Special values $\psi \left(\frac12\right)$ and $\psi \left(\frac13\right)$ I wonder if it is easy to prove that
$$
\begin{align}
\psi \left(\frac12\right) & = -\gamma - 2\ln 2, \\
\psi \left(\frac13\right) & =  -\gamma + \frac\pi6\sqrt{3}- \frac32\ln 3,
\end{align}
$$
where $\psi$ is the digamma function-the logarithmic derivative of $\Gamma$ function- and  $\gamma$ is Euler's constant. 
I started with $\psi \left(\dfrac12\right)=\dfrac{\Gamma'}{\Gamma}\left(\dfrac12\right)$ which is not easy to handle.
Thank you for your help.

You might start from the sum form:
$$ \psi(z) = -\gamma + \sum_{k=1}^\infty \left( \dfrac{1}{k} - \dfrac{1}{k+z-1}\right)$$
For $z=1/2$ the partial sum up to $k=n$ (let's say for convenience that $n$ is even) is
$$ \eqalign{\sum_{k=1}^n \left( \dfrac{1}{k} - \dfrac{2}{2k-1} \right) &= 
\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \ldots  + \dfrac{1}{n} - \dfrac{2}{1} - \dfrac{2}{3} - \ldots - \dfrac{2}{2n-1}\cr
&= \sum_{k=1}^n (-1)^k \dfrac{1}{k} - \sum_{k=n/2+1}^{n} \dfrac{2}{2k-1}\cr}$$
As $n \to \infty$, the first sum approaches $\displaystyle \sum_{k=1}^\infty (-1)^k \dfrac{1}{k} = -\ln 2$, while the second is approximated by
$$ \int_{n/2}^n \dfrac{2 \; dx}{2x-1} =  \ln \left( \dfrac{2n-1}{n-1} \right) \to  \ln 2$$
so the result is
$$ \psi(z) = - \gamma - \ln 2 - \ln 2 = -\gamma - 2 \ln 2$$
