)^{2m+1}}$$

$$-\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+2/3)^{2m+1}}+\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(k+5/6)^{2m+1}}$$

Now use

$$\psi_{n\ge 1}(x)=(-1)^{n+1}n!\sum_{k=0}^{\infty}\frac{1}{(k+x)^{n+1}}$$

to re-write the RHS as:

$$\frac{1}{2}\left(\frac{(-1)^{2m}}{(2m)!}\right)\Bigg\{ \psi_{2m}\left( \frac{1}{6} \right) -\psi_{2m}\left( \frac{1}{3} \right) -\psi_{2m}\left( \frac{2}{3} \right) +\psi_{2m}\left( \frac{5}{6} \right) \Bigg\}=$$

$$\frac{1}{2\,(2m)!}\, \Bigg\{ \psi_{2m}\left( \frac{1}{6} \right) -\psi_{2m}\left( \frac{1}{3} \right) -\psi_{2m}\left( \frac{2}{3} \right) +\psi_{2m}\left( \frac{5}{6} \right) \Bigg\}$$

Next, apply the same process to the two series on the LHS (with the Clausen term):

$$6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\sum_{k=0}^{\infty}\frac{1}{(k+1/2)^{2m+1}}-\sum_{k=0}^{\infty}\frac{1}{(k+1)^{2m+1}}=$$

$$6^{2m+1}\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\frac{1}{(2m)!}\Bigg\{ \psi_{2m}\left(\frac{1}{2}\right)-\psi_{2m}(1)\Bigg\}$$

Multiplying BOTH sides by $$2\, (2m)! \,$$ then gives the identity

$$2 \, (2m)! \, 6^{2m+1} \text{Cl}_{2m+1} \left( \frac{\pi}{3}\right)+2 \psi_{2m}\left(\frac{1}{2}\right)-2 \psi_{2m}(1)=$$

$$\psi_{2m} \left( \frac{1}{6} \right) -\psi_{2m} \left( \frac{1}{3} \right) -\psi_{2m} \left( \frac{2}{3} \right) +\psi_{2m} \left( \frac{5}{6} \right)$$

----------------------------------------------------

Next, repeat ALL of the above, but this time in terms of the Clasuen function with argument $$2\pi/3\,$$

$$\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos (2\pi k/3)}{k^{2m+1}}=$$

$$\cos\left(\frac{2\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+1)^{2m+1}}+ \cos\left(\frac{4\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^{2m+1}}+$$

$$\cos\left(2\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+3)^{2m+1}}+ \cos\left(\frac{8\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^{2m+1}}+$$

$$\cos\left(\frac{10\pi}{3}\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+5)^{2m+1}}+ \cos\left(4\pi\right)\, \sum_{k=0}^{\infty}\frac{1}{(6k+6)^{2m+1}}=$$

$$-\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+1)^{2m+1}} -\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+2)^{2m+1}}+$$

$$\sum_{k=0}^{\infty}\frac{1}{(6k+3)^{2m+1}}- \frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+4)^{2m+1}}+$$

$$-\frac{1}{2}\, \sum_{k=0}^{\infty}\frac{1}{(6k+5)^{2m+1}}+ \sum_{k=0}^{\infty}\frac{1}{(6k+6)^{2m+1}}$$

Continue exactly as before, and you get the second relation

$$2\, (2m)! \, 6^{2m+1} \text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)-2 \psi_{2m}\left(\frac{1}{2}\right)-2 \psi_{2m}(1)=$$

$$-\psi_{2m}\left( \frac{1}{6} \right) -\psi_{2m}\left( \frac{1}{3} \right) -\psi_{2m}\left( \frac{2}{3} \right) -\psi_{2m}\left( \frac{5}{6} \right)$$

----------------------------------------------------

Relative to the arguments 1/3, 2/3, 1/6, and 5/6, the arguments 1 and 1/2 are pretty straightforward, so I'll simply state them now and prove them later.

$$\psi_{2m}(1)=-(2m)!\zeta(2m+1)$$

$$\psi_{2m}\left(\frac{1}{2}\right)=-(2m)!\,(2^{2m+1}-1)\zeta(2m+1)$$

Now, if you add the final forms or relation #1 and relation #2 you get:

$$2\, (2m)! \,6^{2m+1}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]-4\psi_{2m}(1)=$$

$$2\, (2m)! \,6^{2m+1}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+4\,(2m)!\zeta(2m+1)=$$

$$2\Bigg\{ \psi_{2m}\left( \frac{1}{3} \right) +\psi_{2m}\left( \frac{2}{3} \right)\Bigg\}$$

Or

$$(2m)! \,6^{2m+1}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+2\,(2m)!\zeta(2m+1)=$$

$$\psi_{2m}\left( \frac{1}{3} \right) +\psi_{2m}\left( \frac{2}{3} \right)$$

On the other hand, the reflection formula for the polygamma function gives:

$$\psi_{2m}(x)-\psi_{2m}(1-x)=\pi\frac{d^{2m}}{dx^{2m}}\cot\pi x$$

$$\Rightarrow$$

$$\psi_{2m}\left( \frac{1}{3} \right) -\psi_{2m}\left( \frac{2}{3} \right)=\pi\frac{d^{2m}}{dx^{2m}}\cot\pi x\,\Biggr|_{x=1/3}$$

So

$$\displaystyle \psi_{2m}\left( \frac{1}{3} \right)=\frac{(2m)! \,6^{2m+1}}{2}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+$$

$$\displaystyle (2m)!\zeta(2m+1)+\pi\frac{d^{2m}}{dx^{2m}}\cot\pi x\,\Biggr|_{x=1/3}$$

and

$$\displaystyle \psi_{2m}\left( \frac{2}{3} \right)=\frac{(2m)! \,6^{2m+1}}{2}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+$$

$$\displaystyle (2m)!\zeta(2m+1)-\pi\frac{d^{2m}}{dx^{2m}}\cot\pi x\,\Biggr|_{x=1/3}$$

#### DreamWeaver

##### Well-known member
And one final point...

Here's a cut 'n' paste of mine from t'other forum...

Using this duplication formula you express those previous sums of two Clausen functions as a single Clausen function, and thereby simplify your expressions of (even ordered) polygamma functions at 1/3, 2/3, 1/6, and 5/6....

------------------------------------------

Next, let's take the above duplication formula, replace $$\theta\,$$ with the variable x, and integrate both sides:

$$\int_0^{ \varphi}\text{Cl}_{2m}(2x)\,dx=2^{2m-1}\left[\int_0^{ \varphi}\text{Cl}_{2m}(x)\,dx-\int_0^{ \varphi}\text{Cl}_{2m}(\pi-x)\,dx\right]$$

The L.H.S. is equal to

$$\sum_{k=1}^{\infty}\frac{1}{k^{2m}}\,\int_0^{ \varphi}\sin 2kx\,dx=-\frac{1}{2}\,\sum_{k=1}^{\infty}\frac{1}{k^{2m+1}}\Big[\cos 2kx\Big]_0^{\varphi}=$$

$$-\frac{1}{2}\,\sum_{k=1}^{\infty}\frac{\cos 2k\varphi}{k^{2m+1}}+\frac{1}{2}\,\sum_{k=1}^{\infty}\frac{1}{k^{2m+1}}=\frac{1}{2}\,[\zeta(2m+1)-\text{Cl}_{2m+1}(2\varphi)]$$

Whereas the difference of the two integrals on the R.H.S. is

$$2^{2m-1}\,\sum_{k=1}^{\infty}\frac{1}{k^{2m}}\int_0^{\varphi}\left[\sin kx-\sin k(\pi-x)\right]\,dx=$$

$$2^{2m-1}\,\sum_{k=1}^{\infty}\frac{1}{k^{2m+1}}\Big[-\cos kx+\cos k(\pi-x)\Big]_0^{\varphi}=$$

$$2^{2m-1}\,\sum_{k=1}^{\infty}\frac{[\cos k(\pi-\varphi)-\cos k\varphi]}{k^{2m+1}}=2^{2m-1}[\text{Cl}_{2m+1}(\pi-\varphi)-\text{Cl}_{2m+1}(\varphi)]$$

We now have the duplication formula for a CL-type Clausen function of odd order:

Result 2:

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

#### Random Variable

##### Well-known member
MHB Math Helper
DreamWeaver

Correct me if I'm wrong, but basically what you did is generalize what I did for two particular cases.

But to complete the evaluation I need to either show that

$$\displaystyle Cl_{3} \left( \frac{\pi}{3} \right) = \frac{\zeta(3)}{3}$$

or that

$$-\psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) - \psi_{2} \left(\frac{5}{6} \right) = 312 \zeta(3)$$

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#### DreamWeaver

##### Well-known member
DreamWeaver

Correct me if I'm wrong, but basically what you did is generalize what I did for two particular cases.

Errrm... Yes and no. Sort of. My replies were a tad lengthy, so if you skim-read you might have missed the main result in reply #2:

$$\displaystyle \psi_{2m}\left( \frac{1}{3} \right)=\frac{(2m)! \,6^{2m+1}}{2}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+$$

$$\displaystyle (2m)!\zeta(2m+1)+\pi\frac{d^{2m}}{dx^{2m}}\cot\pi x\,\Biggr|_{x=1/3}$$

and

$$\displaystyle \psi_{2m}\left( \frac{2}{3} \right)=\frac{(2m)! \,6^{2m+1}}{2}\left[\text{Cl}_{2m+1}\left(\frac{\pi}{3}\right)+\text{Cl}_{2m+1}\left(\frac{2\pi}{3}\right)\right]+$$

$$\displaystyle (2m)!\zeta(2m+1)-\pi\frac{d^{2m}}{dx^{2m}}\cot\pi x\,\Biggr|_{x=1/3}$$

As I say, sort of the same and also sort of different. The main point I was getting at is that, by using the methods above, you can express any (single) Clausen function of argument $$\displaystyle \pi/3\,$$ or $$\displaystyle 2\pi/3\,$$ in terms of any single polygamma function, of associated order, with argument 1/3, 2/3, 1/6, or 5/6. The reverse is also true.

I was probably wandering a bit off-topic...

#### Random Variable

##### Well-known member
MHB Math Helper
Using the duplication formula for $\psi_{2}(x)$,

$$-\psi_{2} \left(\frac{1}{6} \right) + \psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left(\frac{2}{3} \right) - \psi_{2} \left(\frac{5}{6} \right) = -8 \psi_{2} \left(\frac{1}{3} \right) +\psi_{2} \left(\frac{2}{3} \right) + \psi_{2} \left(\frac{1}{3} \right)$$

$$+\psi_{2} \left(\frac{2}{3} \right) - 8 \psi_{2} \left(\frac{2}{3} \right) + \psi_{2} \left(\frac{1}{3} \right) = -6 \psi_{2} \left(\frac{1}{3} \right) - 6 \psi_{2} \left(\frac{2}{3} \right)$$

Then using the more general multiplication formula,

$$\psi_{2} \left(\frac{1}{3} \right) + \psi_{2} \left( \frac{2}{3} \right) + \psi_{2}(1) = 27 \psi_{2} (1)$$

$$\implies -6 \psi_{2} \left(\frac{1}{3} \right) - 6 \psi_{2} \left(\frac{2}{3} \right) = -156 \psi_{2} (1) = 312 \zeta(3)$$

#### DreamWeaver

##### Well-known member
Nicely done, Sir!