Suppose we seek to show that
$$\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^3} \frac{1}{\sinh(\pi n)}
= \frac{\pi^3}{360}.$$

Using $$\frac{1}{\sinh(x)} = \frac{2}{e^x-e^{-x}}
= 2\frac{e^{-x}}{1-e^{-2x}}$$
this is the same as

$$\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^3}
\frac{e^{-n\pi}}{1-e^{-2n\pi}}
= \frac{\pi^3}{720}.$$

The sum term may be evaluated using harmonic summation techniques.
Since this method has not been presented I will detail this
calculation here.

Let $p$ be a positive integer and introduce

$$S(x;p) =
\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{2p+1}}
\frac{e^{-nx}}{1-e^{-2nx}}.$$

We will evaluate $S(\pi;p)$ using a functional equation for $S(x;p)$
that is obtained by inverting its Mellin transform.

Recall the harmonic sum identity
$$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have
$$\lambda_k = \frac{(-1)^{k+1}}{k^{2p+1}}, \quad
\mu_k = k
\quad \text{and} \quad
g(x) = \frac{e^{-x}}{1-e^{-2x}}.$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which is

$$\int_0^\infty \frac{e^{-x}}{1-e^{-2x}} x^{s-1} dx
= \int_0^\infty
\sum_{q\ge 0} e^{- (2 q+1) x} x^{s-1} dx
= \sum_{q\ge 0} \int_0^\infty e^{-(2q+1)x} x^{s-1} dx
\\= \Gamma(s) \sum_{q\ge 0} \frac{1}{(2q+1)^s}
= \left(1-\frac{1}{2^s}\right) \Gamma(s) \zeta(s)$$
with fundamental strip $\langle 1, \infty\rangle.$

It follows that the Mellin transform $Q(s)$ of the harmonic sum
$S(x;p)$ is given by

$$Q(s) = \left(1-\frac{1}{2^s}\right)
\left(1-\frac{1}{2^{s+2p}}\right)
\Gamma(s) \zeta(s) \zeta(s+2p+1)
\\ \text{because}\quad
\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} =
\sum_{k\ge 1} \frac{(-1)^{k+1}}{k^{2p+1}} \frac{1}{k^s}
= \left(1-\frac{2}{2^{s+2p+1}}\right) \zeta(s+2p+1)$$
for $\Re(s+2p+1) > 1$ or $\Re(s) \gt -2p.$

The Mellin inversion integral here is
$$\frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} Q(s)/x^s ds$$
which we evaluate by shifting it to the left for an expansion about
zero.

Fortunately the trivial zeros of the two zeta function terms cancel
the poles of the gamma function term. The first term cancels those at
$-2m$ where $m\ge 1$ and the second one the odd ones from $-2p-3$ on,
which leaves the poles at $s=1$, and $-2q-1$ with $0\le q\le p.$ It
would appear there is a pole at $s=-2p$ but this is not the case since
we have two simple poles among the five terms but also two zero
values, making for cancelation. The pole at $s=0$ is canceled as
well.

For the residue at $s=1$ we find

$$\frac{1}{2} \frac{2^{2p+1}-1}{2^{2p+1}}
\times 1 \times \zeta(2p+2) \frac{1}{x}
= \frac{2^{2p+1}-1}{2^{2p+2}}
\frac{(-1)^{p} B_{2p+2} (2\pi)^{2p+2}}{2 (2p+2)!} \frac{1}{x}
\\ = (2^{2p+1}-1)
\frac{(-1)^{p} B_{2p+2} \pi^{2p+2}}{2 (2p+2)!} \frac{1}{x}.$$

The negative odd values at $s=-2q-1$ yield

$$\left(1-\frac{1}{2^{-2q-1}}\right)
\left(1-\frac{1}{2^{2p-2q-1}}\right)
\frac{(-1)^{2q+1}}{(2q+1)!} \zeta(-2q-1) \zeta(2p-2q) x^{2q+1}
\\ = (1-2^{2q+1})
\left(1-\frac{1}{2^{2p-2q-1}}\right)
\frac{1}{(2q+1)!} \frac{B_{2q+2}}{2q+2}
\frac{(-1)^{p-q+1} B_{2p-2q} (2\pi)^{2p-2q}}{2(2p-2q)!}
x^{2q+1}
\\ = \frac{1}{2} (1-2^{2q+1})
(2^{2p-2q-1}-1)
\frac{(-1)^{p-q+1}}{(2q+1)!}
\frac{B_{2q+2} B_{2p-2q} \pi^{2p-2q}}{(2p-2q)! (q+1)}
x^{2q+1}.$$

Shifting to $\Re(s) = -2p -3/2$ we get

$$S(x;p) =
(2^{2p+1}-1)
\frac{(-1)^{p} B_{2p+2} \pi^{2p+2}}{2 (2p+2)!} \frac{1}{x}
\\ +
\frac{1}{2} \sum_{q=0}^p
(1-2^{2q+1})
(2^{2p-2q-1}-1)
\frac{(-1)^{p-q+1}}{(2q+1)!}
\frac{B_{2q+2} B_{2p-2q} \pi^{2p-2q}}{(2p-2q)! (q+1)}
x^{2q+1}
\\ + \frac{1}{2\pi i}
\int_{-2p-3/2-i\infty}^{-2p-3/2+i\infty} Q(s)/x^s ds.$$

We will turn this into the promised functional equation.

Substitute $s = -2p - t$ in the remainder integral to get

$$- \frac{1}{2\pi i}
\int_{3/2+i\infty}^{3/2-i\infty}
\left(1-2^{2p+t}\right)
\left(1-2^t\right)
\Gamma(-2p-t) \zeta(-2p-t) \zeta(1-t)
x^{t+2p} dt$$
which is

$$\frac{x^{2p}}{2\pi i}
\int_{3/2-i\infty}^{3/2+i\infty}
\left(1-2^{2p+t}\right)
\left(1-2^t\right)
\Gamma(-2p-t) \zeta(-2p-t) \zeta(1-t)
x^{t} dt.$$

In view of the desired functional equation we now use the functional
equation of the Riemann zeta function on $Q(s)$ to prove that the
integrand of the last integral is in fact $(-1)^p Q(t)/\pi^{2p+2t}.$

Start with the functional equation
$$\zeta(1-s) = \frac{2}{2^s\pi^s}
\cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s)$$
and substitute this into $Q(s)$ to obtain

$$Q(s) =
\left(1-\frac{1}{2^s}\right)
\left(1-\frac{1}{2^{s+2p}}\right)
\frac{\zeta(1-s) 2^s \pi^s}{2\cos\left(\frac{\pi s}{2}\right)}
\zeta(s+2p+1)
\\ = \frac{1}{2}
(2^s-1) \left(1-\frac{1}{2^{s+2p}}\right)
\pi^s \frac{\zeta(s+2p+1)}{\cos\left(\frac{\pi s}{2}\right)}
\zeta(1-s).$$

Apply the functional equation again (this time to $\zeta(s+2p+1)$) to
get

$$Q(s) = \frac{1}{2} \frac{\pi^s}{\cos\left(\frac{\pi s}{2}\right)}
(2^s-1) \left(1-\frac{1}{2^{s+2p}}\right)
\frac{2}{2^{-2p-s} \pi^{-2p-s}}
\cos\left(\frac{\pi (-2p-s)}{2}\right)
\\ \times \Gamma(-2p-s) \zeta(-2p-s) \zeta(1-s)
\\ = \frac{\pi^s}{\cos\left(\frac{\pi s}{2}\right)}
(2^s-1) (2^{2p+s}-1)
\pi^{2p+s}
(-1)^p \cos\left(\frac{-\pi s}{2}\right)
\\ \times \Gamma(-2p-s) \zeta(-2p-s) \zeta(1-s)$$

and we finally get
$$Q(s) = (-1)^p \pi^{2p+2s} (1-2^s) (1-2^{2p+s})
\Gamma(-2p-s) \zeta(-2p-s) \zeta(1-s)$$

thus proving the claim.

Return to the remainder integral and re-write it as follows:

$$(-1)^p \frac{(x/\pi)^{2p}}{2\pi i}
\int_{3/2-i\infty}^{3/2+i\infty}
(-1)^p \pi^{2p+2t}
\left(1-2^{2p+t}\right)
\left(1-2^t\right)
\\ \times \Gamma(-2p-t) \zeta(-2p-t) \zeta(1-t)
(x/\pi^2)^{t} dt.$$

so that the fact of it being a multiple of the defining integral of
$S(\pi^2/x; p)$ becomes readily apparent.

We have established the functional equation

$$\color{green}
{S(x;p) =
(2^{2p+1}-1)
\frac{(-1)^{p} B_{2p+2} \pi^{2p+2}}{2 (2p+2)!} \frac{1}{x}
\\ +
\frac{1}{2} \sum_{q=0}^p
(1-2^{2q+1})
(2^{2p-2q-1}-1)
\frac{(-1)^{p-q+1}}{(2q+1)!}
\frac{B_{2q+2} B_{2p-2q} \pi^{2p-2q}}{(2p-2q)! (q+1)}
x^{2q+1}
\\ + (-1)^p \left(\frac{x}{\pi}\right)^{2p} S(\pi^2/x;p).}$$

Now the value $x=\pi$ is obviously special here (fixed point) and we
get for $p=2r+1$ with $r\ge 0$ ($p$ even yields a Bernoulli number
identity)

$$S(\pi; 2r+1) =
\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{4r+3}}
\frac{e^{-n\pi}}{1-e^{-2n\pi}}
= - \frac{\pi^{4r+3}}{4}
(2^{4r+3}-1)
\frac{B_{4r+4}}{(4r+4)!}
\\ + \frac{\pi^{4r+3}}{4} \sum_{q=0}^{2r+1}
(1-2^{2q+1})
(2^{4r+1-2q}-1)
\frac{(-1)^{q}}{(2q+1)!}
\frac{B_{2q+2} B_{4r+2-2q}}{(4r+2-2q)! (q+1)}.$$

We obtain a rational multiple of $\pi^{4r+3}.$ Scale by two to get
for

$$\sum_{n\ge 1} \frac{(-1)^{n+1}}{n^3} \frac{1}{\sinh(\pi n)},
\quad \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^7} \frac{1}{\sinh(\pi n)},
\\ \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{11}} \frac{1}{\sinh(\pi n)},
\quad \sum_{n\ge 1} \frac{(-1)^{n+1}}{n^{15}} \frac{1}{\sinh(\pi n)},
\quad \ldots$$

the values

$$\bbox[5px,border:2px solid #00A000]{
{\frac {{\pi }^{3}}{360}},\quad
{\frac {13\,{\pi }^{7}}{453600}},\quad
{\frac {4009\,{\pi }^{11}}{13621608000}},\quad
{\frac {13739\,{\pi }^{15}}{4547140416000}},\quad \ldots}$$

These are dominated by the first term

$$\frac{1}{\sinh(\pi)}.$$