Closed form expression for $\psi_{e^{\pi}}^{(3)}(1-i)$ Let $\psi_q(z)$ be the q-DiGamma function defined for a real variable $\Re(z)>0$ as $$\psi_q(z)=\frac{1}{\Gamma_q(z)}\frac{\partial}{\partial z} (\Gamma_q(z))$$
where $\Gamma_q(z)$ is the q-Gamma function defined as $$\Gamma_q(z)=(1-q)^{1-z}\prod_{n=0}^{\infty}\frac{1-q^{n+1}}{1-q^{n+z}}$$
Question I am looking for a closed form for $$\psi_{e^{\pi}}^{(3)}(1-i)$$ where $i=\sqrt{-1}$
Here is a beautiful answer for calculating $$\psi_{e^{\pi}}^{(3)}(1)$$
Wolfram Alpha gives the expansion at $x=\infty$:
$$\psi_x^{(3)}(1)=\ln^4(x)\left(x^{-1}+9x^{-2}+\dots\right)$$
and these match Oeis A$001158$ with divisor $\sigma_v(n)$ and various theta functions after plugging the sum back in here. Use $\vartheta_v(0,x)=\vartheta_v(x)$:
$$\psi_x^{(3)}(1)=\ln^4(x)\sum_{n=1}^\infty\frac{\sigma_3(n)}{x^n}=\frac{\ln^4(x)}{480}\left(\vartheta_2\left(\frac1{\sqrt x}\right)^8+ \vartheta_3\left(\frac1{\sqrt x}\right)^8+ \vartheta_4\left(\frac1{\sqrt x}\right)^8-2\right)$$
Therefore:
$$\psi_{e^\pi}^{(3)}(1)=\frac{\pi^4}{480}\left(\vartheta_2^8\left(e^{-\frac\pi2}\right)+ \vartheta_3^8\left(e^{-\frac\pi2}\right)+ \vartheta_4^8\left(e^{-\frac\pi2}\right)-2\right)$$
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:
$$\boxed{\psi_{e^\pi}^{(3)}(1)=\frac{11\Gamma\left(\frac14\right)^8}{5120\pi^2}-\frac{\pi^4}{240}}$$
shown here
If anyone could please solve this question by hand or mathematica or sage math. I would be highly indebted to you all.
 A: I am not an expert in such sums, this derivation can probably be done in a simpler way. From the link provided by Tyma Gaidash, we have
\begin{align*}
\psi _{{\rm e}^\pi  }^{(3)} (1 - {\rm i}) = & - \pi ^4 \sum\limits_{n = 1}^\infty  {\frac{1}{{{\rm e}^{\pi n}  + 1}}}  + 7\pi ^4 \sum\limits_{n = 1}^\infty  {\frac{1}{{({\rm e}^{\pi n}  + 1)^2 }}} \\ & - 12\pi ^4 \sum\limits_{n = 1}^\infty  {\frac{1}{{({\rm e}^{\pi n}  + 1)^3 }}}  + 6\pi ^4 \sum\limits_{n = 1}^\infty  {\frac{1}{{({\rm e}^{\pi n}  + 1)^4 }}} .
\end{align*}
Re-expanding the fractions in powers of ${\rm e}^{-\pi n}$, changing the order of summation and re-summing again leads to the identities
\begin{align*}
& \sum\limits_{n = 1}^\infty  {\frac{1}{{{\rm e}^{\pi n}  + 1}}}  = -\sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k }}{{{\rm e}^{\pi k}  - 1}}} ,
\\ &\sum\limits_{n = 1}^\infty  {\frac{1}{{({\rm e}^{\pi n}  + 1)^2 }}}  = \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k k}}{{{\rm e}^{\pi k}  - 1}}}  - \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k }}{{{\rm e}^{\pi k}  - 1}}} ,
\\ & \sum\limits_{n = 1}^\infty  {\frac{1}{{({\rm e}^{\pi n}  + 1)^3 }}}  = -\frac{1}{2}\sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k k^2 }}{{{\rm e}^{\pi k}  - 1}}}  + \frac{3}{2}\sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k k}}{{{\rm e}^{\pi k}  - 1}}}  - \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k }}{{{\rm e}^{\pi k}  - 1}}} ,
\\ &\sum\limits_{n = 1}^\infty  {\frac{1}{{({\rm e}^{\pi n}  + 1)^4 }}}  = \frac{1}{6}\sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k k^3 }}{{{\rm e}^{\pi k}  - 1}}}  - \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k k^2 }}{{{\rm e}^{\pi k}  - 1}}}  + \frac{{11}}{6}\sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k k}}{{{\rm e}^{\pi k}  - 1}}}  - \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k }}{{{\rm e}^{\pi k}  - 1}}} .
\end{align*}
Substituting back to the original formula yields the simpler form
$$
\psi _{{\rm e}^\pi  }^{(3)} (1 - {\rm i}) = \pi ^4 \sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k k^3 }}{{{\rm e}^{\pi k}  - 1}}} .
$$
Now we have
$$
\sum\limits_{k = 1}^\infty  {\frac{{( - 1)^k k^3 q^k }}{{1 - q^k }}}  = \frac{{\theta _4^8 (0,q) - 1}}{{16}}
$$
for $|q|<1$ with $\theta_4$ being one of the theta functions. Hence,
$$
\psi _{{\rm e}^\pi  }^{(3)} (1 - {\rm i}) = \pi ^4 \frac{{\theta _4^8 (0,{\rm e}^{ - \pi } ) - 1}}{{16}}.
$$
Using the specific value
$$
\theta _4 (0,{\rm e}^{ - \pi } ) = \frac{{\pi ^{1/4} }}{{2^{1/4} \Gamma (3/4)}},
$$
(see equation $(45)$ here) we finally have
$$\boxed{
\psi _{{\rm e}^\pi  }^{(3)} (1 - {\rm i}) = \frac{{\pi ^6 }}{{64\Gamma ^8 (3/4)}} - \frac{{\pi ^4 }}{{16}}.}
$$
Numerical check.
