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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.

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    $\begingroup$ Why do you expect that there is a closed form? $\endgroup$
    – Gary
    Feb 19, 2023 at 7:20
  • $\begingroup$ @Gary Thanks for your comment. I believe that there is a closed form because it is very close to the value which has been proved in the question. $\endgroup$
    – Max
    Feb 19, 2023 at 7:23
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    $\begingroup$ Well if $1$ is close to $1-\mathrm{i}$, then so is $1/4$ to $1/2$. But unlike $\Gamma(1/2)$, $\Gamma(1/4)$ does not have a simple closed form. $\endgroup$
    – Gary
    Feb 19, 2023 at 8:11
  • $\begingroup$ @Gary I am sorry. Actually by close I meant that can we do this question in a similar way to that given in the post? $\endgroup$
    – Max
    Feb 19, 2023 at 10:22
  • $\begingroup$ Out of curiosity, what methods did your professor use to solve similar questions? $\endgroup$ Feb 19, 2023 at 12:44

1 Answer 1

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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.

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  • $\begingroup$ Thank you for your effort. I really appreciate it. +1 for it. One question I have: How to simplify the following $$ \sum\limits_{n = 1}^\infty {\frac{1}{{{\rm e}^{\pi n} + 1}}} = -\sum\limits_{k = 1}^\infty {\frac{{( - 1)^k }}{{{\rm e}^{\pi k} - 1}}}$$ $\endgroup$
    – Max
    Feb 20, 2023 at 4:03
  • $\begingroup$ You mean how to derive this equality? It is told in my answer. $\endgroup$
    – Gary
    Feb 20, 2023 at 4:05
  • $\begingroup$ Yes you are right. Thank you. $\endgroup$
    – Max
    Feb 20, 2023 at 4:05
  • $\begingroup$ Yes, but sorry I could not follow. Please add a line or two to explain that equality. $\endgroup$
    – Max
    Feb 20, 2023 at 4:07
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    $\begingroup$ There is not closed form in general. See however this. I wonder why you need all this. I can sense from your comments that you are desperate to have a fully rigorous answer (without you doing any of the work). $\endgroup$
    – Gary
    Feb 21, 2023 at 5:00

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