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For which values of $\alpha > 0$ is the following improper integral convergent?

$$\int_{1}^{\infty} \frac{\sin x}{x^{\alpha}}dx$$

I tried to solve this problem by parts method but I am nowhere near to the answer. :(

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  • $\begingroup$ Think about alternating series. $\endgroup$ – David Mitra Jan 3 '15 at 11:03
  • $\begingroup$ @DavidMitra still couldn't figure it out sir!! $\endgroup$ – aavatar Jan 3 '15 at 11:07
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    $\begingroup$ Split the integral into pieces of the form $A_n=\int_{n\pi}^{(n+1)\pi}{\sin x\over x^\alpha}\,dx$ (take $A_0=\int_1^\pi {\sin x\over x^\alpha}\,dx$). The $A_n$ alternate in sign and $|A_n|$ eventually decreases to $0$; so .... $\endgroup$ – David Mitra Jan 3 '15 at 11:16
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    $\begingroup$ You would also need to argue $\Bigl|\int_{n\pi}^{n\pi+a}{\sin x\over x^\alpha}\,dx\Bigr|$ is small for large $n$ and $0\le a\le\pi$ to make the argument precise. This is easy to do by making lazy approximations. $\endgroup$ – David Mitra Jan 3 '15 at 11:59
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As David Mitra mentioned in the comments you can split the integral up. First we do $$\int_{1}^{\infty} \frac{\sin x}{x^{\alpha}}dx = \int_{1}^{\pi} \frac{\sin x}{x^{\alpha}}dx + \int_{\pi}^{\infty} \frac{\sin x}{x^{\alpha}}dx$$

Now the first part converges for sure (you can approxiamate it with $\frac{1}{x^a}$). For the second

$$ \int_{\pi}^{\infty} \frac{\sin x}{x^{\alpha}} dx = \sum_{j=1}^{\infty} \int_{j\pi}^{(j+1)\pi} \frac{\sin x}{x^{\alpha}} dx \le \sum_{j=1}^{\infty} \int_{j\pi}^{(j+1)\pi} \frac{\sin x}{(j\pi)^{\alpha}} dx = \sum_{j=1}^{\infty} \frac{1}{(j\pi)^{\alpha}} \int_{j\pi}^{(j+1)\pi} \sin x dx = \sum_{j=1}^{\infty} 2 (-1)^j \frac{1}{(j\pi)^{\alpha}} $$ and since $\frac{2}{(j\pi)^{\alpha}} \to 0$ for $j \to \infty$ and $(-1)^j$ is alternating, the series converges (and so the integral) for every $\alpha > 0$

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  • $\begingroup$ There is a mistake above: on intervals $[j \pi, (j+1) \pi]$ with $j$ odd (where the function is negative), the first inequality is not true; on such intervals, the denominator $(j \pi)^\alpha$ should be replaced by $((j+1) \pi)^\alpha$, but this ruins everything. In order to save this line of reasoning, one should split the series into two subseries, one containing the "odd" intervals, the other the "even" intervals. But such reshuffling of a series is legal only when the series converges absolutely, and this does not happen here. The correct answer is given below by @Jack D'Aurizio. $\endgroup$ – Alex M. Jun 3 '15 at 16:57
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$\frac{1}{x^\alpha}$ is a continuous decreasing function on $[1,+\infty)$ converging to zero and $\sin x$ is a function with a bounded primitive, hence for any $\alpha >0$ the integral $$ \int_{1}^{+\infty}\frac{\sin x}{x^\alpha}\,dx$$ is converging due to the integral version of Dirichlet's test. Integration by parts leads to: $$\int_{1}^{M}\frac{\sin x}{x^\alpha} = \left.\frac{-\cos x}{x^\alpha}\right|_{1}^{M}-\int_{1}^{M}\frac{\alpha\cos x}{x^{\alpha+1}}\,dx=\cos 1+O\left(\frac{1}{M^\alpha}\right)+O(1).$$

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