justbethlflql

2022-03-31

Studying convergence of ${\int }_{a}^{+\mathrm{\infty }}\frac{\mathrm{sin}\left(x\right)}{{x}^{n}}$ for $a,n\in {\mathbb{R}}^{\mathbb{+}}$?
There is two cases a=0 and $a\ne 0$. This was my attempt. Case a=0:
${\int }_{0}^{+\mathrm{\infty }}\frac{\mathrm{sin}\left(x\right)}{{x}^{n}}={\int }_{0}^{1}\frac{\mathrm{sin}\left(x\right)}{{x}^{n}}+{\int }_{1}^{+\mathrm{\infty }}\frac{\mathrm{sin}\left(x\right)}{{x}^{n}}$
But
${\int }_{0}^{1}\frac{\mathrm{sin}\left(x\right)}{{x}^{n}}={\int }_{0}^{1}\frac{x}{{x}^{n}}={\int }_{0}^{1}\frac{1}{{x}^{n-1}}$
converges for n<2 and diverges for $n\ge 2$. And
${\int }_{1}^{+\mathrm{\infty }}\frac{-1}{{x}^{n}}<{\int }_{1}^{+\mathrm{\infty }}\frac{\mathrm{sin}\left(x\right)}{{x}^{n}}<{\int }_{1}^{+\mathrm{\infty }}\frac{1}{{x}^{n}}$
converges for n>1 and diverges for $n\le 1$. That means that ${\int }_{0}^{+\mathrm{\infty }}\frac{\mathrm{sin}\left(x\right)}{{x}^{n}}$ converges for $n\in \left[1,2\right]$ and diverges for $n\in \mathbb{R}-\left[1,2\right]$.

### Answer & Explanation

Raiden Griffin

You can't make "${\int }_{1}^{+\mathrm{\infty }}\frac{-1}{{x}^{n}}<{\int }_{1}^{+\mathrm{\infty }}\frac{\mathrm{sin}\left(x\right)}{{x}^{n}}<{\int }_{1}^{+\mathrm{\infty }}\frac{1}{{x}^{n}}$ converges for n>1 and diverges for $n\le 1$".
We divide $\left[\pi ,+\mathrm{\infty }\right)$ into $\left[k\pi ,k\pi +\pi \right),k\in N$. Thus,

By Leibniz's test, this integral converges anyway when $\alpha >0$, but does not absolutely converges when $\alpha \le 1$

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