The limit is $\pi$ irrespective of what those sequences are. It is a standard fact that $\lim _{t\to \infty}\int_0^{t} \frac {\sin \, x} x \, dx=\pi/2$ from which my claim follows.
Neither integral is uniformly convergent for values of the parameter in the open interval $(0,\infty)$.
For the first integral, with $y_n = (2n\pi + \pi)^{-1} \in (0,\infty)$ we have
$$\left|\int_{2n\pi}^{2n\pi+\pi} e^{-xy_n} \sin x \, dx\right|\geqslant e^{-(2n\pi+\pi) y_n}\int_{2n\pi}^{2n\pi+\pi} \sin x \, dx = 2 e^{-(2n\pi+\pi)y_n}= 2e^{-1}$$
Since the RHS does not converge to $0$ as $n \to \infty$, the Cauchy criterion for uniform convergence is violated.
For the second integral, with $x_n = 1/n \in (0,\infty)$ we have
$$\left|\int_n^\infty e^{-x_ny} \sin x_n \, dy\right| = \left|\frac{\sin x_n}{x_n} \right|e^{-nx_n} = \frac{\sin \frac{1}{n}}{\frac{1}{n}}e^{-1} \,\,\, \xrightarrow[n \to \infty]{} \,\,e^{-1},$$
and, again, violation of the Cauchy criterion precludes uniform convergence.
Best Answer
You could mimic the same general idea by considering $$\int_0^\infty \frac{(-1)^{\lfloor x\rfloor}}{\lceil x\rceil}\,dx\ .$$