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How to prove using the Abel test that $\int_ {1}^\infty \frac{|\sin x|}{x}$ is divergent?

user402516
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  • The limit of $x^1\times f(x),~~x\to\infty$ is not zero so it diverges. – Mikasa Dec 28 '16 at 20:53
  • Write the integral as a sum of integrals over intervals for which the sine function is of $1$ sign only. Then, apply thW MVT for integrals and compare wih the harmonic series. – Mark Viola Dec 28 '16 at 21:07
  • Related: http://math.stackexchange.com/q/67198/9464 –  Dec 28 '16 at 22:18
  • The Abel's test is a test for convergence of infinite series. I don't understand why you put that in the question while what you want is showing the improper integral being divergent. –  Dec 28 '16 at 22:27
  • @BabakS. There are functions $f\ge 0$ such that $xf(x) \not \to 0$ and yet $\int_1^\infty f < \infty.$ – zhw. Dec 28 '16 at 23:40
  • @zhw.: See this link, page 324 Theorem 1 – Mikasa Dec 29 '16 at 09:40
  • @BabakS. Why did you want me look at that? It has nothing to do with my comment. – zhw. Dec 29 '16 at 16:54

2 Answers2

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For $n>0$, we have

$$\int_{n\pi}^{(n+1)\pi}\frac{|\sin(x)|}{x}dx=$$

$$\int_0^{\pi}\frac{|\sin(t)|}{t+n\pi}\geq \frac{2}{(n+1)\pi}.$$ since $\int_0^{\pi}\sin(t)dt=2$.

Thus,

$$\int_{\pi}^{(n+1)\pi}\frac{|\sin(x)|}{x}dx$$

$$\geq \frac{2}{\pi}\sum_{k=1}^n\frac{1}{(k+1)}$$ which goes to $+\infty$ when $n$ tends to $+\infty$ as an harmonic series. The integral is therefore divergent.

Simply Beautiful Art
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hamam_Abdallah
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\begin{align} &\int_1^M \frac{|\sin x|}xdx>\int_1^M \frac{\sin^2 x}xdx=I(M)\\ \\ I(M)&=\int_{1+\pi/2}^{M+\pi/2} \frac{\sin^2 (x-\pi/2)}{x-\pi/2}dx=\int_{1+\pi/2}^{M+\pi/2} \frac{\cos^2 x}{x-\pi/2}dx\\ &>\int_{1+\pi/2}^{M+\pi/2} \frac{\cos^2 x}{x}dx\\ &=\int_{1}^M \frac{\cos^2 x}{x}dx-\int_{1}^{1+\pi/2} \frac{\cos^2 x}{x}dx+\int_M^{M+\pi/2} \frac{\cos^2 x}{x}dx\\ &>\int_{1}^M \frac{\cos^2 x}{x}dx-\int_{1}^{1+\pi/2} \frac{\cos^2x}{x}dx\\ &=\int_{1}^M \frac{\cos^2 x}{x}dx+K\\ \\ 2I(M)&>\int_1^M \frac{\sin^2 x}xdx+\int_1^M \frac{\cos^2 x}xdx+K\\ &=\int_1^M \frac1xdx+K=\ln M+K\\ \end{align} which means $\lim_{M\to\infty} I(M)$ diverges and so does the original integral.

Kay K.
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  • Manipulating formally a set of divergent integrals lacks rigor. Start with a bounded upper limit and proceed in an analogous fashion. ;-)) – Mark Viola Dec 28 '16 at 22:43
  • @Dr.MV Hi Dr., thanks for your advice. Do you mean it would be better if I assumed that the integral converges and hence had some upper bound and then it leads to some condtradict..? – Kay K. Dec 28 '16 at 22:49
  • If the upper limit is finite, then the integrals exist. Then, the same formal manipulations you used are valid. Wait until the end to take the limit as the upper limit approaches infinity – Mark Viola Dec 28 '16 at 22:52
  • @Dr.MV I got it. Thanks. I also revised the above answer accordingly. – Kay K. Dec 28 '16 at 23:31
  • You forgot to transform the upper limit upon substitution. – Mark Viola Dec 28 '16 at 23:33
  • @Dr.MV Thanks, I fixed it. – Kay K. Dec 28 '16 at 23:37
  • You should consider showing that $K(M)$ is bounded as $M\to \infty$. – Mark Viola Dec 28 '16 at 23:51
  • @Dr.MV The portion of the $K(M)$ that's a function of $M$ is just greater than zero, so I revised so that $K$ is a constant. Please let me know if it's okay now. Thanks! – Kay K. Dec 29 '16 at 00:31