Convergence of $\int_0^\infty \sin(t)/t^\gamma \mathrm{d}t$

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 $(0,2)$.

Informally, $ \sin x \approx x $ near $ 0 $. Hence, $\displaystyle \int_0^1 \frac{\sin x}{x^\gamma} \ dx $ converges if $\displaystyle \int_0^1 \frac{1}{x^{\gamma - 1}} \ dx $ which occurs for $ \gamma < 2 $.

Likewise, for large values of $ x $, $ \sin x $ oscillates between negative and positive. As long as the numerator is increasing, the sum of these areas is an alternating series whose terms are decreasing monotonically to zero in absolute value and so it will converge if $ \gamma > 0 $ by the Alternating Series Test.

Hence, the interval of convergence is $ (0, 2) $.