<s>[INST] Determine that improper integral converge or diverge 
Possible Duplicate:
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. 
 [/INST] 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 $x, \;\; 0<x<\varepsilon$  $$x-\sin{x}=\dfrac{x^3}{3!}+O(x^5), $$ 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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