For $k \in \mathbf{N}^*$ and $t \in \mathbf{R}_+$, we set
$$u_k(t) = \int_{k/2}^{(k+1)/2} \frac{tq(u)}{e^{tu}-1} \mathrm{du} \text{ if } t > 0 \text{, and } u_k(t) = \int_{k/2}^{(k+1)/2} \frac{q(u)}{u} \mathrm{du} \text{ if } t = 0.$$
We admit that the bound $\left|\sum_{k=n}^{+\infty} u_k(t)\right| \leq \frac{1}{2n}$ holds for $t = 0$.
Deduce that
$$\int_{1}^{+\infty} \frac{tq(u)}{e^{tu}-1} \mathrm{~d}u \underset{t \rightarrow 0^+}{\longrightarrow} \frac{\ln(2\pi)}{2} - 1$$