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I often see asymptotics and precise expansion for the gamma $\Gamma$ or the digamma $\psi$ function $\psi$ when the argument goes to $+\infty$, in particular when it stays real (or in a given angle sector towards $+\infty$).

I would like to know the precise asymptotics along the imaginary axis, i.e. asymptotics for $$\psi(x_0 + iy) = \frac{\Gamma'}{\Gamma}(x_0 + iy)$$

when $x_0$ is fixed, say positive, and $y$ goes to $\pm \infty$. Do we know such an expansion, with explicit dependencies in $x_0$?

Typically, the Stirling formula $$\Gamma(z) \sim \sqrt{\frac{2\pi}{z}} \left(\frac{z}{e} \right)^z$$

is valid for all complex number in the angle sector $|\mathrm{arg}(z) - \pi| \geq \delta$ for any $\delta > 0$. This is unfortunately not enough to obtain information on the derivative $\Gamma'$, and therefore on $\psi$. Is there a similar formula for the digamma function?

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  • $\begingroup$ Start with $\frac1{z^2}= \sum_{k=1}^K B_{k-1}(\frac1{z^k}-\frac1{(z+1)^k})+O(z^{-K-1})$, $(\frac{\Gamma'(z)}{\Gamma(z)})' = \sum_{n\ge 0} \frac1{(z+n)^2}$ $\endgroup$
    – reuns
    Oct 21, 2020 at 16:12

1 Answer 1

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$$\psi(x+i y)=\frac{i \pi }{2}+\log (y)+\frac{i(1-2x)}{2y}+\frac{6 (x-1) x+1}{12 y^2}+$$ $$\frac{i (x-1) x (2 x-1)}{6 y^3}-\frac{30 x^4-60 x^3+30 x^2-1}{120 y^4}+O\left(\frac{1}{y^5}\right)$$ seems to be quite decent even for small values of $y$. Trying for $x=\pi$

$$\left( \begin{array}{ccc} y & \text{approximation} & \text{exact} \\ 3 & 1.342042544+0.909681077 \, i & 1.385304688+0.846249307 \, i \\ 4 & 1.557580397+1.002964134 \, i & 1.566566745+0.985468467 \, i \\ 5 & 1.729238357+1.089871573 \, i & 1.731792660+1.083668377 \, i \\ 6 & 1.878793427+1.157957839 \, i & 1.879690031+1.155348063 \, i \\ 7 & 2.011554222+1.210697732 \, i & 2.011920434+1.209455057 \, i \\ 8 & 2.130545425+1.252167991 \, i & 2.130713040+1.251518287 \, i \\ 9 & 2.238060772+1.285412537 \, i & 2.238144592+1.285047147 \, i \\ 10 & 2.335927665+1.312561284 \, i & 2.335972655+1.312343426 \, i \end{array} \right)$$

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