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I encountered following expression

$$\psi ^{(0)}\left(z+\frac{1}{4}\right)-\psi ^{(0)}\left(z-\frac{1}{4}\right)$$

Searching in different resources I managed to find these identities involving the digamma function $\psi^{(0)}(x)=\frac{\mathrm{d}}{\mathrm{d}x}\ln\Gamma(x)=\frac{1}{\Gamma (x)}\frac{\mathrm{d} \Gamma (x)}{\mathrm{d} x}$.

  1. $\psi ^{(0)}(z+1)-\psi ^{(0)}(z-1)=\frac{1}{z}+\frac{1}{z-1}$
  2. $\psi ^{(0)}\left(z+\frac{1}{2}\right)-\psi ^{(0)}\left(z-\frac{1}{2}\right)=\frac{1}{z-\frac{1}{2}}$
  3. $\psi ^{(0)}\left(z+\frac{1}{4}\right)-\psi ^{(0)}\left(z-\frac{1}{4}\right)=?$

Is there any similar expression for the third case? Or are there any simplification of any other form?

user824530
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    According to W|A, it simplifies to $H_{z - 3/4} - H_{z - 5/4}$ where $H_n$ is a generalization of the harmonic numbers to non-integer indices (this might be helpful for further reference). I wonder if there is a better simplification/closed form expression for this... – Prasun Biswas Jan 08 '21 at 05:53
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    It is not a simplification (and I'm sure there's none), it's just renaming. – metamorphy Jan 08 '21 at 06:09
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    @metamorphy: Yes, I suspected that it was just renaming using $H_z:=\psi^{(0)}(z+1)$ but wasn't sure if that was a definition for $H_z$ or equivalence with some other definition of $H_z$. Thanks for the clarification! – Prasun Biswas Jan 08 '21 at 08:22

1 Answers1

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I am not sure that we can simplify more than you did.

However, making the problem more general, we can write $$\psi ^{(0)}\left(z+\frac{1}{n}\right)-\psi ^{(0)}\left(z-\frac{1}{n}\right)=2\sum_{k=1}^\infty \frac{\psi ^{(2 k-1)}(z) }{n^{(2 k-1)}\,(2 k-1)!}$$

If you make $n=4$ and use only $4$ terms in the summation, you have an almost perfect match.

Claude Leibovici
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