Let $\psi_q(x)$ be the q-DiGamma function defined for a real variable $x>0$ as $$\psi_q(x)=\frac{1}{\Gamma_q(x)}\frac{\partial}{\partial x} (\Gamma_q(x))$$ where $\Gamma_q(x)$ is the q-Gamma function defined as $$\Gamma_q(x)=(1-q)^{1-x}\prod_{n=0}^{\infty}\frac{1-q^{n+1}}{1-q^{n+x}}$$

Question I am looking for a closed form for $$\psi_{e^{\pi}}^{(3)}(1)$$

If anyone could please solve this by hand or mathematica or sage math. I would be highly indebted to you all.


1 Answer 1


Wolfram Alpha gives the expansion at $x=\infty$:


and these match Oeis A$001158$ with divisor $\sigma_v(n)$ and various theta functions after plugging the sum back in here. Use $\vartheta_v(0,x)=\vartheta_v(x)$:

$$\psi_x^{(3)}(1)=\ln^4(x)\sum_{n=1}^\infty\frac{\sigma_3(n)}{x^n}=\frac{\ln^4(x)}{480}\left(\vartheta_2\left(\frac1{\sqrt x}\right)^8+ \vartheta_3\left(\frac1{\sqrt x}\right)^8+ \vartheta_4\left(\frac1{\sqrt x}\right)^8-2\right)$$


$$\psi_{e^\pi}^{(3)}(1)=\frac{\pi^4}{480}\left(\vartheta_2^8\left(e^{-\frac\pi2}\right)+ \vartheta_3^8\left(e^{-\frac\pi2}\right)+ \vartheta_4^8\left(e^{-\frac\pi2}\right)-2\right)$$

Clicking “more digits” here shows a smaller error each time implying the result is true.

Now use Dedekind $\eta(z)$ identities for $\vartheta_v\left(e^{-\frac\pi2}\right)$ when $v=2$, $v=3$, and $v=4$

$$\psi_{e^\pi}^{(3)}(1)= \frac{\pi^4}{480}\left(\left(2\frac{\eta^2(i)}{\eta\left(\frac i2\right)}\right)^8+\left(\frac{\eta^5\left(\frac i2\right)}{\eta^2\left(i\right)\eta^2\left(\frac i4\right)}\right)^8+\left(\frac{\eta^2\left(\frac i4\right)}{\eta\left(\frac i2\right)}\right)^8-2\right)$$

Using special values in terms of $\Gamma\left(\frac14\right)$:

$\eta\left(\frac i4\right)=2\eta(4i)=\frac{\sqrt[4]{\sqrt2-1} \Gamma\left(\frac14\right)}{2^\frac{13}{16}\pi^\frac34},\eta\left(\frac i2\right)=\frac{\Gamma\left(\frac14\right)}{2^\frac 78\pi^\frac34},\eta(i)=\frac{\Gamma\left(\frac14\right)}{2\pi^\frac34}$

Finally, substitute and have a form in terms of $\Gamma\left(\frac14\right)$ which has no elementary closed form. Therefore:


shown here

  • $\begingroup$ I am indebted to you for your effort. Thank you. I would really appreciate if a closed form can be found of $\vartheta_v\left(e^{-\frac\pi2}\right)$ $\endgroup$
    – Max
    Commented Feb 18, 2023 at 18:02
  • $\begingroup$ @Anixx Thank you for your edit. One small help please. The link of Dedekind eta function $\eta(z)$ is not working $\endgroup$
    – Max
    Commented Feb 19, 2023 at 1:22

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