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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.

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    $\begingroup$ check $\int \frac {\ln(x)}{x^p} = \frac {x^{-p}\ln(x)}{1-p}- \frac {x^{-p+1}}{(1-p)^2} + C$ it should be $\int \frac {\ln(x)}{x^p} = \frac {x^{1-p}\ln(x)}{1-p}- \frac {x^{-p+1}}{(1-p)^2} + C$ $\endgroup$ – user Mar 29 '18 at 12:25
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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

$$\sum_{n=1}^\infty \frac{\ln(n)}{n^p}=\sum_{n=2}^\infty \frac{\ln(n)}{n^p}\le \int_1^{\infty} \frac {\ln(x)}{x^p}dx$$

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  • $\begingroup$ If I plot in the upper and lower bounds of the integral I get: $(\frac {b^{1-p}ln(b)}{1-p}- \frac {b^{1-p}}{(1-p)^2})-(\frac {a^{1-p}ln(a)}{1-p}- \frac {a^{1-p}}{(1-p)^2}) $ And as $p>1$ we know that ${b^{1-p}}$ approaches $0$ as $b \to \infty $, which means that the limit is $\frac {a^{1-p}ln(a)}{1-p}- \frac {a^{1-p}}{(1-p)^2}$ ? And this can be used to prove that the series is convergent, as the limit $\frac {a^{1-p}ln(a)}{1-p}- \frac {a^{1-p}}{(1-p)^2}$ of the improper integral is less then $\infty$ ? $\endgroup$ – Simbörg Mar 29 '18 at 15:20
  • $\begingroup$ @Simbörg Yes exactly it's a finite quantity $\endgroup$ – user Mar 29 '18 at 15:28
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$$\int^{\infty}_a\frac{\ln x}{x^p}\,dx=\frac{x^{1-p}}{1-p}\ln x\Big|^{\infty}_a-\frac{1}{1-p}\int^{\infty}_a\frac{1}{x^p}\,dx=\frac{x^{1-p}}{1-p}\ln x\Big|^{\infty}_a-\frac{1}{1-p}\cdot\frac{x^{1-p}}{1-p}\Big|^{\infty}_a$$ 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 $a>0$.

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