Integrate $\int_a^\infty\frac{\sqrt{x^2-a^2}}{\sinh x}\,dx$ I wonder whether the following integral ($a\in\mathbb{R}$, $a>0$)
$$S(a)=\int_a^\infty\frac{\sqrt{x^2-a^2}}{\sinh x}\,dx$$
admits a closed form (perhaps using some known special functions).
The integral representation of $K_1(z)$ gives just
$$S(a)=2a\sum\limits_{n=0}^{\infty}\frac{K_1\big((2n+1)a\big)}{2n+1}$$
which doesn't lead me to anything meaningful.
An alternative form comes from contour integration:
$$\frac{S(a\pi)}{\pi^2}=\frac{1-2a}{4}+\sum_{n=1}^{\infty}(-1)^{n-1}(\sqrt{n^2+a^2}-n).$$
For a context, this is what I arrive at in this answer.

Let's denote:
$$F(a)=\frac{1}{2}+\frac{d}{da} \frac{S(a\pi)}{\pi^2} \tag{1}$$
Remember for later that $S(0)= \frac{\pi^2}{4}$.
From the second expression in the OP we obtain:
$$F(a)=a \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\sqrt{n^2+a^2}}  \tag{2}$$
In this answer I found that:
$$\frac{F(a)}{a}=\log 2 +\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1)  \left( \frac{a}{2} \right)^{2k}  \tag{3}$$
and
$$\frac{F(a)}{a}=\log 2 +\int_0^\infty \frac{J_0 (a x)-J_0 (a x/2)}{e^x-1}dx  \tag{4}$$

Integrating (1) from $0$ to $a$, we have:
$$\int_0^a F(a') da'=\frac{a}{2}+\frac{S(a\pi)}{\pi^2}-\frac{1}{4}$$
On the other hand (3) gives us:
$$\int_0^a F(a') da'=\frac{\log 2}{2} a^2+2\sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2 (k+1)}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1)  \left( \frac{a}{2} \right)^{2k+2}$$

This gives us explicit Taylor series for $S(a)$:
$$S(a)= \frac{\pi^2}{4}-\frac{\pi a}{2}+\frac{\log 2}{2} a^2+ \frac{a^2}{2} \sum_{k=1}^\infty \frac{(-1)^k (2k)!}{k!^2 (k+1)}\left(1-\frac{1}{2^{2k}} \right) \zeta(2k+1)  \left( \frac{a}{2 \pi} \right)^{2k} \tag{5}$$

While this is not a closed form, it still may prove useful.

From (4) we obtain:
$$\int_0^a F(a') da'=\frac{\log 2}{2} a^2+a \int_0^\infty \frac{J_1 (a x)-2 J_1 (a x/2)}{x(e^x-1)}dx$$

Which gives us an additional integral form for the function:
$$S(a)= \frac{\pi^2}{4}-\frac{\pi a}{2}+\frac{\log 2}{2} a^2+ \pi a \int_0^\infty \frac{J_1 (a x)-2 J_1 (a x/2)}{x(e^{\pi x}-1)}dx \tag{6}$$

