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.
 A: You can rewrite your sum as $$ \sum_{k=0}^\infty \int_{2k \pi}^{(2k+1)\pi} \sin^x(x)dx $$
Make a substitution $x = y+2k\pi$ in integral $\int_{2k\pi}^{(2k+1)\pi}\sin^x(x)dx$ getting $$ \sum_{k=0}^\infty \int_0^\pi \sin^{2k\pi + y}(y)dy $$
Due to non-negativeness and Monotone Convergence Theorem (or Fubinii) you can interchange sum with integral, getting $$ \int_0^\pi \sin^y(y)\cdot \sum_{k=0}^\infty \sin^{2k\pi}(y) dy = \int_0^\frac{\pi}{2} \sin^y(y) \cdot \frac{1}{1-\sin^\pi(y)}dy + \int_{\frac{\pi}{2}}^\pi \sin^y(y)\frac{1}{1-\sin^{2\pi}(y)}dy $$
Note that I did splited the integral into two parts because at point $y=\frac{\pi}{2}$ our series diverges. However, one point doesn't change the integral at all, so if you're okay with $\int_0^\pi \sin^y(y) \frac{1}{1-\sin^{2\pi}(y)}dy$, then let it be so. All we have to do now is to check whether the above converges or diverges. It is easily seen that the only point where we have $"$problem$"$ is point $y=\frac{\pi}{2}$, because the denominator is $0$ then. However, since $\sin^y(y) \to 1$ as $y \to \frac{\pi}{2}$ and since our integrand is non-negative, equivalently (via limit comparison criterion) we must check the convergence of $$ \int_0^\pi \frac{1}{1-\sin^{2\pi}(y)}dy = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{1-\cos^{2\pi}(x)}dx$$ where we used substitution $y = \frac{\pi}{2} - x$ and equality $\sin(\frac{\pi}{2}-x) = \cos(x)$.
What's left is to use asymptotic of $\cos(x)$ near zero, that is $\cos(x) \sim 1 - \frac{x^2}{2} + o(x^2)$ as $x \to 0$ and $(1+y)^a \sim 1 + ay + o(y)$ as $y \to 0$ (for $a>0$) getting $1-\cos^{2\pi}(x) \sim \pi x^2 + o(x^2)$ as $x \to 0$. Hence, again limit comparison criterion (and non-negativeness of integrand) gives us that convergence of $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{1-\cos^{2\pi}(x)}dx $$ is equivalent with convergence of $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\pi x^2}dx $$ which is known to diverge
EDIT: The part with $"$asymptotic$"$ behaviour of $\cos(y)$ and $(1+y)^a$ was only used to ensure $\lim_{x \to 0} \frac{\pi x^2}{(1-\cos^{2\pi}(x))} = 1$ (to use limit-comparison criterion of convergence). So if you're not so familiar with Taylor expansion, then you can try to compute this limit (via let's say De'Hopital $1$ or $2$ times, depending on your knowledge of $\lim_{x \to 0} \frac{\sin(x)}{x}$) and there will be no need of using asymptotic (however, the asymptotic was really helpful to come up with a function which is $"$comparable$"$ with $\frac{1}{1-\cos^{2\pi}(x)}$ at $x \sim 0$)
A: 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.
A: If your integral converges also
$$\sum_{n=0}^\infty \int_0^1 x^n dx = \sum_{n=1}^\infty \frac{1}{n}= \infty$$
would converge. This is because for $x \in [0,\pi/2]$ we have
$$\frac{2x}{\pi} \le \sin(x).$$
