Trigamma identity $\psi_1\left(\frac{11}{12}\right)-\psi_1\left(\frac{5}{12}\right)=4\sqrt 3 \pi^2-80G$ 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
 A: $$\psi_1(z)=\sum_{n=0}^\infty\frac1{(z+n)^2}.$$
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 $\cot^2$.
A: \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.
