How can I find $\sum\limits_{n=0}^{\infty}\left(\frac{(-1)^n}{2n+1}\sum\limits_{k=0}^{2n}\frac{1}{2n+4k+3}\right)$?

Solution 1:

1. First of all let us introduce the function $$ F(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\sum_{k=0}^{2n}\frac{x^{2n+4k+3}}{2n+4k+3} $$ We need to compute $F(1)$. Let us start by calculating $$ x^{-2}F'(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\sum_{k=0}^{2n}x^{2n+4k}= \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\frac{x^{2n}-x^{4+10n}}{1-x^4}=\frac{\arctan x-\arctan x^5}{x(1-x^4)},$$ where the first step is obtained by straightforward differentiation, the second equality follows by summation of finite geometric series, and the third one is obtained from the series representation $\arctan x=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}$. Since it is clear from the definition that $F(0)=0$, we find $$ F(1)=\int_0^1\frac{x\left(\arctan x-\arctan x^5\right) dx}{1-x^4}$$ Integrate by parts using that $\int\frac{xdx}{1-x^4}=\frac14\ln\frac{1+x^2}{1-x^2}$ (and saying appropriate words about the limits). This gives \begin{align*} F(1)=\frac14\int_0^1\left(\frac{1}{1+x^2}-\frac{5x^4}{1+x^{10}}\right)\ln\left(\frac{1-x^2}{1+x^2}\right)dx=\qquad\qquad\qquad\\ =\frac14\int_0^1\frac{(1+x^2)(1-3x^2+x^4)}{x^8-x^6+x^4-x^2+1}\ln\left(\frac{1-x^2}{1+x^2}\right)dx.\qquad\qquad (\mathrm{A}) \end{align*}

2. Notice that the change of integration variable $x\rightarrow 1/x$ and parity allow to rewrite (A) as $$F(1)=\frac{1}{16}\int_{-\infty}^{\infty}\frac{(1+x^2)(1-3x^2+x^4)}{x^8-x^6+x^4-x^2+1}\ln\left|\frac{1-x^2}{1+x^2}\right|dx $$ Next we write this in terms of complex contour integral: $$F(1)=\frac{1}{16}\mathrm{Re}\left\{\int_C f(z)dz\right\}\qquad\qquad (\mathrm{B})$$ with $$ f(z)=\frac{(1+z^2)(1-3z^2+z^4)}{z^8-z^6+z^4-z^2+1}\ln\left(\frac{1-z^2}{1+z^2}\right) $$ At this point, some comments are necessary:

   • the function $f(z)$ has a number of singularities in the complex $z$-plane. First of all one has 8 simple poles $z=\exp\pm\frac{ik\pi}{10}$, $k=1,3,7,9$. One also has logarithmic branch points $z=\pm1,\pm i$. Accordingly, we introduce 4 branch cuts: $B_{1}=[1,\infty)$, $B_{-1}=(-\infty,-1]$, $B_i=[i,i\infty)$, $B_{-i}=(-i\infty,-i]$.
   • the contour of integration $C$ runs from $-\infty$ to $+\infty$ slightly above the real axis (we could put it slightly below and modify what follows). The logarithms in $f(z)$ will be defined on their main sheets for $z\in(-1,1)$, then $f(z)$ is unambigously defined in the cut plane by analytic continuation.
   • Going slightly above the branch cuts $B_{\pm1}$ produces (irrelevant) imaginary parts in the logarithms. This explains the need to take real part in (B).

3. Now the idea is to pull the contour $C$ to $i\infty$. The integral will then be given by the sum of residues at $z=\exp\frac{ik\pi}{10}$, $k=1,3,7,9$, plus integral of the jump of $f(z)$ on the cut $B_i$. It is not difficult to understand that the latter is real (rational prefactor $\frac{(1+z^2)(1-3z^2+z^4)}{z^8-z^6+z^4-z^2+1}$ is real on the cut, the logarithm jumps by $2\pi i$ and the integral is along $B_i$ so that the integration variable is pure imaginary). Therefore we are left with $$F(1)=\frac{1}{16}\left[2\pi\int_1^{\infty}\frac{(1-x^2)(1+3x^2+x^4)dx}{x^8+x^6+x^4+x^2+1}+\mathrm{Re}\left\{2\pi i \sum_{k=1,3,7,9}\mathrm{res}_{z=\exp\frac{ik\pi}{10}}f(z)\right\}\right]\qquad(\mathrm{C})$$ In the first integral we have a rational function, so it can be calculated by elementary means: $$\int_1^{\infty}\frac{(1-x^2)(1+3x^2+x^4)dx}{x^8+x^6+x^4+x^2+1}=\left[\frac12\ln\frac{1+x+x^2+x^3+x^4}{1-x+x^2-x^3+x^4}\right]_{x=1}^{x=\infty}=-\frac{\ln5}{2}$$ The residues are also relatively easily computed: \begin{align} \mathrm{res}_{z=\exp\frac{i\pi}{10}}f(z)=\frac{i}{2}\left(\ln\tan\frac{\pi}{10}-\frac{i\pi}{2}\right),\\ \mathrm{res}_{z=\exp\frac{3i\pi}{10}}f(z)=-\frac{i}{2}\left(\ln\tan\frac{3\pi}{10}-\frac{i\pi}{2}\right),\\ \mathrm{res}_{z=\exp\frac{7i\pi}{10}}f(z)=-\frac{i}{2}\left(\ln\tan\frac{3\pi}{10}+\frac{i\pi}{2}\right),\\ \mathrm{res}_{z=\exp\frac{9i\pi}{10}}f(z)=\frac{i}{2}\left(\ln\tan\frac{\pi}{10}+\frac{i\pi}{2}\right). \end{align} Substituting this into (C), we finally obtain $$ F(1)=\frac{\pi}{8}\left(\ln\frac{\tan\frac{3\pi}{10}}{\tan\frac{\pi}{10}}-\frac{\ln 5}{2}\right) $$ The statement now follows from the easily verified identity $\frac{\tan\frac{3\pi}{10}}{\tan\frac{\pi}{10}}=\left(\frac{1+\sqrt{5}}{2}\right)^3$. $\blacksquare$