Mac-Laurin-series for $\psi(\frac{2x+1}{2x})-\psi(\frac{x+1}{2x})$? Consider $$f(x)=\psi(\frac{2x+1}{2x})-\psi(\frac{x+1}{2x})$$
$\psi(x)$ is the digamma-function. This function occurs in the calculation of the definite integral
$$\int_0^1 \ln(x^n+1)dx=\ln(2)-\frac{f(n)}{2}$$ for $n>0$. Wolfram alpha gives a series expansion of $f(x)$ for $x\rightarrow\infty$ , but not for $x\rightarrow 0$ and I could not even calculate $$\lim_{x\rightarrow 0} f'(x)$$, which should be $1$ due to numerical calculation.

Does the Mac-Laurin-series for $f(x)$ exist ? If yes, how can I find the series ? And, finally, which convergent radius does it have ?


Let's take a look at the chapter about polygamma functions in Abramowitz and Stegun.
The multiplication formula (6.4.8) shows:
$$\begin{align*}
\psi(2x) &= \ln(2) + \frac{1}{2}(\psi(x) + \psi(x+1/2)) \\
\iff \psi(x + 1/2) &= 2\psi(2x) - \ln(4) - \psi(x)
\end{align*}$$
The recurrence formula (6.4.6) shows:
$$\psi(1 + x) = \psi(x) + \frac{1}{x}.$$
Combining both we get the following:
$$\begin{align*}
f(x) &= \psi(1 + 1/(2x)) - \psi(1/2 + 1/(2x)) = \psi(1/(2x)) + 2x - 2\psi(2/(2x)) + \ln(4) + \psi(1/(2x)) \\
&= 2\big\{\psi(1/(2x)) - \psi(1/x)\big\} + 2x + \ln(4)
\end{align*}$$
Now we can plug in the asymptotic formula (6.3.18):
$$\begin{align*}
\psi(1/(2x)) - \psi(1/x) &= -\ln(2x) - x - \sum \limits_{k = 1}^n  \frac{B_{2k}}{2k} 2^{2k} x^{2k} + \ln(x) + x/2 + \sum \limits_{k = 1}^n \frac{B_{2k}}{2k} x^{2k} + O(x^{2n + 1}) \\
&= -\ln(2) - x/2 - \sum \limits_{k = 1}^n\frac{B_{2k}}{2k}(4^k - 1) x^{2k} + O(x^{2n + 1})
\end{align*}$$
All together:
$$f(x) = x - \sum \limits_{k = 1}^n \frac{B_{2k}}{k} (4^k - 1)x^{2k} + O(x^{2k + 1}).$$
Edit: Since $|B_{2n}| \sim 4 \sqrt{\pi n} \left(\frac{n}{\pi e}\right)^{2n}$ the corresponding series has a convergence radius of $0$.
