# Trigamma identity $\psi_1\left(\frac{11}{12}\right)-\psi_1\left(\frac{5}{12}\right)=4\sqrt 3 \pi^2-80G$

Hello regarding this integral: Integral $\int_0^1 \frac{\sqrt x \ln x} {x^2 - x+1}dx$ The following conjecture comes: $$\psi_1\left(\frac{11}{12}\right)-\psi_1\left(\frac{5}{12}\right)=4\sqrt 3 \pi^2-80G$$ Where $G$ is Catalan's constant. How can we show this? Wolfram-alpha agrees on this: https://www.wolframalpha.com/input/?i=trigamma(11%2F12)-trigamma(5%2F12)%3D4sqrt3pi%5E2-80Catalan

\begin{align} \psi_1(\tfrac{11}{12}) -\psi_1(\tfrac{5}{12}) &=4\sqrt 3 \pi^2-80G \tag{1}\label{1} . \end{align}

The constant $G$ - Catalan's constant has known to appear in relations \begin{align} \psi_1(\tfrac14)&=\pi^2+8G ,\\ \psi_1(\tfrac34)&=\pi^2-8G , \end{align}

so we can try to start from $\psi_1(\tfrac14)$.

Applying triplication identity

\begin{align} 9\psi_1(3x) &= \psi_1(x) +\psi_1(x+\tfrac13) +\psi_1(x+\tfrac23) \tag{2}\label{2} \end{align}

to $\psi_1(\tfrac14)=\psi_1(3\cdot\tfrac1{12})$, we get

\begin{align} 9\psi_1(\tfrac14) &= \psi_1(\tfrac1{12}) +\psi_1(\tfrac5{12}) +\psi_1(\tfrac34) ,\\ \psi_1(\tfrac1{12}) +\psi_1(\tfrac5{12}) &=9\psi_1(\tfrac14)-\psi_1(\tfrac34) \\ &=10\psi_1(\tfrac14)-(\psi_1(\tfrac14)+\psi_1(\tfrac34)) \\ &=10\psi_1(\tfrac14)-2\pi^2 \\ &=10(\pi^2+8G)-2\pi^2 \\ \psi_1(\tfrac1{12}) +\psi_1(\tfrac5{12}) &=8\pi^2+80G , \end{align} and we are almost there.

Now, applying the identity

\begin{align} \psi_1(1-x)&= \frac{\pi^2}{\sin(\pi x)^2} -\psi_1(x) \end{align}

to $\psi_1(\tfrac1{12})=\psi_1(1-\tfrac{11}{12})$,

we get \begin{align} \psi_1(\tfrac1{12}) &=\frac{\pi^2}{\sin(\tfrac\pi{12})^2} -\psi_1(\tfrac{11}{12}) ,\\ &=8\pi^2+4\sqrt3\pi^2 -\psi_1(\tfrac{11}{12}) , \end{align}

and \eqref{1} follows.

• could you give me a link maybe from where to find that triplication formula? – Zacky Jun 3 '18 at 18:57
• @Zacky: Consider $9\sum_{n=0}^\infty\frac1{(3z+n)^2}=\sum_{n=0}^\infty\frac1{(z+\frac{n}3)^2}$ $=\sum_{k=0}^\infty\frac1{(z+\frac{3k}3)^2} +\frac1{(z+\frac{3k+1}3)^2}+\frac1{(z+\frac{3k+2}3)^2}=\dots$. – g.kov Jun 4 '18 at 1:12
• Sir, Is there a problem if I share your solution on other groups? – Zacky Aug 8 '18 at 21:37
• @Zacky: Not sure, what do you mean, this whole site is for sharing solutions (and questions), so if you want to share it - that's great. – g.kov Aug 8 '18 at 22:25

$$\psi_1(z)=\sum_{n=0}^\infty\frac1{(z+n)^2}.$$

Consider the series $$S_1=\sum_{k=0}^\infty\left(\frac1{(12k+1)^2} +\frac1{(12k+5)^2}-\frac1{(12k+7)^2}-\frac1{(12k+11)^2}\right)$$ and $$S_2=\sum_{k=0}^\infty\left(\frac1{(12k+1)^2} -\frac1{(12k+5)^2}-\frac1{(12k+7)^2}+\frac1{(12k+11)^2}\right).$$ Then $$\psi_1(5/12)-\psi_1(11/12)=\frac{12^2(S_1-S_2)}2.$$

The first series is $$S_1=\left(1+\frac1{3^2}\right)\sum_{k=0}^\infty\left(\frac1{(4k+1)^2}-\frac1{(4k+3)^2}\right),$$ a rational multiple of Catalan's constant.

The second series is $L(\chi,2)$ where $\chi$ is the even Dirichlet character of conductor $12$. This can be evaluated using the functional equation of Dirichlet's L-functions (see for instance Washington's book on cyclotomic fields) or else via the infinite series for $\cot^2$.