Does $\int_{0}^{\infty} \sin^x(x) dx$ converge? From what I have found the indefinite integral does not have a closed form solution. Also, the function takes complex values except for in the intervals $[0,\pi],[2\pi,3\pi],$ (and for whole x between those intervals). But if we only considered those intervals where the function takes on real values, i.e.
$$\int_{0}^{\pi}\sin^x(x)dx+\int_{2\pi}^{3\pi}\sin^x(x)dx+\int_{4\pi}^{5\pi}\sin^x(x)dx+{...}$$
Does this infinite sum converge? For the record I have no idea how to go about proving this, but I am curious if anyone does.

We have that
$$\int_{2k\pi}^{(2k+1)\pi}\sin^x(x)dx\ge \int_{2k\pi}^{(2k+1)\pi}\sin^{(2k+1)\pi}(x)dx\ge  2 \int_{0}^{\frac \pi 2 }\left(\frac 2 \pi x\right)^{(2k+1)\pi}dx=\frac{\pi} {2\pi k+\pi+1}$$

therefore
$$\int_{0}^{\pi}\sin^x(x)dx+\int_{2\pi}^{3\pi}\sin^x(x)dx+\ldots=\sum_{k=0}^\infty \int_{2k \pi}^{(2k+1)\pi} \sin^x(x)dx\ge \sum_{k=0}^\infty \frac{\pi}{2\pi k+\pi+1}$$
which diverges.
