Limit with Integration -4

Let $f \colon (0,\infty) \to \mathbb{R},$ Consider the function as $\begin{aligned} f(\alpha) &= \frac{1}{2\alpha} - \int \limits_1^\infty \frac{\mathrm{d} x}{\sinh(\pi \alpha x) \sqrt{x^2 -1}} \stackrel{x = \cosh(t)}{=} \frac{1}{2 \alpha} - \int \limits_0^\infty \frac{\mathrm{d} t}{\sinh(\pi \alpha \cosh(t))} \\ &= \frac{1}{2 \alpha} - \int \limits_0^\infty \left[\frac{1}{\pi \alpha \cosh(t)} + 2 \pi \alpha \cosh(t) \sum \limits_{k=1}^\infty \frac{(-1)^k}{\pi^2 k^2 + \pi^2 \alpha^2 \cosh^2(t)}\right] \mathrm{d} t \\ &\!\!\!\!\!\!\!\!\stackrel{u = \alpha \sinh(t)}{=} \frac{2}{\pi} \int \limits_0^\infty \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^2 + \alpha^2 + u^2} \, \mathrm{d} u \, . \end{aligned}$ 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 $\mathbb{R}$ , so it defines a continuous function of $\alpha$ on $\mathbb{R}$ . We can write, $\lim_{\alpha \to 0^+} f(\alpha) = \lim_{\alpha \to 0^+} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{\sqrt{k^2 + \alpha^2}} = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{\sqrt{k^2 + 0^2}} = \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k} = \log(2)$ Thus, our problem can be solved by observing that $\begin{aligned} \frac{2}{3 \alpha^3} - \frac{4\pi}{3 \alpha} \int \limits_1^\infty \frac{x \cosh(\pi \alpha x)}{\sinh^2(\pi \alpha x) \sqrt{x^2-1}} \, \mathrm{d} x &= - \frac{4}{3 \alpha} f'(\alpha) = \frac{4}{3} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{(k^2 + \alpha^2)^{3/2}} \\ &\!\!\!\stackrel{\alpha \rightarrow 0^+}{\longrightarrow} \frac{4}{3} \sum \limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^3} = \color{#0C6AC7}{\boxed{\frac{4}{3} \operatorname{\eta}(3) }}= \operatorname{\zeta}(3) \, . \end{aligned}$