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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 ?

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  • $\begingroup$ with $\psi(\frac{1}{2}+\frac{1}{2x})-\psi(1+\frac{1}{2x}) = \sum_{n=0}^\infty \frac{1}{1+\frac{1}{2x}+n}-\frac{1}{\frac{1}{2}+\frac{1}{2x}+n} $ you should get what you need $\endgroup$ – reuns Nov 16 '16 at 11:16
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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 $0$.

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I computed the various derivatives (thanks for the fun), computed the limits and obtained $$f(x)=x-\frac{x^2}{2!}+\frac{6x^4}{4!}-\frac{360x^6}{6!}+\frac{85680x^8}{8!}-\frac{56246400x^{10}}{10!}+O(x^{11})$$ that is to say $$f(x)=x-\frac{x^2}{2}+\frac{x^4}{4}-\frac{x^6}{2}+\frac{17 x^8}{8}-\frac{31x^{10}}{2}+O\left(x^{11}\right)$$ hoping no mistake (and there were !).

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  • $\begingroup$ Numerical calculations approve the terms upto $x^6$, but please check the $x^8$-term. $\endgroup$ – Peter Nov 16 '16 at 11:28
  • $\begingroup$ Sure about $85/4?$ If I substitute $y=1/x,\;$ compute the asymptotic expansion for $y\rightarrow \infty$ and backsubstitute I get a factor $17/8$ $\endgroup$ – gammatester Nov 16 '16 at 11:29
  • $\begingroup$ @gammatester This seems to be correct! $\endgroup$ – Peter Nov 16 '16 at 11:30
  • $\begingroup$ @gammatester Did you calculate the derivate numerically ? $\endgroup$ – Peter Nov 16 '16 at 11:31
  • $\begingroup$ @pPeter: No, using Maple 7 which refuses to compute the series for $x=0$ but computes the asymptotic expansion without problems. $\endgroup$ – gammatester Nov 16 '16 at 11:32

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