I am trying to find the Laplace transform of the following equation;
\begin{equation} \frac{X(t1)}{d} = i \frac{t_1}{T_0} + \frac{u_1}{2}t_1^2 - \frac{2}{M}\int_0^{t_1}dt' \int_0^{\infty}\frac{dw}{\pi} \frac{J(w)}{w^2} sin(w(t_1-t'))-\frac{1}{M}\int_0^{\infty}\frac{dw}{\pi}\frac{J(w)}{w^3}(\exp(iwt_1)-iwt_1-1)a_w(t) \end{equation}
, which the text states should be equal to
\begin{equation} \frac{\tilde{X_s}}{d} = \frac{\frac{is}{T_0}+u_1+\frac{1}{M}\int_0^\infty \frac{dw}{\pi}\frac{J(w)}{w}\frac{s*a_w(t)}{s-iw}}{s^3(1+\frac{2}{M}\int_0^\infty \frac{dw}{\pi}\frac{J(w)}{w}\frac{1}{s^2+w^2})} \end{equation}
The text can be seen (https://arxiv.org/abs/cond-mat/9607189), the problem is from equation (37) to equation (40). I have completed all the transformation expect $\frac{2}{M}\int_0^{t_1}dt' \int_0^{\infty}\frac{dw}{\pi} \frac{J(w)}{w^2} sin(w(t_1-t'))$, which I from the solution could see should include a $\tilde{X}$ term, as the laplace tranformation of the sinus term is just $\frac{w}{s(s^2+w^2)}$, but I am clearly missing something. I think it is due to non-interchangeablilty of the integrals. Thank you in advanced