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Prove convergence and absolute convergence of $$ \int_{0}^{\infty}{\sin^{n}\left(x\right)\over x}\,\mathrm{d}x $$

I have seen a related question already, however the answers are not that helpful to me.

My Problem: After using "I.B.P" I get stuck in either using Cauchy criteria or finding a function for direct comparison test.

I was able to show that it is not absolute convergent and that it isn't convergent for even numbers. But I don't know how to do it for the odd numbers.

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  • $\begingroup$ this integral does not converge on the given interval $\endgroup$ – Dr. Sonnhard Graubner Jun 5 '17 at 17:21
  • $\begingroup$ Be careful : this integral does not converge absolutely for $n=1$ $\endgroup$ – Adren Jun 5 '17 at 17:25
  • $\begingroup$ for n=1 it converges, so for sin(x)^n/(x) it must do it too atleast for odd or even numbers $\endgroup$ – MasterPI Jun 5 '17 at 17:28
  • $\begingroup$ and i already could find a proof if n=1 but i dont know how to do it for all n $\endgroup$ – MasterPI Jun 5 '17 at 17:29
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    $\begingroup$ The way you have written the problem it appears we are to prove that the integral converges for all $n.$ $\endgroup$ – zhw. Jun 5 '17 at 19:08
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For $n$ even, the integral diverges since we can look at the interval $\left[\frac{(4k+1)\pi}4,\frac{(4k+3)\pi}4\right]$ $$ \begin{align} \int_0^\infty\frac{\sin^n(x)}{x}\,\mathrm{d}x &\ge\sum_{k=0}^\infty\overbrace{\vphantom{\frac4\pi}\ \ \ \ \ \frac\pi2\ \ \ \ \ }^{\text{width of interval}}\ \ \overbrace{\vphantom{\frac4\pi}\ \ \ 2^{-n/2}\ \ \ }^{\text{min of $\sin^n(x)$}}\ \ \overbrace{\frac4{(4k+3)\pi}}^{\text{min of $1/x$}}\\ &=2^{-n/2}\sum_{k=0}^\infty\frac2{4k+3}\\ \end{align} $$ which diverges by comparison to the Harmonic Series.

Absolutely, for $n$ odd, the integral diverges for the same reason as the integral diverges for even $n$. However, the integral converges conditionally for odd $n$.

$$ \begin{align} &\int_0^\infty\frac{\sin^n(x)}{x}\,\mathrm{d}x\\ &=\int_0^{2\pi}\frac{\sin^n(x)}{x}\,\mathrm{d}x+\sum_{k=1}^\infty\int_0^\pi\sin^n(x)\left(\frac1{x+2k\pi}-\frac1{x+(2k+1)\pi}\right)\,\mathrm{d}x\\ &=\int_0^{2\pi}\frac{\sin^n(x)}{x}\,\mathrm{d}x+\sum_{k=1}^\infty\int_0^\pi\sin^n(x)\frac\pi{(x+2k\pi)(x+(2k+1)\pi)}\,\mathrm{d}x\\ &\le\frac2{\sqrt{n}}+\frac1{4\pi}\sum_{k=1}^\infty\frac1{k^2}\int_0^\pi\sin^n(x)\,\mathrm{d}x\\ &=\frac2{\sqrt{n}}+\frac\pi{24}\int_0^\pi\sin^n(x)\,\mathrm{d}x\\ &\le\frac2{\sqrt{n}}+\frac{\pi^2}{24\sqrt{n}} \end{align} $$


Estimates Used Above $$ \begin{align} \int_0^\pi\frac{\sin^n(x)}{x}\,\mathrm{d}x &\le\int_0^\pi\left(\frac{4x(\pi-x)}{\pi^2}\right)^n\frac{\mathrm{d}x}x\\ &=4^n\int_0^1x^{n-1}(1-x)^n\,\mathrm{d}x\\ &=4^n\frac{\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)}\\ &=\frac{4^n}{n\binom{2n}{n}}\\ &=2\prod_{k=1}^{n-1}\frac{k}{k+\frac12}\\ &\le2\prod_{k=1}^{n-1}\sqrt{\frac{k}{k+1}}\\ &=\frac2{\sqrt{n}} \end{align} $$ $$ \begin{align} \int_0^\pi\sin^n(x)\,\mathrm{d}x &\le\int_0^\pi\left(\frac{4x(\pi-x)}{\pi^2}\right)^n\mathrm{d}x\\ &=4^n\pi\int_0^1x^n(1-x)^n\,\mathrm{d}x\\ &=4^n\pi\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}\\ &=\frac{4^n\pi}{(2n+1)\binom{2n}{n}}\\ &\le\frac\pi{\sqrt{n}} \end{align} $$

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If $n$ is odd, then $\sin^n(t)$ is an odd function. Hence $\int_{-\pi}^\pi \sin^n(t)\, dt = 0.$ It follows by $2\pi$-periodicity that $\int_0^x \sin^n(t)\, dt$ is a bounded function of $x.$ Thus Dirichlet's test implies

$$\int_0^\infty \frac{ \sin^n(t)}{t}\,dt$$

converges.

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