Calculate $\sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} $

Question

question:

how do we find that: $$ S = \sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} $$

I modified the sum

$$\sum\limits_{n = - \infty }^\infty {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}}} = \frac{{\log \left( {{{\left( {\frac{1}{3}} \right)}^2}} \right)}}{{\frac{1}{3}}} + \sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}} + \frac{{\log \left( {{{\left( { - n + \frac{1}{3}} \right)}^2}} \right)}}{{ - n + \frac{1}{3}}}} \right)} $$

$${ = - 3\log \left( 9 \right) + \sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {{{\left( {n + \frac{1}{3}} \right)}^2}} \right)}}{{n + \frac{1}{3}}} - \frac{{\log \left( {{{\left( {n - \frac{1}{3}} \right)}^2}} \right)}}{{n - \frac{1}{3}}}} \right)} = - 3\log \left( 9 \right) + 6\sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {\frac{{3n + 1}}{3}} \right)}}{{3n + 1}} - \frac{{\log \left( {\frac{{3n - 1}}{3}} \right)}}{{3n - 1}}} \right)} = }$$

$${ = - 6\log \left( 3 \right) + 6\sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {3n + 1} \right) - \log 3}}{{3n + 1}} - \frac{{\log \left( {3n - 1} \right) - \log 3}}{{3n - 1}}} \right)} = }$$

$${ = - 6\log \left( 3 \right) + 6\sum\limits_{n = 1}^\infty {\left( {\frac{{\log 3}}{{3n - 1}} - \frac{{\log 3}}{{3n + 1}}} \right)} + 6\sum\limits_{n = 1}^\infty {\left( {\frac{{\log \left( {3n + 1} \right)}}{{3n + 1}} - \frac{{\log \left( {3n - 1} \right)}}{{3n - 1}}} \right)} }$$

Answer 1

welcome to Math SE , lets start by the series you got and denote it by $\Omega$ $$ \Omega=\sum_{n=1}^\infty \left(\frac{\ln\left(\frac{3n+1}{3}\right)}{3n+1}- \frac{\ln\left(\frac{3n-1}{3}\right)}{3n-1}\right)$$ which can be written as $$ \Omega=\frac{2}{\sqrt{3}}\sum_{n=1}^\infty \left(\sin\left(\frac{2\pi}{3}(3n+1) \right)\frac{\ln\left(\frac{3n+1}{3}\right)}{3n+1}+\sin \left(\frac{2\pi}{3}(3n-1) \right) \frac{\ln\left(\frac{3n-1}{3}\right)}{3n-1}\right)$$ now we have $$ \sum_{n=1}^{\infty} \left(\sin\left(\frac{2\pi}{3}(3n+1)\right)f(3n+1)+\sin\left(\frac{2\pi}{3}(3n-1)\right)f(3n-1)\right)=\sum_{n=2}^{\infty} \sin\left(\frac{2\pi}{3}n\right)f(n) $$ So $$ \Omega=\frac{2}{\sqrt{3}}\sum_{n=2}^\infty \sin\left(\frac{2\pi}{3}n\right) \frac{\ln \left(\frac{n}{3} \right)}{n} $$ $$=\frac{2}{\sqrt{3}}\sum_{n=1}^\infty \sin\left(\frac{2\pi}{3}n\right) \frac{\ln n}{n}-\frac{2 \ln3}{\sqrt{3}}\sum_{n=1}^\infty \frac{\sin\left(\frac{2\pi}{3}n\right)}{n}+\ln3$$ for the first series use Fourier expansion of log gamma function $$ \sum_{n=1}^\infty \sin(2\pi x n) \frac{\ln n}{n}=\pi \left(\ln \Gamma(x)-\left(\frac{1}{2}-x \right)(\gamma+\ln 2)+\frac{1}{2} \ln \sin(\pi x)-(1-x) \ln \pi \right)$$ So $$ \sum_{n=1}^\infty \sin\left(\frac{2\pi}{3}n\right) \frac{\ln n}{n}=\pi\ln \Gamma \left(\frac{1}{3} \right)-\frac{\pi}{12} \left(2 \gamma+8\ln (2\pi)-3\ln3 \right) $$ and its easy to show that $$ \sum_{n=1}^\infty \frac{\sin\left(\frac{2\pi}{3}n\right)}{n}=\frac{\pi}{6} $$ finally $$ \Omega=\frac{2\pi}{\sqrt{3}}\ln \Gamma \left(\frac{1}{3} \right)-\frac{\pi}{6\sqrt{3}} \left(2 \gamma+8\ln (2\pi)-\ln3 \right)+\ln3$$ and you will see that $$ S=4\sqrt{3}\pi\ln \Gamma \left(\frac{1}{3} \right)-\frac{\pi}{\sqrt{3}} \left(2 \gamma+8\ln (2\pi)-\ln3 \right) $$

Answer 2

\begin{align*} &\sum_{n = -\infty}^{\infty} \frac{\log \left( (n + \frac{1}{3})^2 \right)}{n + \frac{1}{3}} \\ &= \frac{\log \left( (\frac{1}{3})^2 \right)}{\frac{1}{3}} + \sum_{n = 1}^{\infty} \left( \frac{\log \left( (n + \frac{1}{3})^2 \right)}{n + \frac{1}{3}} + \frac{\log \left( (-n + \frac{1}{3})^2 \right)}{-n + \frac{1}{3}} \right) \\ &= -3\log(9) + \sum_{n = 1}^{\infty} \left( \frac{\log \left( (n + \frac{1}{3})^2 \right)}{n + \frac{1}{3}} - \frac{\log \left( (n - \frac{1}{3})^2 \right)}{n - \frac{1}{3}} \right) \\ &= -3\log(9) + 6\sum_{n = 1}^{\infty} \left( \frac{\log \left( \frac{3n + 1}{3} \right)}{3n + 1} - \frac{\log \left( \frac{3n - 1}{3} \right)}{3n - 1} \right) \\ &= -6\log(3) + 6\sum_{n = 1}^{\infty} \left( \frac{\log(3)}{3n - 1} - \frac{\log(3)}{3n + 1} \right) \\ &\quad+ 6\sum_{n = 1}^{\infty} \left( \frac{\log \left( 3n + 1 \right)}{3n + 1} - \frac{\log \left( 3n - 1 \right)}{3n - 1} \right) \\ &= -6\log(3) + 6\sum_{n = 1}^{\infty} \left( \frac{\log(3)}{3n - 1} - \frac{\log(3)}{3n + 1} \right) \\ &\quad+ 6\sum_{n = 1}^{\infty} \left( \frac{\log \left( 3n + 1 \right)}{3n + 1} - \frac{\log \left( 3n - 1 \right)}{3n - 1} \right) \end{align*}

But \begin{align*} &-6\log(3) + 6\sum_{n = 1}^{\infty} \left( \frac{\log(3)}{3n - 1} - \frac{\log(3)}{3n + 1} \right) \\ &= -6\log(3)\left(1 + \sum_{n = 1}^{\infty} \left( \frac{1}{3n + 1} - \frac{1}{3n - 1} \right) \right) \\ &= -6\log(3)\frac{2}{\sqrt{3}}\sum_{n = 1}^{\infty} \frac{\sin \frac{2n\pi}{3}}{n} \\ &= -\frac{4\sqrt{3}\log(3)}{2i}\sum_{n = 1}^{\infty} \frac{e^{2i\pi n/3} - e^{-2i\pi n/3}}{n} \\ &= -\frac{4\sqrt{3}\log(3)}{2i}\log\frac{1 - e^{-2i\pi/3}}{1 - e^{2i\pi/3}} \\ &= -\frac{4\sqrt{3}\log(3)}{2i}\log\left(-e^{e^{-2i\pi/3}}\right) \\ &= -\frac{4\sqrt{3}\log(3)}{2i}\log(e^{i\pi/3}) \\ &= -\frac{4\sqrt{3}\log(3)}{2i}\left(i\frac{\pi}{3}\right) \\ &= -\frac{2\pi\log(3)}{\sqrt{3}} \end{align*}

So \begin{align*} &\sum_{n = -\infty}^{\infty} \frac{\log \left( (n + \frac{1}{3})^2 \right)}{n + \frac{1}{3}} \\ &= -\frac{2\pi\log(3)}{\sqrt{3}} + 6\sum_{n = 1}^{\infty} \left( \frac{\log \left( 3n + 1 \right)}{3n + 1} - \frac{\log \left( 3n - 1 \right)}{3n - 1} \right) \\ &= -\frac{2\pi\log(3)}{\sqrt{3}} + 4\sqrt{3}\sum_{n = 1}^{\infty} \frac{\log n \cdot \sin \frac{2n\pi}{3}}{n} \end{align*}

From Kummer's formula [^1] we have \begin{align*} &\log \Gamma(x) = \frac{1}{\pi}\sum_{n = 1}^{\infty} \frac{\log n \cdot \sin(2\pi nx)}{n} \\ &\quad+ \frac{\log(2\pi) + \gamma}{\pi}\sum_{n = 1}^{\infty} \frac{\sin(2\pi nx)}{n} + \frac{1}{2}\sum_{n = 1}^{\infty} \frac{\cos(2\pi nx)}{n} + \frac{1}{2}\log(2\pi) \end{align*}

So for $x = \frac{1}{3}$ follows \begin{align*} &\log \Gamma\left(\frac{1}{3}\right) \\ &= \frac{1}{\pi}\sum_{n = 1}^{\infty} \frac{\log n \cdot \sin\left(\frac{2\pi n}{3}\right)}{n} \\ &\quad+ \frac{\log(2\pi) + \gamma}{\pi}\sum_{n = 1}^{\infty} \frac{\sin\left(\frac{2\pi n}{3}\right)}{n} + \frac{1}{2}\sum_{n = 1}^{\infty} \frac{\cos\left(\frac{2\pi n}{3}\right)}{n} + \frac{1}{2}\log(2\pi) \end{align*}

It holds $$\sum_{n = 1}^{\infty} \frac{\sin nx}{n} = \frac{\pi - x}{2}$$, therefore $$\sum_{n = 1}^{\infty} \frac{\sin\left(\frac{2\pi n}{3}\right)}{n} = \frac{\pi - 2\pi/3}{2} = \frac{\pi}{6}$$

as well as $$\sum_{n = 1}^{\infty} \frac{\cos nx}{n} = -\log\left(2\left|\sin\frac{x}{2}\right|\right)$$, i.e. $$\sum_{n = 1}^{\infty} \frac{\cos\left(\frac{2\pi n}{3}\right)}{n} = -\log\left(2\sin\frac{\pi}{3}\right) = -\frac{1}{2}\log 3$$

the above two sums have been proved in Mathematica

so \begin{align*} &\log \Gamma\left(\frac{1}{3}\right) \\ &= \frac{1}{\pi}\sum_{n = 1}^{\infty} \frac{\log n \cdot \sin\left(\frac{2\pi n}{3}\right)}{n} \\ &\quad+ \frac{\log(2\pi) + \gamma}{\pi}\frac{\pi}{6} - \frac{1}{4}\log 3 + \frac{1}{2}\log(2\pi) \end{align*}

i.e. \begin{align*} &\sum_{n = 1}^{\infty} \frac{\log n \cdot \sin\left(\frac{2\pi n}{3}\right)}{n} \\ &= \frac{\pi}{12}\left(12\log\Gamma\left(\frac{1}{3}\right) - 2\gamma + 3\log 3 - 8\log(2\pi)\right) \end{align*}

Finally \begin{align*} &\sum_{n = -\infty}^{\infty} \frac{\log \left( (n + \frac{1}{3})^2 \right)}{n + \frac{1}{3}} \\ &= \frac{\pi}{\sqrt{3}}\left(12\log\left(\Gamma\left(\frac{1}{3}\right)\right) + \log 3 - 2\gamma - 8\log(2\pi)\right) \end{align*}