Ratio of 2 integrals with a limit

Note first that $\int_0^\infty e^{-ax}\sin\,bx\,dx \; = \; \frac{b}{a^2+b^2} \hspace{2cm}a,b > 0$ and so $\begin{aligned} F(m) & = \; \int_0^\infty e^{-\pi mx} \sin(\pi x) \ln\big(1 + e^{-\frac{\pi x}{m}}\big)\,dx \; = \; \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\int_0^\infty e^{-\big(m\pi+\frac{k\pi}{m}\big)x}\sin(\pi x)\,dx \\ & = \; \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \frac{\pi}{\pi^2\big(m + \frac{k}{m}\big)^2 + \pi^2} \; = \; \frac{m^2}{\pi}\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k\big[(k+m^2)^2+m^2\big]} \; = \; \frac{m^2}{\pi}G(m) \end{aligned}$ where $G(m) \; = \; \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k\big[(k+m^2)^2+m^2\big]} \hspace{2cm} m > 0$ Since $\sum_{k=1}^\infty k^{-3}$ is convergent, it is easy to deduce that $\lim_{m\to\infty}G\big(m^{-1}\big) \; = \; \lim_{m \to 0}G(m) \; = \; \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3} \; = \; \sum_{k=1}^\infty \frac{1}{k^3} - 2\sum_{k=1}^\infty \frac{1}{(2k)^3} \; = \; \tfrac34\zeta(3)$ Now $m^4G(m) \; = \; m^4\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k\big[(k+m^2)^2 + m^2\big]} \; = \; \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k\big[(1 + km^{-2})^2 + m^{-2}\big]}$ and a simplistic argument indicates that $\lim_{m \to \infty}m^4G(m) \; = \; \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \; = \; \ln2$ Given that the series for $m^4G(m)$ is not uniformly convergent in $m$ , this deduction is somewhat dodgy. However, a more detailed analysis shows that $m^4G(m) \; = \; \frac{m^2}{2i(m^2+1)}\left[ \begin{array}{l} (m+i)\big[\psi(1 + m^2-im) - \psi(1 + \tfrac12m^2 - \tfrac12im)\big] \\ - (m-i)\big[\psi(1+m^2+im) - \psi(1 + \tfrac12m^2 + \tfrac12im)\big]\end{array}\right]$ where $\psi$ is the digamma function, and the fact that $\psi(z) = \ln z + \mathrm{O}(z^{-1})$ as $|z| \to \infty$ uniformly for $|\mathrm{Arg}(z)| \le \pi-\delta$ for any $\delta > 0$ tells us indeed that $\lim_{m \to \infty}m^4G(m) \; = \; \;\ln2$ Thus $\lim_{m \to \infty} \frac{F(m)}{F(m^{-1})} \; = \; \lim_{m \to \infty} \frac{m^4G(m)}{G(m^{-1})} \; = \; \frac{\ln2}{\frac34\zeta(3)} \; = \; \frac{4\ln2}{3\zeta(3)}$ which makes $T = \boxed{2}$ .