> 
> Check convergence of $\int \_1^{\infty }\:\left(\ln x\right)^p\cdot \frac{\sin x}{x}dx,p>0$
> 
> 
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I tried , $$(\ln x)^p\frac{\sin x}{x}\le(\ln x)^p\frac{1}{x}$$

but $\int \_1^{\infty }\:\left(\ln x\right)^p\cdot \frac{1}{x}dx$ is diverges when $p>0$
.

$\int \_1^{b\:}\:\left(\ln x\right)^p\cdot \frac{1}{x}dx=\left[\dfrac{\left(\ln x\right)^{p+1}}{p+1}\right]\_1^b$

$\int \_1^{\infty }\:\left(\ln x\right)^p\cdot \frac{1}{x}dx=\lim \_{x\to \infty }\dfrac{\left(\ln b\right)^{p+1}}{p+1}=\infty $

any another idea how to check if it convergence?

Thanks!