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.
On
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 $$ \psi(z) = - \gamma - \ln 2 - \ln 2 = -\gamma - 2 \ln 2$$
Using the following representation for the digamma function: $$ \psi(x) = -\gamma+\int_0^1 \frac{1 - t^{x-1}}{1 - t} dt, \,\, x>0, $$ you have
$$ \begin{align} \psi\left(\frac12\right) & = -\gamma+\int_0^1 \frac{1 - t^{-\frac12}}{1 - t} dt \\ & = -\gamma+2\int_0^1 \frac{1 - u^{-1}}{1 - u^2} u\:du,\,\,t=u^2 \\ & = -\gamma-2\int_0^1 \frac{1}{1 + u} du \\ & = -\gamma-2\ln 2 \\ \end{align} $$ and
In the same manner,
$$ \begin{align} \psi\left(\frac13\right) & = -\gamma+\int_0^1 \frac{1 - t^{-\frac23}}{1 - t} dt \\ & = -\gamma+3\int_0^1 \frac{1 - u^{-2}}{1 - u^3} u^2du,\,\,t=u^3 \\ & = -\gamma-3\int_0^1 \frac{1+u}{1 + u+u^2} du \\ & = -\gamma -3\int_0^1 \frac{1+u}{3/4+(u+1/2)^2} du\\ & = -\gamma + \frac\pi6\sqrt{3}- \frac32\ln 3 \\ \end{align} $$ and