# 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.

• Why do you expect that there is a closed form?
– Gary
Feb 19, 2023 at 7:20
• @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.
– Max
Feb 19, 2023 at 7:23
• 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.
– Gary
Feb 19, 2023 at 8:11
• @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?
– Max
Feb 19, 2023 at 10:22
• Out of curiosity, what methods did your professor use to solve similar questions? Feb 19, 2023 at 12:44

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.

• Thank you for your effort. I really appreciate it. +1 for it. One question I have: How to simplify the following $$\sum\limits_{n = 1}^\infty {\frac{1}{{{\rm e}^{\pi n} + 1}}} = -\sum\limits_{k = 1}^\infty {\frac{{( - 1)^k }}{{{\rm e}^{\pi k} - 1}}}$$
– Max
Feb 20, 2023 at 4:03
• You mean how to derive this equality? It is told in my answer.
– Gary
Feb 20, 2023 at 4:05
• Yes you are right. Thank you.
– Max
Feb 20, 2023 at 4:05
• Yes, but sorry I could not follow. Please add a line or two to explain that equality.
– Max
Feb 20, 2023 at 4:07
• 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).
– Gary
Feb 21, 2023 at 5:00