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

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'^7/2' is asked – B11b Dec 12 '12 at 18:14
You must enclose your formulae between dollar signs otherwise it is almost unreadable. – DonAntonio Dec 12 '12 at 18:16
The result is $4\sqrt{2\pi}/15$ :) – Hans Engler Dec 12 '12 at 18:58

Hints: (1) We have potential troubles at $0$ and because the interval is infinite. Break up the integral into two parts, $0$ to (say) $1$ and $1$ to $\infty$.

(2) Deal with the "$\infty$" part. Should not be hard, the function decreases fast.

(3) For the behaviour near $0$, use the Taylor series $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$. For $x$ near $0$, the top is $\lt \dfrac{x^3}{3}$.

(4) So for positive $x$ near $0$, our function is positive but less than a constant times $\dfrac{1}{x^{1/2}}$.

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