Studying convergence of ∫a+∞sin⁡(x)xn for a,n∈R+?There is two cases a=0 and a≠0. This was my attempt. Case a=0:∫0+∞sin⁡(x)xn=∫01sin⁡(x)xn+∫1+∞sin⁡(x)xnBut∫01sin⁡(x)xn=∫01xxn=∫011xn−1converges for n&lt;2 and diverges for n≥2. And∫1+∞−1xn&lt;∫1+∞sin⁡(x)xn&lt;∫1+∞1xnconverges for n&gt;1 and diverges for n≤1. That means that ∫0+∞sin⁡(x)xn converges for n∈[1,2] and diverges for n∈R−[1,2].

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Studying convergence of ∫a+∞sin⁡(x)xn for a,n∈R+?There is two cases a=0 and a≠0. This was my attempt. Case a=0:∫0+∞sin⁡(x)xn=∫01sin⁡(x)xn+∫1+∞sin⁡(x)xnBut∫01sin⁡(x)xn=∫01xxn=∫011xn−1converges for n&amp;lt;2 and diverges for n≥2. And∫1+∞−1xn&amp;lt;∫1+∞sin⁡(x)xn&amp;lt;∫1+∞1xnconverges for n&amp;gt;1 and diverges for n≤1. That means that ∫0+∞sin⁡(x)xn converges for n∈[1,2] and diverges for n∈R−[1,2].

Raiden Griffin · Accepted Answer

You can&#039;t make &quot;∫1+∞−1xn&amp;lt;∫1+∞sin⁡(x)xn&amp;lt;∫1+∞1xn converges for n&amp;gt;1 and diverges for n≤1&quot;.We divide [π,+∞) into [kπ,kπ+π),k∈N. Thus,，2(k+1)απα≤∫kπkπ+πsin⁡(x)xαdx≤2kαπα, x∈[kπ,kπ+π),k is even，−2kαπα≤∫kπkπ+πsin⁡(x)xαdx≤−2(k+1)απα, x∈[kπ,kπ+π),k is odd.By Leibniz&#039;s test, this integral converges anyway when α&amp;gt;0, but does not absolutely converges when α≤1

Studying convergence of \(\displaystyle{\int_{{{a}}}^{{+\infty}}}{\frac{{{\sin{{\left({x}\right)}}}}}{{{x}^{{n}}}}}\) for \(\displaystyle{a},{n}\in{\mathbb{{{R}^{+}}}}\)? There

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