Proving that $\int_0^\infty\sin(x)dx=1$

In your opinion, is it true that $$\lim_{x\to +\infty}\left(\lim_{t\to 0^+} e^{-tx}\sin(x)\right)= \lim_{t\to 0^+}\left( \lim_{x\to +\infty}e^{-tx}\sin(x)\right)\quad ?$$ Switching the order of limits could be dangerous...

See your last line. Are you sure that $$\lim_{t\to 0^+}\left(\lim_{r\to +\infty}\int_0^r e^{-tx}\sin(x)dx\right)=\lim_{r\to +\infty}\left(\lim_{t\to 0^+}\int_0^r e^{-tx}\sin(x)dx\right)\quad ?$$

In your last line what you actually make is

$$ \lim_{t\to 0}\int_{0}^{\infty} e^{-tx}\sin(x) \, dx = \int_{0}^{\infty} \lim_{t\rightarrow 0}e^{-tx}\sin(x) \, dx$$

This last step is only allowed if the convergence is uniform. In a sloppy language this means that the "size" of $sin(x)-e^{-tx} sin(x)$ is "independent" of x. Which is not the case, that's why your result is wrong.