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Find all $\alpha \in \mathbb{R}$ such that : $$I_\alpha = \int_0^{\infty} \frac{\sin t}{t^{\alpha}} \mathrm{d}t$$ converges

My book says the following : We know that $\int_1^\infty \frac{\sin t}{t^{\alpha}} \mathrm{d}t$ converges for all $\alpha > 0$. Moreover near $0^+$ we have : $\frac{\sin t}{t^{\alpha}} \sim 1/t^{\alpha-1}$ thus : $\int_0^1 \frac{\sin t}{t^{\alpha}} \mathrm{d}t$ converges for all $\alpha < 2$. So the answer is $ 0 < \alpha < 2$.

I don't understand this argument and I feel like something is missing. Since we are dealing with non-absolutely convergente integrals we can't seperate cases like this right ? To be more clear the convergence of $\int_a^b f(t) \mathrm{d}t$ with $a, b \in \mathbb{R} \cup \{\pm \infty \}$ doesn't mean that for all $c \in (a,b)$ we have the convergence of : $\int_a^c f(t) \mathrm{d}t$ and the convergence of : $\int_c^b f(t) \mathrm{d}t$ right ? I mean this is true only for absolutely integrable functions ?

So with this solution it only proves that for all $0 < \alpha < 2$ $I_\alpha$ converges but it doesn't prove that these are the only $\alpha$ for which $I_\alpha$ converges.

Thus If I am not mistaken there is a missing argument in the above right ?

I hope my problem is clear, thank you !

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Convergence of the improper integral $\int_0^{\infty} \frac {\sin\, x} {x^{\alpha}} \, dx$ means $\int_a^{M} \frac {\sin\, x} {x^{\alpha}} \, dx$ tends to a limit $L$ as $a \to 0+$ and $M \to \infty$ (independently). Note that $\int_a^{b} \frac {\sin\, x} {x^{\alpha}} \, dx$ exists whenever $0<a<b<\infty$. Hence convergence of $\int_0^{\infty} \frac {\sin\, x} {x^{\alpha}} \, dx$ is equivalent to convergence of $\int_0^{c} \frac {\sin\, x} {x^{\alpha}} \, dx$ and $\int_c^{\infty} \frac {\sin\, x} {x^{\alpha}} \, dx$ for any real number $c$.

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