Prove that $\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac{\pi}{3}$ I was trying to solve the integral $$\int_{0}^{\infty}\frac{x^4}{1+x^6}dx$$
using series. Now I'm stuck at the series below.
How to prove that 
$$\sum_{j=0}^{\infty}(-1)^j\left(\frac{1}{5+6j}+\frac{1}{1+6j} \right) = \frac{\pi}{3}?$$
 A: Denote 
$$
S = \sum_{j=0}^{\infty}(-1)^j \left( \frac{1}{6j+1} + \dfrac{1}{6j+5}\right).
$$
And denote
$$
T= \sum_{j=0}^{\infty} (-1)^j \dfrac{1}{2j+1}.
$$
According to Leibniz formula for $\pi$, $\;T=\dfrac{\pi}{4}$.
Then
$$S - \dfrac{1}{3}T = \\
S - \dfrac{1}{3}\sum_{j=0}^{\infty} (-1)^j \dfrac{1}{2j+1} = \\ S-\sum_{j=0}^{\infty}(-1)^{j}\left(\dfrac{1}{6j+3}\right) = \\
\sum_{j=0}^{\infty}(-1)^j \left( \frac{1}{6j+1} - \frac{1}{6j+3} + \dfrac{1}{6j+5}\right) = \\ 
\sum_{k=0}^{\infty} (-1)^k \dfrac{1}{2k+1} = \\ T.
$$
Therefore $$S-\dfrac{1}{3}T = T,$$
$$S=\dfrac{4}{3}T,$$
$$S=\dfrac{4}{3}\cdot \dfrac{\pi}{4}=\dfrac{\pi}{3}.$$
A: $$
\begin{align}
\sum_{j=0}^\infty(-1)^j\left(\frac1{5+6j}+\frac1{1+6j}\right)
&=\sum_{j=0}^\infty(-1)^j\left(\frac1{-1-6(-j-1)}+\frac1{1+6j}\right)\tag1\\
&=\sum_{j=0}^\infty(-1)^{-j-1}\frac1{1+6(-j-1)}+\sum_{j=0}^\infty(-1)^j\frac1{1+6j}\tag2\\
&=\sum_{j=-\infty}^{-1}(-1)^j\frac1{1+6j}+\sum_{j=0}^\infty(-1)^j\frac1{1+6j}\tag3\\
&=\frac16\sum_{j=-\infty}^\infty(-1)^j\frac1{\frac16+j}\tag4\\[6pt]
&=\frac\pi6\csc\left(\frac\pi6\right)\tag5\\[12pt]
&=\frac\pi3\tag6
\end{align}
$$
Explanation:
$(1)$: $5+6j=-1-6(-j-1)$
$(2)$: split the sum in two
$(3)$: substitute $j\mapsto-1-j$
$(4)$: combine the sums
$(5)$: use $(3)$ from this answer
$(6)$: evaluate
A: $\begin{align}J&=\int_0^{\infty}\frac{x^4}{1+x^6}\,dx\\
&=\int_0^{\infty}\frac{2x^2-1}{3(x^4-x^2+1)}\,dx+\frac{1}{3}\int_0^{\infty}\frac{1}{1+x^2}\,dx\\
&=\int_0^{\infty}\frac{2x^2-1}{3(x^4-x^2+1)}\,dx+\frac{1}{3}\Big[\arctan x\Big]_0^{\infty}\\
&=\frac{1}{3}\int_0^{\infty}\frac{2x^2-1}{x^4-x^2+1}\,dx+\frac{1}{6}\pi
\end{align}$
Perform the change of variable $y=\dfrac{1}{x}$,
$\begin{align}K&=\int_0^{\infty}\frac{2x^2-1}{x^4-x^2+1}\,dx\\
&=\int_0^{\infty}\frac{2-x^2}{x^4-x^2+1}\,dx
\end{align}$
Therefore,
$\begin{align}2K&=\int_0^{\infty}\frac{(2x^2-1)+(2-x^2)}{x^4-x^2+1}\,dx\\
&=\int_0^{\infty}\frac{1+x^2}{x^4-x^2+1}\,dx\\
&=\int_{0}^{+\infty}\frac{\frac{1}{x^2}+1}{x^2-1+\frac{1}{x^2}}\,dx\\
&=\int_{0}^{+\infty}\frac{\frac{1}{x^2}+1}{\left(x-\frac{1}{x}\right)^2+1}\,dx
\end{align}$
Perform the change of variable $y=x-\dfrac{1}{x}$,
$\begin{align}2K&=\int_{-\infty}^{+\infty}\frac{1}{1+x^2}\,dx\\
&=\Big[\arctan x\Big]_{-\infty}^{+\infty}\\
&=\pi
\end{align}$
Therefore,
$\begin{align}K&=\frac{1}{2}\pi\end{align}$
Therefore,
$\begin{align}J&=\frac{1}{3}K+\frac{1}{6}\pi\\
&=\frac{1}{3}\times \frac{1}{2}\pi++\frac{1}{6}\pi\\
&=\boxed{\frac{1}{3}\pi}
\end{align}$
