SOME SPECIAL VALUES OF POLYGAMMA FUNCTIONS

Special values of of Polygamma Function

We have already seen from previous posts that many infinite series and integrals can be evaluated with techniques that relies on Gamma and Polygamma functions. In view of this fact I decided to proof a few special values for some of these functions to serve primarily as a reference when we encounter them in future computations. It will be a pretty long post but lots of techniques involved, therefore a fun one. Needless to mention their own beauty with expressions involving lots of well known mathematical constants, justifying their derivation just for their own sake.

In previous posts we have proved the following relations for the Gamma Function:

The recurrence equation of the Gamma Function

$\Gamma\left(x+1 \right)=x\Gamma\left(x \right)$ (1)

The reflection formula

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

The duplication formula

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

The triplication formula

$\Gamma (3x)=\frac{3^{3x-1/2}}{2 \pi}\Gamma(x)\Gamma \left(x+\frac{1}{3}\right)\Gamma\left(x+\frac{2}{3}\right)$ (4)

Additionally, we have already proved the following infinite series representation for the Digamma function

$\psi(x+1) &=-\gamma+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+x}\right)$ (5)

Which gives rise the following integral representation:

$\psi(x+1) &=-\gamma+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+x}\right)$

$=-\gamma+\sum_{n=1}^{\infty} \int_{0}^{1}\left(t^{n-1}-t^{n+x-1}\right) d t$

$=-\gamma+\int_{0}^{1} \sum_{n=1}^{\infty}\left(t^{n-1}-t^{n+x-1}\right) d t$

$\psi(x+1) &=-\gamma+\int_{0}^{1} \frac{1-t^{x}}{1-t} d t$ (6)

Or $x \longmapsto x-1$ in (6)

$\psi(x) &=-\gamma+\int_{0}^{1} \frac{1-t^{x-1}}{1-t} d t$ (7)

Equations (1) to (7) are the basic equations needed to find some special values of polygamma functions. From them, we will now derive another sets of relations required in the proofs.

If we take log on both sides of (1) and differentiate w.r. to we get

$\psi(x+1)=\frac{1}{x}+\psi(x)$ (8)

differentiating (8) with respect to x we obtain

$\psi^{\prime}(x+1)=-\frac{1}{x^2}+\psi^{\prime}(x)$ (9)

differentiating (9) with respect to x we obtain

$\psi^{\prime \prime}(x+1)=\frac{2}{x^3}+\psi^{\prime \prime}(x)$ (10)

Taking log on both sides of (2) and differentiate w.r. to we get

$\psi(x)-\psi(1-x)=-\pi \cot(\pi x)$ (11)

Differentiating (11) with respect to x we get

$\psi^{\prime}(x)+\psi^{\prime}(1-x)=\frac{\pi^2} {\sin^2(\pi x)}$ (12)

Differentiating (12) with respect to x we get

$\psi^{\prime \prime}(x)-\psi^{\prime \prime}(1-x)=-2\pi^3\frac{\cot(\pi x)} {\sin^2(\pi x)}$ (13)

Taking log on both sides of (3) and differentiate w.r. to we get

$\ln \Gamma\left(2x \right)=(2x-1)\ln 2 + \ln \Gamma\left( x\right) + \ln \Gamma\left( x+\frac{1}{2}\right) - \frac{1}{2} \ln \pi$ (14)

differentiating (14) with respect to x we get

$2 \psi \left(2x \right)=2 \ln 2+ \psi(x)+\psi \left( x+\frac{1}{2}\right)$ (15)

differentiating (15) with respect to x we get

$4\psi^{\prime} \left(2x \right)= \psi^{\prime}(x)+\psi^{\prime} \left( x+\frac{1}{2}\right)$ (16)

differentiating (16) with respect to x we get

$8\psi^{\prime \prime} \left(2x \right)= \psi^{\prime \prime}(x)+\psi^{\prime \prime} \left( x+\frac{1}{2}\right)$ (17)

Taking log on both sides of (4) and differentiate w.r. to we get

$\ln \Gamma\left(3x \right)=\left(3x-\frac{1}{2} \right) \ln 3- \ln 2\pi+ \ln \Gamma\left(x \right)+\ln \Gamma\left(x+\frac{1}{3} \right)+\ln \Gamma\left(x+\frac{2}{3} \right)$ (18)

$3\psi\left(3x \right)=3 \ln 3+\psi\left(x \right)+\psi\left(x+\frac{1}{3} \right)+\psi\left(x+\frac{2}{3} \right)$ (19)

differentiating (19) with respect to x we get

$9\psi^{\prime}\left(3x \right)= \psi^{\prime}\left(x \right)+\psi^{\prime}\left(x+\frac{1}{3} \right)+\psi^{\prime}\left(x+\frac{2}{3} \right)$ (20)

differentiating (20) with respect to x we get

$27\psi^{\prime \prime}\left(3x \right)= \psi^{\prime \prime}\left(x \right)+\psi^{\prime \prime}\left(x+\frac{1}{3} \right)+\psi^{\prime \prime}\left(x+\frac{2}{3} \right)$ (21)

differentiating (5) with respect to x we obtain

$\psi^{\prime}(x+1)=\sum_{n=1}^{\infty}\frac{1}{(n+x)^2}$ (22)

differentiating (22) with respect to x we obtain

$\psi^{\prime \prime}(x+1)=-2\sum_{n=1}^{\infty}\frac{1}{(n+x)^3}$ (23)

differentiating (7) with respect to x we obtain

$\psi^{\prime}(x) &=-\int_{0}^{1} \frac{t^{x-1} \ln t}{1-t} d t$ (24)

differentiating (24) with respect to x we obtain

$\psi^{\prime \prime}(x) &=-\int_{0}^{1} \frac{t^{x-1} \ln^2 t}{1-t} d t$ (25)

Special values of the Digamma function

Setting x=0 in (5) we immediately obtain

$\boxed{\psi(1)=-\gamma}$ (26)

To compute $\psi\left(\frac{1}{2}\right)$ , let´s do the following. Start with (7) and set $s=\frac{1}{2}$

$\psi\left(\frac{1}{2}\right) &=-\gamma+\int_{0}^{1} \frac{1-x^{-1/2}}{1-x} d x$

let x=t^2

$\psi\left(\frac{1}{2}\right) &=-\gamma+2\int_{0}^{1} \frac{1-u^{-1}}{1-u^2} u d u$

$\psi\left(\frac{1}{2}\right) &=-\gamma-2\int_{0}^{1} \frac{(1-u)}{(1-u)(1+u)} u d u$

$\psi\left(\frac{1}{2}\right) &=-\gamma-2\int_{0}^{1} \frac{1}{1+u} u d u$

$\boxed{\psi\left(\frac{1}{2}\right) &=-\gamma-2\ln(2)}$ (27)

To find $\psi\left(\frac{1}{3}\right)$ lets first start from (19) and set $x=\frac{1}{3}$

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

$\psi\left(\frac{1}{3} \right)+\psi\left(\frac{2}{3} \right)=-2\gamma-3 \ln 3$ (28)

On the other hand setting $x=\frac{1}{3}$ in (11)

$\psi\left(\frac{1}{3} \right)-\psi\left(\frac{2}{3} \right)=-\pi \cot\left(\frac{\pi}{3} \right)=-\frac{\pi}{\sqrt{3}}$ (29)

Adding (28) and (29) we get

$2\psi\left(\frac{1}{3} \right)=-\frac{\pi}{\sqrt{3}}-2\gamma-3 \ln 3$

$\boxed{\psi\left(\frac{1}{3} \right)=-\frac{\pi}{2\sqrt{3}}-\frac{3 \ln 3}{2}-\gamma}$ (30)

Subtracting (29) and (28) we get

$2\psi\left(\frac{2}{3} \right)=-2\gamma-3 \ln 3+\frac{\pi}{\sqrt{3}}$

$\boxed{\psi\left(\frac{2}{3} \right)=\frac{\pi}{2\sqrt{3}}-\frac{3 \ln 3}{2}-\gamma}$ (31)

An alternative way to compute $\psi\left(\frac{1}{3} \right)$ is through the integral representation (7) which is more involving. I left it´s computation in the appendix for those interested.

To find $\psi\left(\frac{1}{4}\right)$ lets first start from (15) and set $x=\frac{1}{4}$

$2 \psi \left(\frac{1}{2} \right)=2 \ln 2+ \psi\left(\frac{1}{4}\right)+\psi \left( \frac{3}{4}\right)$

From (27) we get

$\psi\left(\frac{1}{4}\right)+\psi \left( \frac{3}{4}\right)=-2\gamma-6\ln(2)$ (32)

On the other hand setting $x=\frac{1}{4}$ in (11)

$\psi\left(\frac{1}{4}\right)-\psi \left( \frac{3}{4}\right)=-\pi$ (33)

Adding (32) and (33) we get

$\boxed{\psi\left(\frac{1}{4}\right)=-\frac{\pi}{2}-3\ln(2)-\gamma}$ (34)

Subtracting (33) from (32) we get

$\boxed{\psi\left(\frac{3}{4}\right)=\frac{\pi}{2}-3\ln(2)-\gamma}$ (34)

Letting $x=\frac{1}{6}$ in (19)

$3\psi\left(\frac{1}{2} \right)=3 \ln 3+\psi\left(\frac{1}{6} \right)+\psi\left(\frac{1}{2} \right)+\psi\left(\frac{5}{6} \right)$

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

$\psi\left(\frac{1}{6} \right)+\psi\left(\frac{5}{6} \right)=-2\gamma-4\ln 2-3 \ln 3$ (35)

Letting $x=\frac{1}{6}$ in (11)

$\psi\left(\frac{1}{6} \right)-\psi\left(\frac{5}{6} \right)=-\pi \cot\left(\frac{\pi}{6} \right)$

$\psi\left(\frac{1}{6} \right)-\psi\left(\frac{5}{6} \right)=-\pi \sqrt{3}$ (36)

Adding (35) and (36) we get

$\boxed{\psi\left(\frac{1}{6}\right)=-\frac{\pi \sqrt{3}}{2}-2 \ln 2 -\frac{3 \ln 3}{2}-\gamma}$ (37)

Subtracting (33) and (32) we obtain

$\boxed{\psi\left(\frac{5}{6}\right)=\frac{\pi \sqrt{3}}{2}-2 \ln 2 -\frac{3 \ln 3}{2}-\gamma}$ (38)

Special Values of the Trigamma

letting x=0 in (22)

$\psi^{\prime}(1) =\sum_{n=1}^{\infty}\frac{1}{n^2}=\zeta(2)$

$\boxed{\psi^{\prime}(1) =\frac{\pi^2}{6}}$ (39)

For $\psi^{\prime}\left(\frac{1}{2}\right)$ set $s=\frac{1}{2}$ in (16)

$4\psi^{\prime}\left(1\right) =\psi^{\prime}\left(\frac{1}{2}\right)+\psi^{\prime}\left(\frac{1}{2}+\frac{1}{2}\right)$ (40)

$\psi^{\prime}\left(\frac{1}{2}\right)=3\psi^{\prime}\left(1\right)$

plugging (39) in (40) we get

$\boxed{\psi^{\prime}\left(\frac{1}{2}\right) =3\zeta(2)}$ (41)

To find $\psi^{\prime}\left( \frac{1}{3}\right)$ we start from (24)

$\psi^{\prime}\left( \frac{1}{3}\right)=-\int_0^1\frac{x^{\frac{1}{3}-1}\ln x}{1-x}dx$

let $x=t^3\, \Rightarrow \, dx=3t^{2}dt$

$\psi^{\prime}\left( \frac{1}{3}\right)=-9\int_0^1\frac{\ln t}{1-t^3}dt$

$\psi^{\prime}\left( \frac{1}{3}\right)=9\int_0^1\frac{\ln t}{t^3-1}dt$ (42)

On the other hand

$\psi^{\prime}\left( \frac{2}{3}\right)=-\int_0^1\frac{x^{-\frac{1}{3}}\ln x}{1-x}dx$

let x=t^3

$\psi^{\prime}\left( \frac{2}{3}\right)=-3\int_0^1\frac{t^{-1}\ln t^3}{1-t^3}t^2dt$

$\psi^{\prime}\left( \frac{2}{3}\right)=9\int_0^1\frac{t\ln t}{t^3-1}dt$ (43)

Subtracting (43) from (42)

$\psi^{\prime}\left( \frac{1}{3}\right)-\psi^{\prime}\left( \frac{2}{3}\right)=-9\int_0^1\frac{\left(t-1\right)\ln t}{t^3-1}dt$

we have

$t^3-1=\left(t-1 \right)\left(t^2+t+1 \right)$

$\Rightarrow \, \frac{t-1}{t^3-1}=\frac{1}{t^2+t+1 }$

Therefore

$\psi^{\prime}\left( \frac{1}{3}\right)-\psi^{\prime}\left( \frac{2}{3}\right)=-9\int_0^1\frac{\ln t}{t^2+t+1}dt$

$\int_0^1\frac{\ln t}{t^2+t+1}dt=-\frac{1}{9}\left(\psi^{\prime}\left( \frac{1}{3}\right)-\psi^{\prime}\left( \frac{2}{3}\right) \right)$ (44)

From the apendix we have

$Cl_{2}\left(\frac{2 \pi}{3} \right)=-\frac{\sqrt{3}}{2}\int_0^1\frac{\ln t}{t^2+t+1}dt$

Where $Cl_{2}\left(x\right)$ is the Clausen function

$\int_0^1\frac{\ln t}{t^2+t+1}dt=-\frac{2}{\sqrt{3}}Cl_{2}\left(\frac{2 \pi}{3} \right)$ (45)

Equating (44) with (45)

$-\frac{2}{\sqrt{3}}Cl_{2}\left(\frac{2 \pi}{3} \right)=-\frac{1}{9}\left(\psi^{\prime}\left( \frac{1}{3}\right)-\psi^{\prime}\left( \frac{2}{3}\right) \right)$

$Cl_{2}\left(\frac{2 \pi}{3} \right)=\frac{\sqrt{3}}{2 \cdot 3 \cdot 3}\left(\psi^{\prime}\left( \frac{1}{3}\right)-\psi^{\prime}\left( \frac{2}{3}\right) \right)$

$\boxed{\psi^{\prime}\left( \frac{1}{3}\right)-\psi^{\prime}\left( \frac{2}{3}\right)=6 \sqrt{3} \cdot Cl_{2}\left(\frac{2 \pi}{3} \right)}$ (46)

From (12), letting $x=\frac{1}{3}$ we obtain

$\psi^{\prime}\left( \frac{1}{3}\right)+\psi^{\prime}\left( \frac{2}{3}\right)=\frac{\pi^2}{\sin^2\left( \frac{\pi}{3}\right)}$

$\boxed{\psi^{\prime}\left( \frac{1}{3}\right)+\psi^{\prime}\left( \frac{2}{3}\right)=\frac{4\pi^2}{3}}$ (47)

Adding (46) and (47)

$2\psi^{\prime}\left( \frac{1}{3}\right)=\frac{4\pi^2}{3}+6 \sqrt{3} \cdot Cl_{2}\left(\frac{2 \pi}{3} \right)$

$\boxed{\psi^{\prime}\left( \frac{1}{3}\right)=\frac{2\pi^2}{3}+3 \sqrt{3} \cdot Cl_{2}\left(\frac{2 \pi}{3} \right)}$ (48)

Subtracting (47) from (46)

$2\psi^{\prime}\left( \frac{2}{3}\right)=\frac{4\pi^2}{3}-6 \sqrt{3} \cdot Cl_{2}\left(\frac{2 \pi}{3} \right)$

$\boxed{\psi^{\prime}\left( \frac{2}{3}\right)=\frac{2\pi^2}{3}-3 \sqrt{3} \cdot Cl_{2}\left(\frac{2 \pi}{3} \right)}$ (49)

To find $\psi^{\prime}\left( \frac{1}{4}\right)$ , we again use (24)

$\psi^{\prime}\left( \frac{1}{4}\right)=-\int_0^1\frac{x^{\frac{1}{4}-1}\ln x}{1-x}dx$

Let $x=t^4\, \Rightarrow \, dx=4t^3dt$ , then

$\psi^{\prime}\left( \frac{1}{4}\right)=16\int_0^1\frac{\ln t}{t^4-1}dt$

$=16\int_0^1\frac{\ln t}{(t^2+1)(t^2-1)}dt$

applying partial fractions

$\frac{1}{(t^2+1)(t^2-1)}=\frac12 \left(\frac1{t^2-1}-\frac{1}{t^2+1}\right)$

$\Rightarrow \,\psi^{\prime}\left( \frac{1}{4}\right)=8\int_0^1\frac{\ln t}{t^2-1}dt-8\int_0^1\frac{\ln t}{t^2+1}dt$

These two integrals are evaluated in the appendix below, we therefore obtain

$\boxed{\psi^{\prime}\left( \frac{1}{4}\right)=\pi^2+8G}$ (50)

From (12) we may obtain

$\psi^{\prime}\left( \frac{1}{4}\right)+\psi^{\prime}\left( \frac{3}{4}\right)=\frac{\pi^2} {\sin^2\left(\frac{\pi}{4} \right)}$

$\psi^{\prime}\left( \frac{1}{4}\right)+\psi^{\prime}\left( \frac{3}{4}\right)=2\pi^2}$

$\boxed{\psi^{\prime}\left( \frac{3}{4}\right)=\pi^2-8G}$ (51)

Special Values of the Quadrigamma

If we let x=0 in (23) we immediately obtain

$\psi^{\prime \prime}(1)=-2\sum_{n=1}^{\infty}\frac{1}{n^3}$

$\boxed{\psi^{\prime \prime}(1)=-2 \zeta(3)}$ (52)

Now, if we let $x=\frac{1}{2}$ in(23), we get

$8\psi^{\prime \prime} \left(1 \right)= \psi^{\prime \prime}\left(\frac{1}{2} \right)+\psi^{\prime \prime} \left( \frac{1}{2}+\frac{1}{2}\right)$

$\psi^{\prime \prime}\left(\frac{1}{2} \right)=7 \psi^{\prime \prime}(1)$

$\boxed{\psi^{\prime \prime}\left(\frac{1}{2} \right)=-14 \zeta(3)}$ (53)

let $x=\frac{1}{3}$ in (21)

$27\psi^{\prime \prime}\left(1\right)= \psi^{\prime \prime}\left(\frac{1}{3} \right)+\psi^{\prime \prime}\left(\frac{2}{3} \right)+\psi^{\prime \prime}\left(1 \right)$

$\psi^{\prime \prime}\left(\frac{1}{3} \right)+\psi^{\prime \prime}\left(\frac{2}{3} \right)=-52\zeta(3)$ (54)

On the other hand setting $x=\frac{1}{3}$ in (13) we obtain

$\psi^{\prime \prime}\left(\frac{1}{3}\right)-\psi^{\prime \prime}\left(\frac{2}{3}\right)=-2\pi^3\frac{\cot\left(\frac{\pi}{3}\right)} {\sin^2\left(\frac{\pi}{3}\right)}$

$\psi^{\prime \prime}\left(\frac{1}{3}\right)-\psi^{\prime \prime}\left(\frac{2}{3}\right)=-2\pi^3\frac{\frac{1}{\sqrt{3}}} {\left(\frac{\sqrt{3}}{2}\right)^2}$

$\psi^{\prime \prime}\left(\frac{1}{3}\right)-\psi^{\prime \prime}\left(\frac{2}{3}\right)=-\frac{8\pi^3} {3\sqrt{3}}}$ (55)

Adding (54) and (55)

$\boxed{\psi^{\prime \prime}\left(\frac{1}{3}\right)=-\frac{4\pi^3} {3\sqrt{3}}-26\zeta(3)}$ (56)

Subtracting (55) from (54)

$\boxed{\psi^{\prime \prime}\left(\frac{2}{3}\right)=\frac{4\pi^3} {3\sqrt{3}}-26\zeta(3)}$ (57)

Now let $x=\frac{1}{4}$ in (17)

$8\psi^{\prime \prime} \left(\frac{1}{2} \right)= \psi^{\prime \prime}\left(\frac{1}{4} \right)+\psi^{\prime \prime} \left( \frac{3}{4}\right)$

$\psi^{\prime \prime}\left(\frac{1}{4} \right)+\psi^{\prime \prime} \left( \frac{3}{4}\right)=-112 \zeta(3)$ (58)

On the other hand setting $x=\frac{1}{4}$ in (13) we obtain

$\psi^{\prime \prime}\left(\frac{1}{4}\right)-\psi^{\prime \prime}\left(\frac{3}{4}\right)=-2\pi^3\frac{\cot\left(\frac{\pi}{4}\right)} {\sin^2\left(\frac{\pi}{4}\right)}$

$\psi^{\prime \prime}\left(\frac{1}{4}\right)-\psi^{\prime \prime}\left(\frac{3}{4}\right)=-4\pi^3$ (59)

Adding (58) and (59)

$\boxed{\psi^{\prime \prime}\left(\frac{1}{4}\right)=-2 \pi^3 -56 \zeta(3)}$ (60)

Subtracting (59) from (58) we get

$\boxed{\psi^{\prime \prime}\left(\frac{3}{4}\right)=2 \pi^3 -56 \zeta(3)}$ (61)

let $x=\frac{1}{6}$ in (21)

$27\psi^{\prime \prime}\left(\frac{1}{2}\right)= \psi^{\prime \prime}\left(\frac{1}{6} \right)+\psi^{\prime \prime}\left(\frac{1}{2} \right)+\psi^{\prime \prime}\left(\frac{5}{6} \right)$

$\psi^{\prime \prime}\left(\frac{1}{6} \right)+\psi^{\prime \prime}\left(\frac{5}{6} \right)=26\psi^{\prime \prime}\left(\frac{1}{2}\right)$

$\psi^{\prime \prime}\left(\frac{1}{6} \right)+\psi^{\prime \prime}\left(\frac{5}{6} \right)=-364 \zeta(3)$ (62)

On the other hand setting $x=\frac{1}{6}$ in (13) we obtain

$\psi^{\prime \prime}\left(\frac{1}{6}\right)-\psi^{\prime \prime}\left(\frac{5}{6}\right)=-2\pi^3\frac{\cot\left(\frac{\pi}{6}\right)} {\sin^2\left(\frac{\pi}{6}\right)}$

$\psi^{\prime \prime}\left(\frac{1}{6}\right)-\psi^{\prime \prime}\left(\frac{5}{6}\right)=-2\pi^3\frac{\sqrt{3}} {\left(\frac{1}{2}\right)^2}$

$\psi^{\prime \prime}\left(\frac{1}{6}\right)-\psi^{\prime \prime}\left(\frac{5}{6}\right)=-8\sqrt{3}\pi^3$ (63)

Adding (62) and (63)

$\boxed{\psi^{\prime \prime}\left(\frac{1}{6}\right)=-182\zeta(3)-4\sqrt{3}\pi^3}$ (64)

subtracting (63) from (62)

$\boxed{\psi^{\prime \prime}\left(\frac{5}{6}\right)=-182\zeta(3)+4\sqrt{3}\pi^3}$ (65)

Apendix

A.1 Integral representation Clausen Function

First note the following fact

$\int_0^1x^{k-1}\ln x dx=-\frac{1}{k^2}$

proof: Integrating by parts

$\int_0^1x^{k-1}\ln x dx=\frac{x^k}{k}\ln x \Big|_0^1-\frac{1}{k}\int_0^1x^{k-1}dx$

$\int_0^1x^{k-1}\ln x dx=-\frac{1}{k^2}$ (A.1)

The Clausen function is defined as following

$Cl_{m}(\theta)=\begin{cases}\sum_{k=1}^{\infty}\frac{\sin(k\theta)}{k^m} & \text{when m is even}\\\sum_{k=1}^{\infty}\frac{\cos(k\theta)}{k^m} &\text{when m is odd}\end{cases}$

For m=2 we have

$Cl_{2}(\theta)=\sum_{k=1}^{\infty}\frac{\sin(k\theta)}{k^2}$ (A.2)

Plugging (A.1) in (A.2) we obtain

$Cl_{2}(x)=-\sum_{k=1}^{\infty}\sin(k\theta)\int_0^1x^{k-1}\ln x dx$

$=-\int_0^1\ln x \left(\sum_{k=1}^{\infty}\sin(k\theta) x^{k-1}\right)dx$

$=-\frac{1}{2i}\int_0^1\ln x \left(\sum_{k=1}^{\infty} x^{k-1}e^{ik\theta}-\sum_{k=1}^{\infty} x^{k-1}e^{-ik\theta}\right)dx$

$=-\frac{1}{2i}\int_0^1\ln x \left(\sum_{k=0}^{\infty} x^{k}e^{i\theta(k+1)}-\sum_{k=0}^{\infty} x^{k}e^{-i\theta(k+1)}\right)dx$

$=-\frac{1}{2i}\int_0^1\ln x \left(\frac{e^{i\theta}}{1-xe^{i\theta}}-\frac{e^{-i\theta}}{1-xe^{-i\theta}}\right)dx$

$=-\frac{1}{2i}\int_0^1\ln x \left(\frac{1}{e^{-i\theta}-x}-\frac{1}{e^{i\theta}-x}\right)dx$

$=-\frac{1}{2i}\int_0^1\ln x \left(\frac{(-1)}{x-e^{-i\theta}}-\frac{(-1)}{x-e^{i\theta}}\right)dx$

$=-\frac{1}{2i}\int_0^1\ln x \left(\frac{(-1)\left(e^{-i\theta}-e^{i\theta} \right)}{(x-e^{i\theta})(x-e^{-i\theta})}\right)dx$

$=-\frac{\left(e^{i\theta}-e^{-i\theta} \right)}{2i}\int_0^1 \frac{\ln x}{(x-e^{i\theta})(x-e^{-i\theta})}dx$

$\boxed{Cl_{2}(\theta)=-\sin(\theta) \int_0^1 \frac{\ln x}{x^2-2x \cos(\theta)+1}dx}$ (A.3)

Example: $Cl_{2}\left(\frac{2 \pi}{3}\right)$

From the values

$\sin\left(\frac{2 \pi}{3} \right)=\sin\left(\frac{ \pi}{3} \right)=\frac{\sqrt{3}}{2}$

$\cos\left(\frac{2 \pi}{3} \right)=-\cos\left(\frac{ \pi}{3} \right)=-\frac{1}{2}$

we obtain

$\boxed{Cl_{2}\left(\frac{2 \pi}{3} \right)=-\frac{\sqrt{3}}{2}\int_0^1\frac{\ln t}{t^2+t+1}dt}$ (A.4)

A.2 Evaluation of $\,\int_0^1\frac{\ln t}{t^2-1}dt\,\,\,\text{and}\,\,\,\int_0^1\frac{\ln t}{1+t^2}dt$

$\int_0^1\frac{\ln t}{t^2-1}dt=-\int_0^1\frac{\ln t}{1-t^2}dt$

$=-\int_0^1\ln t\sum_{k=0}^{\infty}t^{2k}dt$

$=-\sum_{k=0}^{\infty}\int_0^1 t^{2k}\ln tdt$

Integrating by parts

$\int_0^1\frac{\ln t}{t^2-1}dt=\sum_{k=0}^{\infty}\frac{1}{(2k+1)}\int_0^1 t^{2k}dt$

$\int_0^1\frac{\ln t}{t^2-1}dt=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}=\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}$

Now observe the following:

$\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}$

splitting in even and odd terms we get:

$\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{(2n)^2}+\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^2}$

$\zeta(2)=\frac{1}{4}\zeta(2)+\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^2}$

$\sum_{n=1}^{\infty}\frac{1}{\left(2n-1\right)^2}=\frac{3}{4}\zeta(2)$

Therefore

$\boxed{\int_0^1\frac{\ln t}{t^2-1}dt=\frac{3}{4}\zeta(2)}$ (A.5)

To compute the second integralwe follow a similar procedure

$\int_0^1\frac{\ln t}{1+t^2}dt=\int_0^1 \ln x\sum_{k=0}^{\infty}(-1)^kt^{2k}dt$

$=\sum_{k=0}^{\infty}(-1)^k\int_0^1 t^{2k}\ln xdt$

Integrating by parts

$\int_0^1\frac{\ln t}{1+t^2}dt=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)}\int_0^1 t^{2k}dt$

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

$\boxed{\int_0^1\frac{\ln t}{1+t^2}dt=-G}$ (A.6)

Where is Catalan´s constant

A.3

An alternative way to compute $\psi\left(\frac13\right)\,$ is through the integral representation (7) as following:

$\psi\left(\frac13\right) & = -\gamma+\int_0^1 \frac{1 - t^{-\frac23}}{1 - t} dt$

let t=u^3

$\psi\left(\frac13\right) = -\gamma+3\int_0^1 \frac{1 - u^{-2}}{1 - u^3} u^2du$

$& = -\gamma-3\int_0^1 \frac{1+u}{1 + u+u^2} du$

$= -\gamma -3\int_0^1 \frac{1+u}{3/4+(u+1/2)^2} du$

$\psi\left(\frac13\right) = -\gamma -3\int_0^1 \frac{1/2+u}{3/4+(u+1/2)^2} du-\frac{3}{2}\int_0^1 \frac{1}{3/4+(u+1/2)^2} du$

let 1/2+u=t in the first integral

$= -\gamma -3\int_{1/2}^{3/2} \frac{t}{3/4+t^2} dt-\frac{3}{2}\int_0^1 \frac{1}{3/4+(u+1/2)^2} du$

$= -\gamma -\frac{3}{2}\ln\left(3/4+t^2\right)\Big|_{1/2}^{3/2}-\frac{3}{2}\int_0^1 \frac{1}{3/4+(u+1/2)^2} du$

$= -\gamma - \frac32\ln 3-\frac{3}{2}\int_0^1 \frac{1}{3/4+(u+1/2)^2} du$

$= -\gamma - \frac32\ln 3-2\int_0^1 \frac{1}{1+\frac{3}{4}(u+1/2)^2} du$

$= -\gamma - \frac32\ln 3-2\int_0^1 \frac{1}{1+\left(\frac{2}{\sqrt{3}}u+\frac{1}{\sqrt{3}}\right)^2} du$

$= -\gamma - \frac32\ln 3-2\int_0^1 \frac{1}{1+\left(\frac{1+2u}{\sqrt{3}}\right)^2} du$

let $\frac{1+2u}{\sqrt{3}}=t$

$\psi\left(\frac13\right) = -\gamma - \frac32\ln 3-\sqrt{3}\int_{1/\sqrt{3}}^{3/\sqrt{3}} \frac{1}{1+t^2} dt$

$\psi\left(\frac13\right) = -\gamma - \frac32\ln 3-\sqrt{3}\left( \arctan(t)\right)\Big|_{1/\sqrt{3}}^{\sqrt{3}}$

$\psi\left(\frac13\right) = -\gamma - \frac32\ln 3-\sqrt{3}\left(\frac{\pi}{3}-\frac{\pi}{6} \right)$

$\boxed{\psi\left(\frac{1}{3} \right)=-\frac{\pi}{2\sqrt{3}}-\frac{3 \ln 3}{2}-\gamma }$

Which is equal to (30)

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