Twist on a well-known integral It is well known that $\int_{\mathbb{R}^+} \sin (x^n)\,dx$ converges and has a closed form when $n>1$. Does $$\int_{\mathbb{R}^+}|\sin (x^n)|\,dx$$ converge for $n>1?$ My guess is that it does not, but I am completely incapable of proving this: the function is too erratic for my toolbox. Any ideas?

Let's consider $n > 0$ first. You can substitute $t = x^n$ here too. Let's write $\alpha = \frac1n$, then

$$\int_0^R \lvert \sin (x^n)\rvert\,dx = \frac1n \int_0^{R^n} \frac{\lvert \sin t\rvert}{t^{1-\alpha}}\,dt.$$

Now, for an interval $[k\pi,(k+1)\pi]$, you have

$$\int_{k\pi}^{(k+1)\pi} \frac{\lvert \sin t\rvert}{t^{1-\alpha}}\,dt > \frac{1}{(k+1)^{1-\alpha}\pi^{1-\alpha}}\int_{k\pi}^{(k+1)\pi} \lvert \sin t\rvert\,dt = \frac{2}{(k+1)^{1-\alpha}\pi^{1-\alpha}},$$

and since

$$\sum_{m=1}^\infty \frac{1}{m^{1-\alpha}} = \infty$$

for all $\alpha > 0$, the integral diverges (is infinite) for all $n > 0$.

For $n < 0$, you have $\lvert \sin (x^n)\rvert \sim x^n$ for large $x$, and the integral converges if and only if $n < -1$.