Board index Computation of Series Calculation of $\sum_{n=1}^{\infty}\frac{\psi_1(n)}{2^nn^2}$

## Calculation of $\sum_{n=1}^{\infty}\frac{\psi_1(n)}{2^nn^2}$

Post your questions related to Computation of Series here.

Moderators: galactus, sos440, zaidalyafey

### Calculation of $\sum_{n=1}^{\infty}\frac{\psi_1(n)}{2^nn^2}$

Thu Apr 07, 2016 3:59 am

Posts: 852
Location: Jaipur, India

\begin{align*}
\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n^2} &= -\sum_{n=1}^\infty\psi_1(n)\left(\frac{\log(2)}{2^n n}+\int_0^{\frac{1}{2}} x^{n-1}\log(x)dx\right) \\
&= \color{green}{-\log(2)\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n}}\color{blue}{-\int_0^{1\over 2}\frac{\log(x)}{1-x}\left(\zeta(2)-\text{Li}_2(x) \right)dx }\tag{1}
\end{align*}

Evaluation of $\int_0^{1\over 2}\frac{\log(x)}{1-x}\left(\zeta(2)-\text{Li}_2(x) \right)dx$

Using the identity $\text{Li}_2(x)+\text{Li}_2(1-x)=\zeta(2)-\log(x)\log(1-x)$, we have

\begin{align*}
\int_0^{1\over 2}\frac{\log(x)}{1-x}\left(\zeta(2)-\text{Li}_2(x) \right)dx &= \color{red}{\int_0^{1\over 2}\frac{\log^2(x)\log(1-x)}{1-x}dx}+\color{orange}{\int_0^{1\over 2}\frac{\log(x)\text{Li}_2(1-x)}{1-x}dx} \\
&=\color{red}{-\frac{\pi^4}{360}-\frac{\log^4(2)}{4}}+\color{orange}{\left(\frac{1}{2}\text{Li}_2^2(1-x) \right)\Big|_0^{1\over 2}} \\
&= \boxed{-\frac{19\pi^4}{1440}-\frac{\pi^2}{24}\log^2(2)-\frac{\log^4(2)}{8}}\tag{2}
\end{align*}

The red integral was evaluated using equation 12 on page 310 of Leonard Lewin - Polylogarithms and associated functions. The proof is discussed on page 203 and 204.

Evaluation of $\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n}$

\begin{align*}
\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n} &= \sum_{n=1}^\infty \psi_1(n)\int_0^{1\over 2}x^{n-1}dx \\
&= \int_0^{\frac{1}{2}}\frac{\zeta(2)-\text{Li}_2(x)}{1-x}dx \\
&= \color{brown}{\int_0^{1\over 2}\frac{\log(x)\log(1-x)}{1-x}dx}+\color{violet}{\int_0^{1\over 2}\frac{\text{Li}_2(1-x)}{1-x}dx} \\
&= \color{brown}{\frac{\log^3(2)}{2}+\frac{1}{2}\int_0^{\frac{1}{2}}\frac{\log^2(1-x)}{x}dx }+\color{violet}{\zeta(3)-\text{Li}_3\left(\frac{1}{2}\right)} \\
&= \color{brown}{\frac{\log^3(2)}{2}+\frac{1}{2}\left( -\log^3(2)-2\log(2)\text{Li}_2\left(\frac{1}{2}\right)+2\zeta(3)-2\text{Li}_3\left( \frac{1}{2}\right)\right) }+\color{violet}{\zeta(3)-\text{Li}_3\left(\frac{1}{2}\right)} \\
&= \boxed{\frac{\pi^2}{12}\log(2)+\frac{\log^3(2)}{6}+\frac{\zeta(3)}{4}}\tag{3}
\end{align*}

The brown integral was evaluated using the generalized result found in this thread.

Substitute (2) and (3) into (1) to get $$\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n^2}=\boxed{\frac{19\pi^4}{1440}-\frac{\pi^2}{24}\log^2(2)-\frac{\log^4(2)}{24}-\frac{\zeta(3) \log(2)}{4}}$$