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Why does $\displaystyle\int_1^\infty\dfrac{\sin^2(x)}{x}$ diverge ?

Why does Dirichlet-Test not work ?

Define $f(x)=g(x)=\dfrac{\sin(x)}{x^{1/2}}$, Then $\forall b>1$ and $a>0;$ $$\Big\lvert\displaystyle\int_1^b\dfrac{\sin(x)}{x^{a}}\Big\rvert<+\infty$$

also for $a=1/2$

and since $\lim\limits_{x\to\infty}\dfrac{\sin(x)}{x^{1/2}}=0$

The integral should be convergent, did I overlook something ?

but with the same method of Dirichlet, one can show the divergence


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$\sin$ changes signs, $\sin^2$ does not. – Thomas Apr 20 '14 at 12:51
@Thomas but in the statement I couldn't find anything about the signs. – derivative Apr 20 '14 at 12:52
For Dirichlet's criterion, you need one factor with bounded integrals, and one monotonic. You have no monotonic factor here, when you use $$\frac{\sin x}{\sqrt{x}}\cdot \frac{\sin x}{\sqrt{x}}.$$ If you take a different decomposition with a monotonic factor, the integrals of the other aren't bounded. – Daniel Fischer Apr 20 '14 at 12:52
@Daniel Fischer Yes that was the point. – derivative Apr 20 '14 at 12:55

For example because, for every $n\geqslant1$, $$ \int_{n\pi+\pi/4}^{n\pi+3\pi/4}\frac{\sin^2x}x\mathrm dx\geqslant\int_{n\pi+\pi/4}^{n\pi+3\pi/4}\frac{\frac12}{(n+1)\pi}\mathrm dx=\frac1{4(n+1)}$$

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awesome, but my question was why Dirichlet fails, now it is clear. – derivative Apr 20 '14 at 13:00

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