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Convergence/divergence of $\int_0^{\infty}\frac{x-\sin x}{x^{7/2}}\ dx$

Determine the improper integral $\int_0^{\infty} \frac{x-\sin x}{x^{7/2}}dx$ converge or diverge. Prove that please.

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    @icurays1 x^(7/2)2012-12-12

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Write for $0< \varepsilon<1$ $\int\limits_{0}^{+\infty}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}=\int\limits_{0}^{\varepsilon}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}+\int\limits_{\varepsilon}^{1}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}+\int\limits_{1}^{+\infty}{\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\, dx}.$ Since for small $x, \;\; 0 $x-\sin{x}=\dfrac{x^3}{3!}+O(x^5), $ then $\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\sim {\dfrac{1}{3!}}x^{3-\frac{7}{2}}={\dfrac{1}{3!}}x^{-\frac{1}{2}},$ therefore the first integral in RHS converges.

The second integral is proper integral, therefore it is finite.

The third integral converges, since $\left|\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\right| \leqslant \dfrac{x+1}{x^{\frac{7}{2}}} \leqslant \dfrac{2x}{x^{\frac{7}{2}}}=2x^{-\frac{5}{2}}.$

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    @Antonio Vargas Thanks! Edited.2012-12-12