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.
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
$$ \psi\left(\frac12\right)=-\gamma-2\ln 2. $$
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
$$ \psi\left(\frac13\right)=-\gamma + \frac\pi6\sqrt{3}- \frac32\ln 3. $$
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 $$ \psi(z) = - \gamma - \ln 2 - \ln 2 = -\gamma - 2 \ln 2$$
Here's another approach for $\displaystyle{\psi\left(\frac{1}{3}\right)}$. The conventional harmonic number is:
$$ H_n=\sum\limits_{k=1}^n\frac{1}{k} $$
In this case we must have $n\in\mathbb{N}$. Now define generalized harmonic number such that:
$$ \mathbf{H}_n=\sum\limits_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+n}\right) $$
where $n\in\mathbb{R}$. Clearly when $n\in\mathbb{N}$, the values match, so $\lim\limits_{n\rightarrow\infty}\mathbf{H}_n=\gamma+\lim\limits_{n\rightarrow\infty}\ln{n}$. From the definition we can write:
\begin{alignat}{2} \mathbf{H}_{mn}-\mathbf{H}_n &=\sum\limits_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+mn}\right)-\mathbf{H}_n \\ &= \sum\limits_{k=1}^\infty\sum\limits_{j=0}^{m-1}\left(\frac{1}{km-j}-\frac{1}{km-j+mn}\right)-\mathbf{H}_n & ~~~~~~\color{blue}{(k\rightarrow km-j)}\\ &=\frac{1}{m}\sum\limits_{k=1}^\infty\sum\limits_{j=0}^{m-1}\left(\frac{1}{k-j/m}-\frac{1}{k-j/m+n}\right)-\mathbf{H}_n\\ &=\frac{1}{m}\sum\limits_{j=0}^{m-1}\sum\limits_{k=1}^\infty\left[\left(\frac{1}{k}-\frac{1}{k-j/m+n}\right)-\left(\frac{1}{k}-\frac{1}{k-j/m}\right)\right]-\mathbf{H}_n\\ &=\frac{1}{m}\sum\limits_{j=0}^{m-1} (\mathbf{H}_{n-j/m}-\mathbf{H}_{-j/m})-\mathbf{H}_{n}\\ &= \frac{1}{m}\sum\limits_{j=0}^{m-1} (\mathbf{H}_{n-j/m}-\mathbf{H}_{-j/m}-\mathbf{H}_{n}) \end{alignat}
Let $n\rightarrow \infty$, then
$$ \lim\limits_{n\rightarrow\infty}\mathbf{H}_{mn}-\mathbf{H}_n=\ln{m} $$
$$ \lim\limits_{n\rightarrow\infty}\frac{1}{m}\sum\limits_{j=0}^{m-1} (\mathbf{H}_{n-j/m}-\mathbf{H}_{-j/m}-\mathbf{H}_{n})=-\frac{1}{m}\sum\limits_{j=0}^{m-1} \mathbf{H}_{-j/m} $$
Plus, obviously $\mathbf{H}_0=0$, we can conclude:
\begin{equation} \sum\limits_{j=1}^{m-1} \mathbf{H}_{-j/m}=-m\ln{m} \tag{*} \end{equation}
Recall
$$ \pi\cot(\pi n)=\sum\limits_{k= -\infty}^\infty \frac{1}{n+k} $$
Rewrite this as:
\begin{alignat}{2} \pi\cot(\pi n) &=\sum\limits_{k=1}^\infty \left(\frac{1}{k+n-1}-\frac{1}{k-n}\right)\\ &= \sum\limits_{k=1}^\infty \left[\left(\frac{1}{k}-\frac{1}{k-n}\right)-\left(\frac{1}{k}-\frac{1}{k+n-1}\right)\right]\\ &= \mathbf{H}_{-n}-\mathbf{H}_{n-1} \end{alignat}
Hence, with $n=j/m$,
\begin{equation} \mathbf{H}_{-j/m}-\mathbf{H}_{-(m-j)/m}=\pi\cot\left(\frac{\pi j}{m}\right) \tag{**} \end{equation}
Particularly for $m=3$, $(*)$ becomes
\begin{equation} \mathbf{H}_{-1/3}+ \mathbf{H}_{-2/3}=-3\ln{3} \tag{1} \end{equation}
For $m=3$ and $j=1$, $(**)$ becomes
\begin{equation} \mathbf{H}_{-1/3}- \mathbf{H}_{-2/3}=\pi\cot\left(\frac{\pi }{3}\right)=\frac{\pi}{\sqrt{3}} \tag{2} \end{equation}
Recall the series definition for the digamma function:
$$ \psi(z)=-\gamma-\frac{1}{z}+\sum\limits_{k= 1}^\infty \left(\frac{1}{k}-\frac{1}{k+z}\right)=-\gamma-\frac{1}{z}+\mathbf{H}_z $$
Thus, solve $(1)$ and $(2)$ simultaneously,
$$ \psi\left(-\frac{2}{3}\right)=-\gamma+\frac{3}{2}-\frac{3}{2}\ln{3}-\frac{\pi}{2\sqrt{3}} $$
At the end, use the recurrence relationship:
$$ \psi(z+1)=\psi(z)+\frac{1}{z} $$
the required value can be found:
$$ \psi\left(\frac{1}{3}\right)=-\gamma-\frac{3}{2}\ln{3}-\frac{\pi}{2\sqrt{3}} $$
The result you provided, as well as Nash's, was wrong as you both missed a minus sign, but just to clarify, Nash's working was commendable, only the last line is wrong.
