What does it mean for the first term to "give" the condition $s \leq 1/2$ and why?

Question

What does it mean for the first term to "give" the condition $s \leq 1/2$ and why? Also the second term "gives" $s \lt2$? I don't understand what this means at all. If someone could clarify

Answer 1

The author considers under what condition for $s$ the limit of the integral for $a\to 0^+$ exists.

Now, looking at the first term and using $\sin \sqrt x \stackrel{x\to 0^+}{\sim}\sqrt x$ you see $$\frac{\sin \sqrt x}{x^s}\stackrel{x\to 0^+}{\sim}x^{\frac 12 -s}$$

So, convergence happens for $$\frac 12 -s\geq 0\Leftrightarrow s\leq \frac 12$$.

Looking at the second term, you see (limit comparison) $$\int_a^b\frac{\sin \sqrt x}{x^{s+1}}dx \stackrel{a\to 0^+}{\sim} \int_a^b\frac{1}{x^{s+\frac 12}}dx$$

So, convergence happens for

$$s+\frac 12 < 1 \Leftrightarrow s<\frac 12$$

So, considering these terms while $a\to 0^+$ under the condition that they should have a finit limit, these terms "give" the listed conditions for $s$.

Additional info after comments:

The question is for which $s$ does the improper integral exists for $a\to 0^+$.

First term:

Since $$\left. \frac{\sin \sqrt x}{x^s}\right|_{a}^b=\frac{\sin \sqrt b}{b^s}- \frac{\sin \sqrt a}{a^s}$$ we are only interested in the behaviour of

$$\frac{\sin \sqrt a}{a^s}= \frac{\sin \sqrt a}{\sqrt a}\cdot\frac{\sqrt a}{a^s}$$ $$= \frac{\sin \sqrt a}{\sqrt a}\cdot a^{\frac 12 -s}\stackrel{a\to 0^+}{\longrightarrow}\begin{cases} 1 & \text{ for } s=\frac 12 \\ 0 & \text{ for } s < \frac 12 \\ +\infty & \text{ for } s > \frac 12\end{cases}$$

Second term:

For the integrand you have

$$\frac{\sin \sqrt x}{x^{s+1}} =\frac{\sin \sqrt x}{\sqrt x}\cdot \frac 1{x^{s+\frac 12}}$$ or $$\frac{\frac{\sin \sqrt x}{x^{s+1}}}{\frac 1{x^{s+\frac 12}}} \stackrel{x\to 0^+}{\longrightarrow}1$$ So, according to the limit comparison test for improper integrals $\int_0^b\frac{\sin \sqrt x}{x^{s+1}}dx$ is convergent ($a\to 0^+$) iff $\int_0^b\frac{1}{x^{s+\frac 12}}dx$ is convergent.

This happens, if and only if $s+\frac 12 < 1$. Otherwise it is divergent to $+\infty$.