$$\int^{+\infty}_{0} \frac{ |\sin x| }{x(1+ \sqrt x)} dx$$
I try to set a upper boundary 1 for $| \sin x |$, but it doesn't work well
This is initially to prove the absolute convergence
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Try using $|\sin(x)| \leq x$ for $x < 1$ and $|\sin(x)| \leq 1$ for $x > 1$. – Zarrax May 31 '14 at 01:56
2 Answers
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Using your observation that $|\sin x| \le 1$ for all $x$ together with the fact that
$$1 + \sqrt x > \sqrt x$$
we have
$$\frac{|\sin x|}{x(1 + \sqrt x)} \le \frac{1}{x \sqrt x} = x^{-3/2}$$
Now it's easy to check that $\int_1^{\infty} x^{-3/2} dx$ is convergent, dealing with the integral on $[1,\infty)$.
On the other hand, we have that
$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$
implying (as a rather weak corollary) that $\sin x / x$ remains bounded near the origin. Hence the integrand is bounded on $(0,1]$, and the result follows.
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Hint
$\lim_{x\to 0} \frac{ |\sin x| }{x(1+ \sqrt x)}=1$ since $\lim_{x\to 0} \frac{\sin x}{x}=1$
Since the function is positive, you need to prove that the integral is above bounded
So try considering the function equal to 1 on [0,1] and equal to $\frac{1}{x\sqrt x}$ on $]1,\infty[$
Fabien
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