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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.

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up vote 2 down vote accepted

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 $t\lt2$ by comparison with $\dfrac{t}{t^\gamma}=\dfrac1{t^{\gamma-1}}$.

This shows that the interval of convergence is $(0,2)$.

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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) $.

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