For what values of $\gamma\geq 0$ does the improper integral $$\int_0^\infty \frac{\sin(t)}{t^\gamma} \mathrm{d}t$$ converge?
In order to avoid two "critical points" $0$ and $+\infty$ I've thought that it would be easier to test the convergence of the sum (is this coherent?):
$$
\int_0^1 \frac{\sin(t)}{t^\gamma}\mathrm{d}t +
\int_1^\infty \frac{\sin(t)}{t^\gamma}\mathrm{d}t.
$$
For the second integral, it converges if $\gamma > 1$ (comparision) and also converges if $0 <\gamma \leq 1$. I'm stuck on proving the last part and the fact that the first integral converges for $\gamma < 2$. Any help would be appreciated. Thanks in advance.
PD: I've checked the answers for this question but I would not like to solve this integral using $(n\pi,(n+1)\pi)$ intervals.
Best Answer
I was not going to answer, but the previous answers left me a bit anxious for $t$ near $\infty$.
Integrate by parts to get $$ \int_1^\infty\frac{\sin(t)}{t^\gamma}\,\mathrm{d}t =\left.\frac{-\cos(t)}{t^\gamma}\right]_1^\infty -\gamma\int_1^\infty\frac{\cos(t)}{t^{\gamma+1}}\,\mathrm{d}t $$ and both converge at $\infty$ when $\gamma\gt0$.
Of course, as the previous answers have said $$ \int_0^1\frac{\sin(t)}{t^\gamma}\,\mathrm{d}t $$ converges when $\gamma\lt2$ by comparison with $\dfrac{t}{t^\gamma}=\dfrac1{t^{\gamma-1}}$.
This shows that the interval of convergence is $(0,2)$.