For $\lambda > -1$, use Monotone Convergence Theorem on $t^\lambda \chi_{[\frac{1}{n},1]}$, where $\chi$ is the characteristic function.
For $\lambda \leq -1 $, assume that the integral is finite. Then it is finite for each $t^\lambda \chi_{[\frac{1}{n},1]}$. But this integral gets arbitrarily large as $n$ goes to infinity which is a contradiction.
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
$$\int^{\pi}_0 \frac{\sin x}{(x+k\pi)^a} dx > \frac{1}{\sqrt{2}} \int^{3\pi/4}_{\pi/4} \frac{1}{( x+k\pi)^a} dx > \frac{1}{\sqrt{2}} \cdot \frac{\pi}{2} \frac{1}{( 3\pi/4 + k\pi)^a}= \frac{1}{2\sqrt{2}\pi^{a-1}}\frac{1}{(k+3/4)^a} .$$
Best Answer
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:
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).