How to solve $\int_0^{2\pi} $ $(\int_1^\infty$ $\sin(s-t) \over t^3$ $dt$ $)ds$? I already tried to solve it myself and to type it into the calculator, but it gives out some weird stuff with an endless long solution.
Since $\left|\sin\right|\leq 1$, Fubini's theorem applies, hence:
$$ I=\int_{0}^{2\pi}\int_{1}^{+\infty}\frac{\sin(s-t)}{t^3}\,dt\,ds = \int_{1}^{+\infty}\frac{1}{t^3}\int_{0}^{2\pi}\sin(s-t)\,ds\,dt = 0. $$
But you don't even need Fubini's theorem to state the same:
$$ \int_{1}^{+\infty}\frac{\sin(s-t)}{t^3}\,dt = \sin(s)\int_{1}^{+\infty}\frac{\cos(t)}{t^3}\,dt - \cos(s)\int_{1}^{+\infty}\frac{\sin(t)}{t^3}\,dt $$
where both $\int_{1}^{+\infty}\frac{\cos(t)}{t^3}\,dt$ and $\int_{1}^{+\infty}\frac{\sin(t)}{t^3}\,dt$ are converging integrals, bounded by $\int_{1}^{+\infty}\frac{dt}{t^3}=\frac{1}{2}$ in absolute value. It follows that:
$$ I = \int_{0}^{2\pi}\left(A \sin(s)+B\cos(s)\right)\,ds = 0.$$
Zero is the correct result.

Switching the integrals you have

$$\int_1^{+\infty}\frac{\text{d}t}{t^3}\int_0^{+2\pi} \sin(s-t)\ \text{d}s = \int_1^{+\infty}\frac{\text{d}t}{t^3}\left(-\cos(s-t)\bigg|_{0}^{+2\pi}\right)$$

But

$$\left(-\cos(s-t)\bigg|_{0}^{+2\pi}\right) = 0$$

Hence

$$\int_1^{+\infty} \frac{\text{d}t}{t^3}\cdot 0 = 0$$