Show $\int_o^\infty \frac{|sin(u)|}{u^{1+a}}du$ Convergence for $a>0$

Question

Show $\int_o^\infty \frac{|sin(u)|}{u^{1+a}}du$ Convergence for $a>0$.

and show that for all $h\in R$: $$\int_o^\infty \frac{|sin(\frac{hs}{2})|}{|h|^a(a+s^{a+1})}ds\le\int_o^\infty \frac{|sin(u)|}{u^{1+a}}du$$

I know to show $\int_b^\infty \frac{|sin(u)|}{u^{1+a}}du$ Convergence for any $b>0$. but how to show $\int_0^b \frac{|sin(u)|}{u^{1+a}}du$ Convergence?

Answer 1

Note that

$$\int_0^\infty \frac{|\sin(u)|}{u^{1+a}}du=\int_0^1 \frac{|\sin(u)|}{u^{1+a}}du+\int_1^\infty \frac{|\sin(u)|}{u^{1+a}}du$$

converges for $a\in(0,1)$.

Indeed

$$\int_0^1 \frac{|\sin(u)|}{u^{1+a}}du$$

converges for $a\in(0,1)$ and diverges for $a\ge1$ for comparison test with $\frac{1}{u^{a}}$ since

$$u\to0 \quad \frac{\frac{|\sin(u)|}{u^{1+a}}}{\frac{1}{u^a}}=\frac{|\sin(u)|}{u}\to1$$

And

$$\int_1^\infty \frac{|\sin(u)|}{u^{1+a}}du$$

converges for $a> 0$ since

$$\int_1^\infty \frac{|\sin(u)|}{u^{1+a}}du<\int_1^\infty \frac{1}{u^{1+a}}du$$