Convergence of improper integral and infinite series As $p>1$ is a real number, the function $f$ is defined as 
$$ f(x)= \frac {\ln(x)}{x^p}\,,x>0$$
$1)$Show that the improper integral 
$$ \int_a^\infty \frac {\ln(x)}{x^p} \, dx$$
is convergent for $ a>0 $, and determine its value. 
$2)$ Show that the infinite series 
$$\sum_{n=1}^\infty \frac{\ln(n)}{n^p}$$ is convergent.

$1)$ I know that the improper integral $ \int_a^\infty f(x) \, dx$ is convergent if and only if the function $F(b) = $$ \int_a^b f(x)\,dx$ is bounded. I have found the indefinite integral of $$ \int \frac {\ln(x)}{x^p} = \frac {x^{1-p} \ln(x)}{1-p}- \frac {x^{-p+1}}{(1-p)^2} + C  $$ 
How do I prove that is bounded and determine its value?  
$2)$ I know that I can make use of the integral criteria: The infinite series  $\sum_{n=1}^\infty = f(n)$ converges if and only if the improper integral $ \int_1^\infty f(x)\,dx$ is convergent. 
I am uncertain as to how I can show this. 
 A: Note that
$$\int_a^{\infty} \frac {\ln(x)}{x^p}dx=\lim_{b\to \infty} \int_a^{b} \frac {\ln(x)}{x^p} dx$$
then since
$$\int_a^b \frac {\ln(x)}{x^p} dx= F(b)-F(a)$$
$$\int_a^{\infty} \frac {\ln(x)}{x^p}dx=\lim_{b\to \infty} \left(F(b)-F(a)\right)$$
For the series note that
$$\sum_{n=1}^\infty \frac{\ln(n)}{n^p}=\sum_{n=2}^\infty \frac{\ln(n)}{n^p}\le \int_1^{\infty} \frac {\ln(x)}{x^p}dx$$
A: $$\int^{\infty}_a\frac{\ln x}{x^p}\,dx=\frac{x^{1-p}}{1-p}\ln x\Big|^{\infty}_a-\frac{1}{1-p}\int^{\infty}_a\frac{1}{x^p}\,dx=\frac{x^{1-p}}{1-p}\ln x\Big|^{\infty}_a-\frac{1}{1-p}\cdot\frac{x^{1-p}}{1-p}\Big|^{\infty}_a$$
Notice that 
$$\lim_{x\to\infty}x^{1-p}\ln x=0\hspace{0.2cm}\text{and}\hspace{0.2cm}\lim_{x\to\infty}x^{p-1}=0$$
since $p>1$ and $\ln x$ grows slower than any positive power of $x$ i.e. $\ln x=o(x^r)$ where $r>0$. You can also use L'Hospital rule if you want. Therefore
$$\int^{\infty}_a\frac{\ln x}{x^p}\,dx=-\frac{a^{1-p}}{1-p}\ln a+\frac{a^{1-p}}{(1-p)^2}=\frac{a^{1-p}}{p-1}\Big(\frac{1}{p-1}+\ln a\Big)<+\infty$$ 
Then the convergence of your series follows from the integral test since the function $f(x):=\ln x/x^p$ is monotone decreasing on $[a,\infty)$ for all $a>0$. 
