Let $n$ be a positive integer, and $s\in \mathbb{C}\;,\Re(s)>0$. I want to compute the integral : $$\int_{0}^{\infty}\sin\left(2\pi ne^{x}\right)\left[\frac{s}{e^{sx}-1}-\frac{1}{x}\right]dx$$
I tried using the integral presentation found here :
$$\sin(2\pi ne^{x})=\frac{\sqrt{\pi}}{2\pi i }\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\Gamma(z)}{\Gamma\left(\frac{3}{2}-z \right )}\left(\pi n \right )^{-2z+1}e^{-(2z-1)x}dz\;\;\;\;\;0<\gamma<1$$ in conjunction with : $$r+\log(r)+\psi\left(\frac{1}{r}\right)=-\int_{0}^{\infty}e^{-y}\left(\frac{r}{e^{ry}-1}-\frac{1}{y}\right)dy$$ Where $\psi\left(\cdot\right)$ is the digamma function. But that made the problem even more difficult. Any help is highly appreciated.
EDIT :
Using the Fourier reciprocity : $$\frac{s}{e^{sx}-1}-\frac{1}{x}=2\int_{0}^{\infty}\left(\frac{1}{e^{2\pi \theta/s}-1}-\frac{s}{2\pi \theta} \right )\sin(x\theta)d\theta$$ Our integral reads : $$2\int_{0}^{\infty} g(\theta)\left(\frac{1}{e^{2\pi \theta/s}-1}-\frac{s}{2\pi \theta} \right )d\theta$$ Where : $$g(\theta)=\int_{0}^{\infty}\sin(2\pi ne^{x})\sin(x\theta)dx$$ $$=\frac{\sqrt{\pi}}{2\pi i }\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\Gamma(z)}{\Gamma\left(\frac{3}{2}-z \right )}\left(\pi n \right )^{-2z+1}\frac{\theta}{\theta^{2}+(2z-1)^{2}}dz$$