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Let $\psi_q(z)$ be the q-DiGamma function defined for a complex variable $z$ with $\Re(z)>0$ 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 $$\int_{0}^\infty x^2( \psi_{e^{-2\pi}}(1+x)+\log(1-e^{-2\pi})) dx$$

My try Let $$I=\int_{0}^\infty x^2( \psi_{e^{-2\pi}}(1+x)+\log(1-e^{-2\pi})) dx$$ Integrating by parts taking first function as $u= \psi_{e^{-2\pi}}(1+x)+\log(1-e^{-2\pi}) $ and second function as $v=x^2$, we get $$I=-\frac{1}{3} \int_{0}^\infty x^3\psi_{e^{-2\pi}}^{(1)}(1+x) dx$$ The above integral converges (see:- https://www.wolframalpha.com/input?i=integral+x%5E3+QPolygamma%5B1%2C1%2Bx%2Ce%5E%28-2+pi%29%5D+from+0+to+infinity).

We have the following Stieltjes integral representation of $\psi_q(x)$ (see:- https://arxiv.org/abs/1301.1749) $$\psi_q(x)=-\log(1-q)-\int_{0}^{\infty}\frac{e^{-xt}}{1-e^{-t}} d\gamma_q(t)$$ where $0<q<1$, $x>0$ and $d\gamma_q(t)$ is a discrete measure with positive masses $-\log q$ at the positive points $-k\log q$, $k=1,2,...$. Also for $0<q<1$,$$\gamma_q(t)=-\log q\sum_{k=1}^{\infty}\delta(t+k\log q)$$ Hence by the above Stieltjes integral of $\psi_q(x)$ we have for $q=e^{-2 \pi}$ $$\psi_{e^{-2\pi}}(1+x)+\log(1-e^{-2\pi})=-\int_{0}^{\infty}\frac{e^{-(1+x)t}}{1-e^{-t}} d\gamma_{e^{-2\pi}}(t)$$ So we obtain $$\int_{0}^{\infty}x^2(\psi_{e^{-2\pi}}(1+x)+\log(1-e^{-2\pi})) dx=-\int_{0}^{\infty}\int_{0}^{\infty}\frac{x^2 e^{-(1+x)t}}{1-e^{-t}} d\gamma_{e^{-2\pi}}(t) dx$$ Changing the order of integration (I am not sure if that is allowed in this question) we get $$\int_{0}^{\infty}x^2(\psi_{e^{-2\pi}}(1+x)+\log(1-e^{-2\pi})) dx=-\int_{0}^{\infty} \frac{ e^{-t}}{1-e^{-t}} \left(\int_{0}^{\infty} x^2 e^{-t x} dx\right) \ d\gamma_{e^{-2\pi}}(t) $$ So we obtain $$\int_{0}^{\infty}x^2(\psi_{e^{-2\pi}}(1+x)+\log(1-e^{-2\pi})) dx=-2\int_{0}^{\infty} \frac{ e^{-t}}{t^3(1-e^{-t})} \ d\gamma_{e^{-2\pi}}(t) $$ So finally we have $$\int_{0}^{\infty}x^2(\psi_{e^{-2\pi}}(1+x)+\log(1-e^{-2\pi})) dx=-2\int_{0}^{\infty} \frac{1}{t^3(e^{t}-1)} \ d\gamma_{e^{-2\pi}}(t) $$ If anyone could please solve this question by hand or mathematica or sage math, I would be highly indebted to you all.

Max
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1 Answers1

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All series/integrals below converge absolutely, therefore a change in the order of summation/integration is justified.

By definition

$$\psi_{q}(z) + \ln(1-q) = \ln q \sum_{n = 0}^{\infty}\frac{q^{n+z}}{1-q^{n+z}}$$

Then, your integral is equal to

\begin{align} \mathfrak{I} &= -2\pi \sum_{n = 0}^{\infty}\int_{0}^{\infty}\frac{x^{2}}{e^{2\pi(n+x+1)} - 1}\,\mathrm{d}x \\ &= -2\pi \sum_{n = \color{red}{1}}^{\infty}\int_{0}^{\infty}\frac{x^{2}}{e^{2\pi(n+x)} - 1}\,\mathrm{d}x \\ &= -2\pi \sum_{n = 1}^{\infty} \sum_{k = 1}^{\infty}\int_{0}^{\infty}x^{2}e^{-2\pi k(n+x)}\,\mathrm{d}x \\ &= -2\pi \sum_{n = 1}^{\infty} \sum_{k = 1}^{\infty} e^{-2\pi kn} \int_{0}^{\infty}x^{2}e^{-2\pi kx}\,\mathrm{d}x \\ &= -\frac{1}{2\pi^2} \sum_{n = 1}^{\infty} \sum_{k = 1}^{\infty} \frac{1}{k^3e^{2\pi kn}} \\ &= -\frac{1}{2\pi^2} \sum_{k = 1}^{\infty} \frac{1}{k^3(e^{2\pi k} - 1)} \\ \end{align}

For the last series Wolfram Alpha gives the following answer: $$ \sum_{k = 1}^{\infty} \frac{1}{k^3(e^{2\pi k} - 1)} = \frac{7}{360}\pi^3 - \frac{1}{2}\zeta(3) $$

The final answer is: $$ \mathfrak{I} = \frac{1}{4}\frac{\zeta(3)}{\pi^2}-\frac{7}{720}\pi$$

Knas
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