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For which values of ''$a$'', does the integral $$\int_{0}^{\infty} \frac{\sin x}{x^a}\;\mathrm dx$$ converges?

I have shown that $$\left| \int_{0}^{\infty} \frac{\sin x}{x^a}\;\mathrm dx \right|\le \lim_{n\to\infty} \int_{0}^{n} \frac 1{x^a}\;\mathrm dx,$$ which converges if $a > 1$. Are there other values of $a$ for which the integral converges?

Chinnapparaj R
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user398623
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  • What makes you think $\displaystyle \lim_{n\to\infty} \int_{0}^{n} \frac{1}{x^a} dx$ converges when $a \lt 1$? (You have problems as $n$ increases) – Henry May 14 '18 at 07:38
  • Sorry, I have edited the question. – user398623 May 14 '18 at 07:45
  • In fact, $\int_0^\infty \frac {\sin x}{x^a};\mathrm dx$ converges for $a>0$ and has the limit $\int_0^\infty \frac {\sin x}{x^a};\mathrm dx \xrightarrow{a\searrow 0} 1$. – Kanu Kim May 14 '18 at 07:46
  • Can you explain how I can show that? – user398623 May 14 '18 at 07:47
  • Your edit does not help much as $\displaystyle \lim_{n\to\infty} \int_{0}^{n} \frac{1}{x^a} dx$ diverges near $x=0$ for $a \ge 1$. But note that $\sin x \approx x$ for $x$ near $0$ while you have an alternating sum for large $x$ – Henry May 14 '18 at 07:52
  • More precisely, $\frac 2 \pi x \le \sin x \le x$ on $(0,\frac\pi 2)$. – Kanu Kim May 14 '18 at 07:54
  • Fyi: https://math.stackexchange.com/q/2771577/512032 – Szeto May 14 '18 at 08:04
  • Can we show just the convergence of the integral using less advanced techniques, I mean without calculating the value? – user398623 May 14 '18 at 08:19
  • @KanuKim I don't think that's true. As KaviRamaMurthy points out in his answer, the integral converges for 1 < a < 2. And for other values of a doesn't. – user398623 May 14 '18 at 09:03
  • Since this question is a better question than the question of which it is closed as a duplicate, I think this one could be reopened. The other has been closed for being asked poorly. – robjohn Nov 19 '20 at 10:10
  • Integration by parts gives $$\int_0^L\frac{\sin(x)}{x^\alpha},\mathrm{d}x=\frac{1-\cos(L)}{L^\alpha}+\alpha\int_0^L\frac{1-\cos(x)}{x^{\alpha+1}},\mathrm{d}x$$ which handles $0\lt\alpha\le1$. For $1\lt\alpha\lt2$, the integral near $0$ is okay since $|!\sin(x)|\le|x|$. This is, if by $\int_0^\infty\frac{\sin(x)}{x^\alpha},\mathrm{d}x$, you mean $\lim\limits_{L\to\infty}\int_0^L\frac{\sin(x)}{x^\alpha},\mathrm{d}x$. If you mean Lebesgue integrability, then for $\alpha\le1$, the integral does not exist. – robjohn Nov 19 '20 at 10:25

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Both integrability near 0 and integrability near $\infty$ have to be taken into account. The integral exists for $1<a<2$ and the improper integral $\int_0 ^{\infty} \frac {\sin (x)} x \, dx$ exists and the value is $\frac {\pi} 2$. [ This is proved in almost all texts on Complex Analysis]. For other values of $a$ the integral does not exist.

Kavi Rama Murthy
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  • Can't this be proved without using complex analysis, I mean just the existence part? – user398623 May 14 '18 at 08:59
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    @user398623 you have different proof here: https://math.stackexchange.com/questions/1175042/evaluating-int-0-infty-frac-sin-xx-dx-with-fubini-theorem – Kavi Rama Murthy May 14 '18 at 10:43
  • How can it be shown that integral exist for $1<a<2$ ? – Shak Jun 08 '18 at 05:25
  • If you allow $\int_0^\infty\frac{\sin(x)}x,\mathrm{d}x=\frac\pi2$, then you are considering the improper integral as a limit, and not considering Lebesgue integrability. In that case, the integral exists for $0\lt\alpha\lt2$. – robjohn Nov 19 '20 at 10:30