# Evaluating $\int_0^{\infty} \frac{\sin t}{t^{\alpha}} \mathrm{d}t$

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 !

## 1 Answer

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 $$0. Hence convergence of $$\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$$.