Challenging limit: $\lim_{\alpha\to0^{+}}\left(\frac{1}{2\alpha}-\int_1^\infty\frac{dx}{\sinh(\pi\alpha x)\sqrt{x^2-1}}\right)$ Here is a challenging limit proposed  by a friend:

$$\lim_{\alpha\to0^{+}}\left(\frac{1}{2\alpha}-\int_1^\infty\frac{dx}{\sinh(\pi\alpha x)\sqrt{x^2-1}}\right)$$

and he claims that the closed form for this limit is really pleasant.
I am not good at limits so I am not going to show any work and just leave it to those who find it interesting.

Addendum: A similar problem proposed by the same person:

$$\lim_{\alpha\to0^{+}}\left(\frac{2}{3\alpha^3}-\frac{4\pi}{3\alpha}\int_1^\infty\frac{x\cosh(\pi\alpha x)}{\sinh^2(\pi\alpha x)\sqrt{x^2-1}}dx\right)$$

 A: Let $f \colon (0,\infty) \to \mathbb{R},$
\begin{align}
f(\alpha) &= \frac{1}{2\alpha} - \int \limits_1^\infty \frac{\mathrm{d} x}{\sinh(\pi \alpha x) \sqrt{x^2 -1}} \stackrel{x = \cosh(t)}{=} \frac{1}{2 \alpha} - \int \limits_0^\infty \frac{\mathrm{d} t}{\sinh(\pi \alpha \cosh(t))} \\
&= \frac{1}{2 \alpha} - \int \limits_0^\infty \left[\frac{1}{\pi \alpha \cosh(t)} + 2 \pi \alpha \cosh(t) \sum \limits_{k=1}^\infty \frac{(-1)^k}{\pi^2 k^2 + \pi^2 \alpha^2 \cosh^2(t)}\right] \mathrm{d} t \\
&\!\!\!\!\!\!\!\!\stackrel{u = \alpha \sinh(t)}{=} \frac{2}{\pi} \int \limits_0^\infty \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^2 + \alpha^2 + u^2} \, \mathrm{d} u \, .
\end{align}
Here we have used the pole expansion of $\operatorname{csch}$ and the elementary integral $\int_0^\infty \operatorname{sech}(t) \, \mathrm{d} t = \frac{\pi}{2}$. Since the partial sums of the remaining alternating series (with terms decreasing in absolute value) are bounded by the first term, i. e. the integrable function $u \mapsto \frac{1}{1 + \alpha^2 + u^2}$ , the dominated convergence theorem allows us to interchange summation and integration. We obtain
$$ f(\alpha) = \frac{2}{\pi} \sum \limits_{k=1}^\infty (-1)^{k-1}\int \limits_0^\infty \frac{\mathrm{d} u}{k^2 + \alpha^2 + u^2} = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{\sqrt{k^2 + \alpha^2}} $$
for $\alpha > 0$. The series on the right-hand side converges uniformly on $\mathbb{R}$ (see this question), so it defines a continuous function of $\alpha$ on $\mathbb{R}$. In particular, as predicted by Claude Leibovici in the comments,
$$ \lim_{\alpha \to 0^+} f(\alpha) = \lim_{\alpha \to 0^+} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{\sqrt{k^2 + \alpha^2}} = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{\sqrt{k^2 + 0^2}} = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k} = \log(2)$$
holds. The additional problem can be solved by noting that
\begin{align} \frac{2}{3 \alpha^3} - \frac{4\pi}{3 \alpha} \int \limits_1^\infty \frac{x \cosh(\pi \alpha x)}{\sinh^2(\pi \alpha x) \sqrt{x^2-1}} \, \mathrm{d} x &= - \frac{4}{3 \alpha} f'(\alpha) = \frac{4}{3} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{(k^2 + \alpha^2)^{3/2}} \\
&\!\!\!\stackrel{\alpha \rightarrow 0^+}{\longrightarrow} \frac{4}{3} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^3} = \frac{4}{3} \operatorname{\eta}(3) = \operatorname{\zeta}(3) \, .
\end{align}
