TITLE: Riemann-integrable (improperly) but not Lebesgue-integrable
QUESTION [10 upvotes]: Show that for any $\alpha>1$ the function 
    $$
g_{\alpha}(x):=\sin(x^{\alpha}), x\geq 0
$$
    is improperly Riemann-integrable but not Lebesgue-integrable on $\mathbb{R}_+$.


Could you please give me a hint how to show that?
Concerning the Riemann-integrability I have to show that
$$
\lim\limits_{R\to\infty}\int\limits_0^R\sin(x^{\alpha})\, dx<\infty,
$$
but I do not know how to show that.
Concerning the Lebesgue-integrability I have to show that
$$
\int\limits_0^{\infty}\lvert\sin(x^{\alpha})\rvert\, dx=\infty,
$$
but again I have no idea how to do so.

REPLY [2 votes]: I felt the counterexamples mentioned before are complicated and not straightforward to justify/remember. Instead, we can consider the following relatively simpler counterexample.
Consider the function $f(x)=\chi_{[n,n+1]}(x)\left(\frac{(-1)^n}{n}\right).$
$f$ is improper Riemann integrable and $\int\limits_{\mathbb{R}}f(x)dx=\sum\limits_{n \in \mathbb{N}}\frac{(-1)^n}{n} < \infty,$ but not Lebesgue integrable as $\int\limits_{\mathbb{R}} \left|f(x)\right|dx=\sum\limits_{n \in \mathbb{N}}\frac{1}{n}=\infty.$