How can I prove that the following improper integral is convergent? How can I prove that the following improper integral is convergent
$$
\int_0^{\infty}\frac{\sin(ax)}{ax}\log^n(\frac{b}{x})\,dx
$$
where $a,b>0$
 A: If we apply De L' Hospital' rule, we get that $\displaystyle{\lim _\limits{u\rightarrow +\infty }\frac{\log ^n(u)}{\sqrt{u}}=0}$ and so $\displaystyle{\lim _\limits{x\rightarrow 0^+ }\frac{\log ^n(\frac{b}{x})}{\sqrt{\frac{b}{x}}}=0}$ and so for $\displaystyle{M>0, \exists δ>0:\forall x \in (0,δ),\ \left|\log ^n\left(\frac{b}{x}\right)\right|<\frac{M_b}{\sqrt{x}}\ (1)}$,
where $M_b:=M\cdot \sqrt{b}$.
Moreover $\left|\frac{\sin (u)}{u}\right|\leq 1,$ for $u \in \mathbb R$. We combine this inequality with $(1)$ and obtain:
$$\displaystyle{\left| \int_{0}^{δ}\frac{\sin (ax)}{ax}\log ^n\left(\frac{b}{x}\right)dx\right|\leq \int_{0}^{δ}\left|\frac{\sin (ax)}{ax}\log ^n\left(\frac{b}{x}\right)\right|dx<M_b\int_{0}^{δ}\frac{1}{\sqrt{x}}dx=2M_b\sqrt{δ}}$$ 
Now we will use the inequalities $|\sin (ax)|\leq 1$ and$\ \log x < x$, for $a,x \in \mathbb R$. Then if $n\geq 1$ we have the following:
$$\displaystyle{\left| \int_{δ}^{\infty}\frac{\sin (ax)}{ax}\log ^n\left(\frac{b}{x}\right)dx\right|\leq \frac{1}{a}\int_{δ}^{\infty}\left|\frac{\sin (ax)}{x}\log ^n\left(\frac{b}{x}\right)\right|dx <\frac{b^n}{a} \int_{δ}^{\infty}\frac{1}{x^{n+1}}dx=\frac{b^n}{anδ^n}}$$
Finally by the triangular inequality we have:
$$\displaystyle{\left| \int_{0}^{\infty}\frac{\sin (ax)}{ax}\log ^n\left(\frac{b}{x}\right)dx\right|=\left| \int_{0}^{δ}\frac{\sin (ax)}{ax}\log ^n\left(\frac{b}{x}\right)dx+\int_{δ}^{\infty}\frac{\sin (ax)}{ax}\log ^n\left(\frac{b}{x}\right)dx\right|\leq \left|\int_{0}^{δ}\frac{\sin (ax)}{ax}\log ^n\left(\frac{b}{x}\right)dx\right|+\left|\int_{δ}^{\infty}\frac{\sin (ax)}{ax}\log ^n\left(\frac{b}{x}\right)dx\right|<2M_b\sqrt{δ}+\frac{b^n}{anδ^n}<+\infty}$$
and we are done.
Note: When $n=0$ we can be reduced to the known integral of $\displaystyle{\frac{\sin x}{x}}$, which is convergent.
