The convergence of $\int_0^{\infty} \sin(x)(x-[x])/x^{\alpha}$ I need to evaluate the following integral
$$\int_0^{\infty} \sin(x)(x-[x])/x^{\alpha}dx$$
where $\alpha\in (0,1)$ and $[x]$ is the floor function.
Without $x-[x]$, we can evaluate the integral easily using analytical continuation.
$$\int_0^{\infty} \sin(x)/x^{\alpha}dx=\Im \int_0^{\infty} e^{ix}/x^{\alpha}dx=\Im [\Gamma(1+\alpha)/i^{\alpha-1}]$$
In the presence of $x-[x]$, my idea is that one could use inequality to upper bound the integral if it converges or lower bound the integral if it diverges. But I'm not sure what kind of inequality I should use.
I tried Hölder's inequality, but it will square the sin function in the upper bound, which makes it diverge.
 A: Let's settle (as in the title of the OP) just the convergence of the integral. The obvious choice is Dirichlet's test: we only have to show that $\int^N_0\sin(x)(x-[x])\,dx$ is uniformly bounded. Since $\sin(x)(x-[x])$ is bounded, it's sufficient to show that for integer $N$. But then,
$$\int^N_0\sin(x)(x-[x])\,dx=\sum^{N-1}_{n=0}\int^{n+1}_n\sin(x)(x-[x])\,dx=\int^1_0y\,\sum^{N-1}_{n=0}\sin(n+y)\,dy,$$ and
$$\sum^{N-1}_{n=0}\sin(n+y)=\cos y\,\sum^{N-1}_{n=0}\sin n+\sin y\,\sum^{N-1}_{n=0}\cos n.$$
Since $$\left|\sum^{N-1}_{n=0}\sin n\right|=\left|\frac{\cos(1/2)-\cos(N-1/2)}{2\,\sin(1/2)}\right|\le\frac1{\sin(1/2)}$$ and $$\left|\sum^{N-1}_{n=0}\cos n\right|=\left|\frac{\sin(N-1/2)+\sin(1/2)}{2\,\sin(1/2)}\right|\le\frac1{\sin(1/2)},$$ this is uniformly bounded. $x^{-\alpha}\to0$ as $x\to\infty$ guarantees the convergence of the integral.
BONUS: If $\alpha>0$, $e^{-bx}\,x^{-\alpha}\to0$ as $x\to\infty$ uniformly in $b\ge0$, so $\displaystyle\int_0^{\infty} e^{-bx}\,\sin(x)(x-[x])\,x^{-\alpha}\,dx$ converges uniformly in $b\ge0$.
A: To answer to @ProfessorVector
For $a\in (1,2)$, $$\int_0^{\infty} \sin(x)(x-[x])x^{-a}dx= \lim_{b\to 0} \int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$$ $\int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$ is analytic for $a < 2$.
For $a \in (0,1/2)$

*

*$e^{-bx/2}(\frac{\sin(x)}2-\sum_{n\ge 1} \frac{\cos((2\pi n-1)x)-\cos((2\pi n+1)x)}{2\pi n})$  converges in $L^2(0,\infty)$


*and $x^{-a}e^{-bx/2}$ is $L^2(0,\infty)$
Thus no problem to switch $\int,\sum$
$$\int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$$ $$= \int_0^{\infty} e^{-bx}(\frac{\sin(x)}2-\sum_{n\ge 1} \frac{\cos((2\pi n-1)x)-\cos((2\pi n+1)x)}{2\pi n})x^{-a}dx$$
$$ = \Gamma(1-a)\Re(\frac{(b+i)^{a-1}}{2i}-\sum_{n\ge 1} \frac{(b+i(2\pi n-1))^{a-1}-( b+i(2\pi n+1))^{a-1}}{2\pi n})$$
The latter series extends analytically to $\Re(a)\in (0,2)$  so it must stay equal to $\int_0^{\infty} e^{-bx}\sin(x)(x-[x])x^{-a}dx$.
Thus for $a\in (1,2)$ we can let $b\to 0$ to obtain
$$\int_0^{\infty} \sin(x)(x-[x])x^{-a}dx = \Gamma(1-a)\Re(\frac{i^{a-1}}{2i}-\sum_{n\ge 1} \frac{(i(2\pi n-1))^{a-1}-(i(2\pi n+1))^{a-1}}{2\pi n})$$
