@phymath7 You asked if there is any integral representation of the series that could facilitate its evaluation. In my notes, I provided a complex integral representation of the sum that can be easily evaluated. There is also real integral representations available, and they can be evaluated.
The sum has the following essentially equivalent integral representations:
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} & = \frac{1}{4 \cdot 6!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^{x} - 1)^2} dx + \frac{1}{4 \cdot 6!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^{x} + 1)^2} dx
\end{align*}
and
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} = (1 - 2^{-6}) \frac{1}{2 \cdot 6!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x - 1)^2} dx
\end{align*}
which I derive below. Comparing them you find
\begin{align*}
\int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x - 1)^2} dx = \frac{2^5}{2^5-1} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^{x} + 1)^2} dx
\end{align*}
This relation can also be proven directly. Substituting this into the second integral representation you get another integral representation
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} = \frac{2^6-1}{2^6-2} \frac{1}{2 \cdot 6!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x + 1)^2} dx
\end{align*}
This integral can be evaluated with complex contour integration techniques. More on this later.
Deriving integral representations of the sum
I now derive these integral representations. We have
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} & = \frac{1}{2} \sum_{n=1}^\infty \dfrac{1 - \cos \pi n }{n^6}
\nonumber \\
& = \frac{1}{2 \cdot 5!} \sum_{n=1}^\infty \dfrac{1 - \cos \pi n}{n^6} \int_0^\infty e^{-y} y^5 dy
\nonumber \\
& = \frac{1}{2 \cdot 5!} \sum_{n=1}^\infty [1 - \cos \pi n] \int_0^\infty e^{-nx} x^5 dx
\nonumber \\
& = \frac{1}{2 \cdot 5!} \sum_{n=1}^\infty \int_0^\infty [1 - \cos \pi n] e^{-nx} x^5 dx
\nonumber \\
& = \frac{1}{4 \cdot 5!} \sum_{n=1}^\infty \int_0^\infty (2 e^{-nx} - e^{-nx + i \pi n} - e^{-nx + -i \pi n} ) x^5 dx
\nonumber \\
& = \frac{1}{4 \cdot 5!} \int_0^\infty \left( \dfrac{2}{e^{x} - 1} - \dfrac{1}{e^{x - i \pi} - 1} - \dfrac{1}{e^{x + i\pi} - 1} \right) x^5 dx
\nonumber \\
& = \frac{1}{2 \cdot 5!} \int_0^\infty \left( \dfrac{1}{e^x - 1} + \dfrac{1}{e^{x} + 1} \right) x^5 dx
\nonumber \\
& = \frac{1}{2 \cdot 5!} \int_0^\infty \dfrac{x^5}{e^{x} - 1} dx + \frac{1}{2 \cdot 5!} \int_0^\infty \dfrac{x^5}{e^{x} + 1} dx
\end{align*}
Integrating by parts and then extending the range of integration:
\begin{align*}
\int_0^\infty \dfrac{x^5}{e^{x} - 1} dx & = \frac{1}{6} \int_0^\infty \dfrac{x^6 e^x}{(e^{x} - 1)^2} dx = \frac{1}{12} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^{x} - 1)^2} dx
\nonumber \\
\int_0^\infty \dfrac{x^5}{e^{x} + 1} dx & = \frac{1}{6} \int_0^\infty \dfrac{x^6 e^x}{(e^x + 1)^2} = \frac{1}{12} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x + 1)^2} dx
\end{align*}
Using these in the previous equation, we obtain the first integral representation of the sum quoted at the beginning.
Recall the identity
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} = (1 - 2^{-6}) \sum_{n=1}^\infty \dfrac{1}{n^6}
\end{align*}
We have
\begin{align*}
\sum_{n=1}^\infty \dfrac{1}{n^6} & = \frac{1}{5!} \sum_{n=1}^\infty \dfrac{1}{n^6} \int_0^\infty e^{-y} y^5 dy
\nonumber \\
& = \frac{1}{5!} \sum_{n=1}^\infty \int_0^\infty e^{-nx} x^5 dx
\nonumber \\
& = \frac{1}{5!} \int_0^\infty \dfrac{x^5}{e^{x} - 1} dx
\nonumber \\
& = \frac{1}{6 \cdot 5!} \int_0^\infty \dfrac{x^6 e^x}{(e^x - 1)^2} dx
\nonumber \\
& = \frac{1}{12 \cdot 5!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x - 1)^2} dx
\end{align*}
where in the second to last step we integrated by parts, followed by extending the range of integration. Substituting this into the identity I just mentioned results in the second integral representation of the sum quoted at the beginning.
As noted at the beginning, another representation follows from the first two:
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^6} = \frac{2^6-1}{2^6-2} \frac{1}{2 \cdot 6!} \int_{-\infty}^\infty \dfrac{x^6 e^x}{(e^x + 1)^2} dx
\end{align*}
I will illustrate a method of evaluating integrals of the form ##\int_{-\infty}^\infty \dfrac{x^{2k} e^x}{(e^x + 1)^2} dx## with the simpler example of the sum ##\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2}##.
Evaluating the integral: with simpler example
We look at the example:
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2}
\end{align*}
and evaluate the integral of the integral representation. This is simpler to do with this example. We have
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \frac{1}{2} \sum_{n=1}^\infty \dfrac{1 - \cos \pi n}{n^2} \int_0^\infty e^{-y} y dy
\nonumber \\
& = \frac{1}{2} \sum_{n=1}^\infty [1 - \cos \pi n] \int_0^\infty e^{-nx} x dx
\nonumber \\
& = \frac{1}{2} \sum_{n=1}^\infty \int_0^\infty [1 - \cos \pi n] e^{-nx} x dx
\nonumber \\
& = \frac{1}{4} \sum_{n=1}^\infty \int_0^\infty (2 e^{-nx} - e^{-nx + i\pi n} - e^{-nx + -i\pi n} ) x dx
\nonumber \\
& = \frac{1}{4} \int_0^\infty \left( \dfrac{2}{e^{x} - 1} - \dfrac{1}{e^{x - i \pi} - 1} - \dfrac{1}{e^{x + i\pi} - 1} \right) x dx
\nonumber \\
& = \frac{1}{2} \int_0^\infty \dfrac{x}{e^x - 1} dx + \frac{1}{2} \int_0^\infty \dfrac{x}{e^{x} + 1} dx
\nonumber \\
& = \frac{1}{4} \int_0^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx + \frac{1}{4} \int_0^\infty \dfrac{x^2 x^x}{(e^x + 1)^2} dx
\nonumber \\
& = \frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx + \frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx
\end{align*}
So
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = \frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx + \frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx \qquad (*)
\end{align*}
Alternatively, we can write
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} = (1 - 2^{-2}) \sum_{n=0}^\infty \dfrac{1}{n^2}
\end{align*}
We have
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{n^2} & = \sum_{n=1}^\infty \dfrac{1}{n^2} \int_0^\infty e^{-y} y dy
\nonumber \\
& = \sum_{n=1}^\infty \int_0^\infty e^{-nx} x dx
\nonumber \\
& = \int_0^\infty \dfrac{x}{e^{x} - 1} dx
\nonumber \\
& = \frac{1}{2} \int_0^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx
\nonumber \\
& = \frac{1}{4} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx
\end{align*}
So
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} & = (1 - 2^{-2}) \sum_{n=0}^\infty \dfrac{1}{n^2}
\nonumber \\
& = \frac{3}{16} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx \qquad (**)
\end{align*}
Comparing ##(*)## and ##(**)##, we have
\begin{align*}
\frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx + \frac{1}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx = \frac{3}{16} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx
\end{align*}
From which we get
\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x - 1)^2} dx = 2 \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx
\end{align*}
Substituting this into ##(**)## gives
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} = \frac{3}{8} \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx \qquad (***)
\end{align*}
We now evaluate this integral using complex analysis. Consider the rectangular contour, ##C##, in the figure
and the complex contour integral:
\begin{align*}
\oint_C \dfrac{e^z z^3}{(e^z + 1)^2} dz
\end{align*}
The integral along the vertical edges vanishes as:
\begin{align*}
f(z) = \dfrac{(x+iy)^3 e^{(x+iy)}}{(e^{x+iy} + 1)^2} =
\begin{cases}
(x+iy)^3 e^{-(x+iy)} & x \rightarrow \infty \\
(x+iy)^3 e^{(x+iy)} & x \rightarrow - \infty \\
\end{cases}
\end{align*}
So that
\begin{align*}
\oint_C \dfrac{e^z z^3}{(e^z + 1)^2} dz & = \int_{-\infty}^\infty \dfrac{e^x x^3}{(e^x + 1)^2} dx - \int_{-\infty+2 \pi i}^{\infty + 2 \pi i} \dfrac{e^x (x+ 2 \pi i)^3}{(e^x - 1)^2} dx
\nonumber \\
& = - \int_{-\infty}^\infty \dfrac{e^x (3 x^2 (2 \pi i) + 3x (2 \pi i)^2 + (2\pi i)^3)}{(e^x + 1)^2} dx
\nonumber \\
& = - 6 \pi i \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx + 12 \pi^2 \int_{-\infty}^\infty \dfrac{x e^x}{(e^x + 1)^2} dx
\nonumber \\
& i 8 \pi^3 \int_{-\infty}^\infty \dfrac{e^x}{(e^x + 1)^2} dx
\end{align*}
The function ##\dfrac{e^z z^3}{(e^z + 1)^2}## has residue ##3 \pi^2## at ##z=i \pi## (we calculate this in a moment). So that
\begin{align*}
6 \pi^3 = - 6 \pi \int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx + 8 \pi^3 \int_{-\infty}^\infty \dfrac{e^x}{(e^x + 1)^2} dx
\end{align*}
Which rearranged is
\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx = - \pi^2 + \frac{4}{3} \pi^2 \int_{-\infty}^\infty \dfrac{e^x}{(e^x + 1)^2} dx
\end{align*}
The integral on the RHS is straightforward to do:
\begin{align*}
\int_{-\infty}^\infty \dfrac{e^x}{(e^x + 1)^2} dx & = \int_{-\infty}^\infty \left( - \frac{d}{dx} \dfrac{1}{e^x + 1} \right) dx
\nonumber \\
& =\left[ - \dfrac{1}{e^x + 1} \right]_{-\infty}^\infty = 1
\end{align*}
So,
\begin{align*}
\int_{-\infty}^\infty \dfrac{x^2 e^x}{(e^x + 1)^2} dx = \frac{\pi^2}{3}
\end{align*}
Plugging this into ##(***)##, we finally have
\begin{align*}
\sum_{n=0}^\infty \dfrac{1}{(2n+1)^2} = \frac{\pi^2}{8} .
\end{align*}
Calculating the residue:
We now calculate
\begin{align*}
\text{Res}_{z = \pi i} \dfrac{e^z z^3}{(e^z + 1)^2}
\end{align*}
We have, where ##z_0## is the position of the pole ##\pi i##:
\begin{align*}
\frac{1}{(e^z + 1)^2} & = \frac{1}{(e^z - e^{z_0})^2}
\nonumber \\
& = \frac{e^{-2 z_0}}{(e^{z-z_0} - 1)^2}
\nonumber \\
& = \frac{e^{-2 z_0}}{[(z-z_0) + \frac{1}{2!} (z-z_0)^2 + \cdots]^2}
\nonumber \\
& = \dfrac{e^{-2 z_0}}{(z-z_0)^2 [1 + \frac{1}{2!} (z-z_0) + \cdots]^2}
\nonumber \\
& = \dfrac{e^{-2z_0}}{(z-z_0)^2} [1 - (z-z_0) + \cdots]
\nonumber \\
& = \dfrac{e^{-2z_0}}{(z-z_0)^2} - \dfrac{e^{-2z_0}}{z-z_0} + \cdots
\end{align*}
We use this in extracting the residue of ##f(z)## at ##z_0##:
\begin{align*}
\frac{e^z z^3}{(e^z + 1)^2} & = \frac{e^z z^3}{(e^z - e^{z_0})^2}
\nonumber \\
& = e^{-2z_0} \frac{e^{z_0} e^{z-z_0} [z_0 +(z-z_0)]^3}{(z-z_0)^2} - e^{-2z_0} \dfrac{z_0^3 e^{z_0}}{z-z_0} + \cdots
\nonumber \\
& = e^{-z_0} \dfrac{[1 + (z-z_0) + \cdots] [z_0^3 + 3z_0^2 (z-z_0) + \cdots]}{(z-z_0)^2} - e^{-z_0} \dfrac{z_0^3}{z-z_0} + \cdots
\nonumber \\
& = e^{-z_0} \dfrac{z_0^3}{z-z_0} + e^{-z_0} \dfrac{3 z_0^2}{z-z_0} + \cdots
\end{align*}
So
\begin{align*}
\text{Res}_{z = \pi i} \dfrac{e^z z^3}{(e^z + 1)^2} & = 3 \pi^2 .
\end{align*}