First substitute $u= x^{\alpha}$ into the integral to see your problem reduces to asking the same question about the integral $\displaystyle \int^{\infty}_0 \dfrac{\sin x}{x^a} dx$ for $a\in (0,1).$
To show that is Riemann integrable, we consider the integral over $(0,1)$ and $[1,\infty)$ separately. To show the integral over $[1,\infty)$ exists, integrate by parts. You'll have
$$ \int^R_1 \frac{\sin x}{x^a} dx = -x^{-a} \cos x \mid^R_1 - a \int^R_1 \frac{\cos x}{x^{a+1}} dx$$
and it should be easy to finish from there.
To see $\displaystyle \int^1_0 \frac{\sin x}{x^a} dx$ exists note that $\sin x \sim x$ and $\displaystyle \int^1_0 \frac{x}{x^a} dx$ is finite.
To show it is not Lebesgue integrable, it suffices to show $$\int^{\infty}_0 \frac{|\sin x|}{x^a} dx = \sum_{k=0}^{\infty} (-1)^k \int^{(k+1)\pi}_{k\pi} \frac{\sin x}{x^a} dx = \sum_{k=0} \int^{\pi}_0 \frac{\sin x}{(x+k\pi)^a} dx$$ diverges, which is not hard with a basic estimate on the last integral. Note that the integral over $[0,\pi]$ is certainly greater than the integral over $[\pi/4,3\pi/4]$ and on that interval we have $\sin x \geq \frac{1}{\sqrt{2}}$ so
$$\int^{\pi}_0 \frac{\sin x}{(x+k\pi)^a} dx > \frac{1}{\sqrt{2}} \int^{3\pi/4}_{\pi/4} \frac{1}{( x+k\pi)^a} dx > \frac{1}{\sqrt{2}} \cdot \frac{\pi}{2} \frac{1}{( 3\pi/4 + k\pi)^a}= \frac{1}{2\sqrt{2}\pi^{a-1}}\frac{1}{(k+3/4)^a} .$$
The main difference between integrability in the sense of Lebesgue and Riemann is the way we measure 'the area under the curve'.
The Riemann integral asks the question what's the 'height' of $f$ above a given part of the domain of the function. The Lebesgue integral on the other hand asks, for a given part of the range of $f$, what's the measure of the $x$'s which contribute to this 'height'.
The following is taken from the wikipedia page for Lebesgue integration, and most instructive (Riemann in blue on the top, Lebesgue in red on the bottom):
(Taken from https://en.wikipedia.org/wiki/Lebesgue_integration#/media/File:RandLintegrals.png, CC BY-SA 3.0)
Or, another way to explain this:
I have to pay a certain sum, which I have collected in my pocket. I take the bills and coins out of my pocket and give them to the creditor in the order I find them until I have reached the total sum. This is the Riemann integral. But I can proceed differently. After I have taken all the money out of my pocket I order the bills and coins according to identical values and then I pay the several heaps one after the other to the creditor. This is my integral.
This due to Reinhard Siegmund-Schultze, (2008), "Henri Lebesgue", in Timothy Gowers, June Barrow-Green, Imre Leader, Princeton Companion to Mathematics, Princeton University Press.
As a result of the different definitions, different classes of functions are integrable:
By definition, $f$ is Lebesgue-integrable iff $|f|$ is Lebesgue-integrable.
And, indeed, as basket noted, we can integrate a lot of functions in the Lebesgue sense which can't be integrated in the Riemann sense (e.g. the so-called Dirichlet function which is $1$ on the rational numbers and $0$ on the irrational ones). We even have the nice result that if $f$ is bounded and defined on a compact set and Riemann integrable, then it is Lebesgue integrable.
On the other hand, if the domain isn't bounded, then $f$ might be integrable in the improper Riemann sense, but not integrable in the Lebesgue sense (e.g. $f(x)=\frac{\sin(x)}{x}$ on $[1,\infty)$ - for this function, even $|f|$ is not integrable in the improper Riemann sense).
Best Answer
The answer is yes to Riemann and Lebesgue integrals, though for different reasons: Your reasoning works for the Lebesgue integral since (with a measurability assumption as background) integrability is all about controlling the size of the function. On the other hand, for the Riemann integral you also have to worry about irregularities: For instance the function $$ f(x)= \begin{cases} 1 & x\in \mathbb{Q},\\ -1 & x\in \mathbb{R}\setminus\mathbb{Q}, \end{cases} $$ is not Riemann integrable, but $|f|$ is. So it's not all about the size for the Riemann integral.
Nonetheless you have nice properties of the Riemann integral such as: If $f,g$ are Riemann integrable, then so is $f\cdot g$. One way to see this is to write $2f\cdot g= (f+g)^2-f^2-g^2$, and then prove the result for $f=g$. Once we have this the result you want is immediate since $g(x)=e^{-inx}$ is continuous (in particular integrable).
So to answer your questions:
Yes, this is true. This follows from the definition of integral as the limit of upper and lower sums and the triangle inequality.
Yes, and as I mentioned above, your proof goes through in that case.
Again see my first paragraph.
No, see above for a counterexample.