# Determine that improper integral converge or diverge [duplicate]

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.

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 "infinity" should be "\infty" and do you mean $x^{7/2}$ or $\frac{x^7}{2}$? (if its the first, enclose $7/2$ in curly braces: x^{7/2} – icurays1 Dec 12 '12 at 18:30 You need to analyze the badness near $0$. My approach would be the Taylor series for sine. Informally, near $0$ the top behaves like $x^3/6$. – André Nicolas Dec 12 '12 at 18:31 Why did you have to ask the same question twice? – Nameless Dec 12 '12 at 18:31 @icurays1 x^(7/2) – B11b Dec 12 '12 at 18:32

## marked as duplicate by rschwieb, oen, draks ..., Micah, Austin MohrDec 12 '12 at 19:54

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 third integral converges, since $\left|\dfrac{x-\sin{x}}{x^{\frac{7}{2}}}\right| \leqslant \dfrac{x+1}{x^{\frac{7}{2}}} \leqslant \dfrac{2x}{x^{\frac{7}{2}}}=2x^{-\frac{5}{2}}.$
 I think you mean to write $x - \sin x = x^3/3! + O(x^5)$, since it does not make sense to say that something is $o(\cdots)$ on a fixed interval. – Antonio Vargas Dec 12 '12 at 19:48 @Antonio Vargas Thanks! Edited. – M. Strochyk Dec 12 '12 at 19:52