TITLE: The closed form of $\lim_{x\to\frac{4}{3}}\frac{\partial}{\partial x}\left[\,_2{\rm{F}}_1\left(\frac{1}{3},1;x;-1\right)\right]$
QUESTION [12 upvotes]: Do you think the following limit might have a closed form? Some hints or clues?
$$\lim_{x\to\frac{4}{3}}\frac{\partial}{\partial x}\left[\,_2{\rm{F}}_1\left(\frac{1}{3},1;x;-1\right)\right]$$

REPLY [8 votes]: \begin{align}
\small\frac{\partial}{\partial x}\left.{}_2\mathrm{F}_1\left(\left.\begin{matrix}\frac{1}{3},1\\x\end{matrix}\right|-1\right)\right|_{x=\frac{4}{3}}
&\small=\frac{\partial}{\partial x}\left.(x-1)\int^1_0\frac{(1-t)^{x-2}}{(1+t)^\frac{1}{3}}dt\right|_{x=\frac{4}{3}}\tag{1a}\\
&\small=\int^1_0\frac{dt}{(1-t)^{\frac{2}{3}}(1+t)^{\frac{1}{3}}}+\frac{1}{3}\int^1_0\frac{\ln(1-t)}{(1-t)^{\frac{2}{3}}(1+t)^{\frac{1}{3}}}dt\tag{1b}\\
&\small=\left(1+\frac{1}{3}\ln{2}\right)\int^1_0\frac{t^{-\frac{2}{3}}}{1+t}dt+\frac{1}{3}\int^1_0\frac{t^{-\frac{2}{3}}\ln{t}}{1+t}dt-\frac{1}{3}\int^1_0\frac{t^{-\frac{2}{3}}\ln(1+t)}{1+t}dt\tag{1c}\\
&\small=\frac{\pi}{\sqrt{3}}+\ln{2}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{\ln^2{2}}{3}+\frac{2\pi^2}{9}-\frac{1}{2}\psi_1\left(\frac{1}{3}\right)-\int^1_0\frac{\ln(1+t^3)}{1+t^3}dt\tag{1d}
\end{align}
Explanation:
$(\text{1a})$: Used the integral representation of ${}_2\mathrm{F}_1$. 
$(\text{1b})$: Applied the product rule. 
$(\text{1c})$: Applied the substitution $t\mapsto\frac{1-t}{1+t}$. 
$(\text{1d})$: Used the fact that $\int^1_0\frac{t^a}{1+t}dt=\frac{1}{2}\left(\psi_0\left(\frac{a}{2}+1\right)-\psi_0\left(\frac{a+1}{2}\right)\right)$, substituted $t\mapsto t^3$.

\begin{align}
\small{\int^1_0\frac{\ln(1+t^3)}{1+t^3}dt}
=&\ \small{4\int^1_{-1}\frac{\ln\left(\frac{1}{8}(x+3)(x^2+3)\right)}{(x+3)(x^2+3)}dx\tag{2a}}\\
=&\ \small{-\frac{\pi\ln{2}}{\sqrt{3}}-\frac{\ln^2{2}}{2}+\int^1_{-1}\frac{\ln(x+3)}{x^2+3}dx-\frac{1}{3}\int^1_{-1}\frac{x\ln(x+3)}{x^2+3}dx+\frac{1}{3}\int^1_{-1}\frac{\ln(x^2+3)}{x+3}dx}\\
&\small{+2\int^1_0\frac{\ln(x^2+3)}{x^2+3}dx\tag{2b}}\\
=&\ \small{-\frac{\pi\ln{2}}{\sqrt{3}}-\frac{5\ln^2{2}}{6}+\int^1_{-1}\frac{\ln(x+3)}{x^2+3}dx+\frac{1}{2}\int^1_{-1}\frac{\ln(x^2+3)}{x+3}dx+2\int^1_0\frac{\ln(x^2+3)}{x^2+3}dx}\\
&\tag{2c}
\end{align}
Explanation:
$(2\text{a})$ Used the substitution $x=2t-1$. 
$(2\text{b})$ Laws of logarithms, Partial Fractions. 
$(2\text{c})$ Integrated by parts.

\begin{align}
\small{2\int^1_0\frac{\ln(x^2+3)}{x^2+3}dx}
&\small=\frac{2}{\sqrt{3}}\int^\frac{\pi}{6}_0\left(\ln{3}-2\ln(\cos{x})\right).dx\tag{3a}\\
&\small=\frac{\pi\ln{3}}{3\sqrt{3}}+\frac{2\pi\ln{2}}{3\sqrt{3}}-\frac{4}{\sqrt{3}}\sum^\infty_{n=1}\frac{(-1)^{n-1}}{n}\int^\frac{\pi}{6}_0\cos(2nx)dx\tag{3b}\\
&\small=\frac{\pi\ln{3}}{3\sqrt{3}}+\frac{2\pi\ln{2}}{3\sqrt{3}}-\frac{2}{\sqrt{3}}\sum^\infty_{n=1}\frac{(-1)^{n-1}\sin(\pi n/3)}{n^2}\\
&\small=\frac{\pi\ln{3}}{3\sqrt{3}}+\frac{2\pi\ln{2}}{3\sqrt{3}}-\sum^\infty_{n=0}\frac{1}{(6n+1)^2}+\sum^\infty_{n=0}\frac{1}{(6n+2)^2}-\sum^\infty_{n=0}\frac{1}{(6n+4)^2}+\sum^\infty_{n=0}\frac{1}{(6n+5)^2}\\
&\small=\frac{\pi\ln{3}}{3\sqrt{3}}+\frac{2\pi\ln{2}}{3\sqrt{3}}+\frac{1}{36}\left(-\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)\right)\tag{3c}\\
&\small=\frac{\pi\ln{3}}{3\sqrt{3}}+\frac{2\pi\ln{2}}{3\sqrt{3}}+\frac{4\pi^2}{27}-\frac{2}{9}\psi_1\left(\frac{1}{3}\right)\tag{3d}\\
\end{align}
Explanation:
$(3\text{a})$ Used the substitution $x\mapsto\sqrt{3}\tan{x}$. 
$(3\text{b})$ Used the Fourier series of $\ln(\cos{x})$. 
$(3\text{c})$ Recognised the series representation of $\psi_1$. 
$(3\text{d})$ Used the duplication & reflection formulae. 

\begin{align}
\small\frac{1}{2}\int^1_{-1}\frac{\ln(x^2+3)}{x+3}dx
&\small=\Re\int^1_{-1}\frac{\ln(x+i\sqrt{3})}{x+3}dx\\
&\small=\Re\int^\frac{i}{\sqrt{3}}_{-\frac{1}{2}+\frac{i}{2\sqrt{3}}}\frac{\ln\left((3-i\sqrt{3})x\right)}{1+x}dx\tag{4a}\\
&\small=\Re\Bigg{[}\ln(1+x)\ln\left((3-i\sqrt{3})x\right)+\mathrm{Li}_2(-x)\Bigg{]}^\frac{i}{\sqrt{3}}_{-\frac{1}{2}+\frac{i}{2\sqrt{3}}}\tag{4b}\\
&\small=\ln^2{2}+\frac{\pi^2}{18}+\Re\left[\mathrm{Li}_2\left(-\frac{i}{\sqrt{3}}\right)-\mathrm{Li}_2\left(\frac{1}{2}-\frac{i}{2\sqrt{3}}\right)\right]\\
&\small=\ln^2{2}+\frac{\ln^2{3}}{8}-\frac{\pi^2}{72}+\frac{1}{4}\mathrm{Li}_2\left(-\frac{1}{3}\right)\tag{4c}
\end{align}
Explanation:
$(4\text{a})$ Used the substitution $x\mapsto(3-i\sqrt{3})x-i\sqrt{3}$. 
$(4\text{b})$ Integrated by parts. 
$(4\text{c})$ Used dilogarithm duplication & reflection formulae. 

\begin{align}
\small\int^1_{-1}\frac{\ln(x+3)}{x^2+3}dx
&\small=\frac{1}{\sqrt{3}}\int^\frac{\pi}{6}_{-\frac{\pi}{6}}\ln\left(\frac{2\sqrt{3}\sin\left(x+\frac{\pi}{3}\right)}{\cos{x}}\right)dx\tag{5a}\\
&\small=\frac{\pi\ln{3}}{6\sqrt{3}}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{2\pi^2}{27}-\frac{1}{9}\psi_1\left(\frac{1}{3}\right)+\frac{1}{\sqrt{3}}\int^\frac{\pi}{2}_\frac{\pi}{6}\ln(2\sin{x})dx\tag{5b}\\
&\small=\frac{\pi\ln{3}}{6\sqrt{3}}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{2\pi^2}{27}-\frac{1}{9}\psi_1\left(\frac{1}{3}\right)-\frac{1}{\sqrt{3}}\sum^\infty_{n=1}\frac{1}{n}\int^\frac{\pi}{2}_\frac{\pi}{6}\cos(2nx)dx\tag{5c}\\
&\small=\frac{\pi\ln{3}}{6\sqrt{3}}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{2\pi^2}{27}-\frac{1}{9}\psi_1\left(\frac{1}{3}\right)+\frac{1}{2\sqrt{3}}\sum^\infty_{n=1}\frac{\sin(\pi n/3)}{n^2}\\
&\small=\frac{\pi\ln{3}}{6\sqrt{3}}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{2\pi^2}{27}-\frac{1}{9}\psi_1\left(\frac{1}{3}\right)+\frac{1}{144}\left(12\psi_1\left(\frac{1}{3}\right)-8\pi^2\right)\tag{5d}\\
&\small=\frac{\pi\ln{3}}{6\sqrt{3}}+\frac{\pi\ln{2}}{3\sqrt{3}}+\frac{\pi^2}{54}-\frac{1}{36}\psi_1\left(\frac{1}{3}\right)\\
\end{align}
Explanation:
$(5\text{a})$ Used the substitution $x\mapsto\sqrt{3}\tan{x}$. 
$(5\text{b})$ Used $3\text{d}$, substituted $x\mapsto x-\frac{\pi}{3}$. 
$(5\text{c})$ Used the Fourier seires of $\ln(2\sin{x})$. 
$(5\text{d})$ Same explanation as $3\text{b}$ to $3\text{d}$. 

Putting all of this together,
$$\small\int^1_0\frac{\ln(1+t^3)}{1+t^3}dt=\frac{\pi\ln{3}}{2\sqrt{3}}+\frac{\ln^2{2}}{6}+\frac{\ln^2{3}}{8}+\frac{11\pi^2}{72}-\frac{1}{4}\psi_1\left(\frac{1}{3}\right)+\frac{1}{4}\mathrm{Li}_2\left(-\frac{1}{3}\right)$$
and
\begin{align}
\small\frac{\partial}{\partial x}\left.{}_2\mathrm{F}_1\left(\left.\begin{matrix}\frac{1}{3},1\\x\end{matrix}\right|-1\right)\right|_{x=\frac{4}{3}}
&\small\color{red}{=\frac{\pi}{\sqrt{3}}+\ln{2}+\frac{\pi\ln{2}}{3\sqrt{3}}-\frac{\pi\ln{3}}{2\sqrt{3}}+\frac{\ln^2{2}}{6}-\frac{\ln^2{3}}{8}+\frac{5\pi^2}{72}-\frac{1}{4}\psi_1\left(\frac{1}{3}\right)-\frac{1}{4}\mathrm{Li}_2\left(-\frac{1}{3}\right)}
\end{align}