Does that improper integral converge or diverge? Does the improper integral $$\int_0^\infty |\sin(x)|^x\,dx$$ converge or does the one diverge? Calculations suggest its convergence.
 A: It actually looks like it should diverge, since $$\int_0^{2\pi} |\sin x|^n d x \sim \frac1{\sqrt{n}}.$$ Writing your integral as a sum of integrals from $2\pi n$ to $2\pi(n+1),$ you seem to get a divergent series.
A: I know this is late, but I think it's a cool different approach.
First of all, note that $|\sin x|$ has period $\pi$. Consider some interval $[(n-1)\pi,n\pi]$ with $n>0$ an integer. Then for  $x\in [(n-1)\pi,n\pi]$, $|\sin x|^{x}\geq|\sin x|^{n\pi}$, because $|\sin x|$ is between $0$ and $1$.
Also note that on $[0,\frac{\pi}{2}]$, $|\sin x|\geq \frac{2x}{\pi}$. Thus for any natural number $m$, $|\sin x|^{m\pi}\geq (\frac{2x}{\pi})^{m\pi}$. So $\int_0^{\frac{\pi}{2}} |\sin x|^{m\pi}dx\geq\int_0^{\frac{\pi}{2}}(\frac{2x}{\pi})^{m\pi}dx=\frac{\pi}{2m\pi+2}$.
Finally, about the line $x=\frac{\pi}{2}$, $|\sin x|^{m\pi}$ is symmetric, so $\int_0^{\pi} |\sin x|^{n\pi}=2\int_0^{\frac{\pi}{2}} |\sin x|^{n\pi}$.
Putting everything together, $\int_0^\infty |\sin x|^x dx=\sum_{n=1}^\infty\int_0^{\pi} |\sin x|^{x+(n-1)\pi}dx\geq\sum_{n=1}^\infty\int_0^{\pi} |\sin x|^{n\pi}dx=\sum_{n=1}^\infty2\int_0^{\frac{\pi}{2}} |\sin x|^{n\pi}dx\geq\sum_{n=1}^\infty2\int_0^{\frac{\pi}{2}} (\frac{2x}{\pi})^{n\pi}dx=\sum_{n=1}^\infty2\frac{\pi}{2n\pi+2}\rightarrow\infty$
A: It also looks to me as if it diverges:
Consider $\displaystyle \int_{x=2^{n-1}\pi}^{2^n\pi} |\sin x|^x dx \ge  \int_{x=2^{n-1}\pi}^{2^n\pi} (\sin x)^{2^n} dx .$
It is possible to do the integrations on the right-hand side into closed forms.  For small $n$ it gives the following values:
n    rhs
1    pi / 2
2    3 pi / 4
3    35 pi / 32 
4    6435 pi / 4096 
5    300540195 pi / 134217728
6    916312070471295267 pi / 288230376151711744

and so the lower bound is increasing with $n$, and likely to go on doing so, so the sum of the lower bounds will increase to infinity, and so will the the original expression. 
