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Let $s$ be a real number greater than 1. Let the function $f$ be defined as

$f(x)=\frac{ln(x)}{x^s}, x>0$.

a) Show that $\int_{a}^{\infty}\frac{ln(x)}{x^s}dx$ converges for $a>0$ and find its value.

b) Show that the infinite series $\sum_{n=1}^{\infty} \frac{ln(x)}{n^s}$ converges.


I believe I can show that (a) converges but I'm not certain how to find it's value. Any help is appreciated.

For (a) I note that for $f(s)=\frac{ln(x)}{x^s}, x>0$ we have that f is a positive valued decreasing function because $f'(s)=\frac{1}{x^{s+1}}-\frac{s\cdot ln(x)}{x^{s+1}}<0$ for $x>0$.

Therefore

$\int_{a}^{\infty}f(x)dx=\lim_{t \to \infty}\int_{a}^{t}\frac{ln(x)}{x^s}dx=\frac{ln(a)}{a^{a-s}(s-1)}+\frac{1}{a^{a-s}(s-1)^2}=\frac{(s-1)ln(a)+1}{a^{s-1}(s-1)^2}<\infty$.

Which implies that $\int_{a}^{\infty}\frac{ln(x)}{x^s}dx$ is convergent.

But I'm uncertain how to find it's value.

for (b). I will argue that it converges because f is positive and decreasing with $\lim_{x \to \infty}f(x)=0$. Meaning that as $\int_{a}^{\infty}\frac{ln(x)}{x^s}dx$ converges $=>$ $\sum_{n=1}^{\infty} \frac{ln(x)}{n^s}$ converges.

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    $\begingroup$ to calculate $\int_a^{\infty}x^{-s}\log x dx$ simply integrate by parts. $\endgroup$ – Lorenzo Quarisa Apr 7 '18 at 13:02
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b. Use Cauchy's Condensation Test to transform the general term into $$\frac{2^n \ln(x)}{2^{ns}} = \frac{\ln(x)}{(2^{s-1})^n},$$ so that it becomes that of a geometric series for each $x$.

Perhaps you mean to prove the convergence of $\displaystyle \sum_{n=1}^{\infty} \frac{\ln(n)}{n^s}$. Use the same trick again. $$\frac{2^n \ln(2^n)}{2^{ns}} = \frac{n\ln2}{(2^{s-1})^n},$$ The Root/Ratio Test shows the convergence of $\displaystyle \sum_{n=1}^{\infty} \frac{n\ln2}{(2^{s-1})^n}$.

$$\frac{(n+1)\ln2}{(2^{s-1})^{n+1}} \cdot \frac{(2^{s-1})^n}{n\ln2} = \frac{n+1}{n} \cdot \frac{1}{2^{s-1}} \xrightarrow[n \to \infty]{} \frac{1}{2^{s-1}} < 1$$

This gives the desired conclusion.

a. You can show this without calculating $f'(x)$.

$f$ is not decreasing on $(0,\infty)$, but that doesn't affect your argument in (b) because $f$ is decreasing for $x$ sufficiently large. To determine whether $f$ is decreasing, you need to solve $$f'(x)=\frac{1}{x^{s+1}}-\frac{s\cdot \ln(x)}{x^{s+1}}<0 \iff x > e^{1/s}$$

The substitution $n \mapsto 2^n$ suggests us to consider the substitution $x = e^t$, $dx = e^t\,dt = x\,dt$. As $y \to \infty$, $\ln y \to \infty$. The inegral becomes \begin{align} \int_a^{y} \frac{\ln(x)}{x^s} dx &= \int_{\ln a}^{\ln y} \frac{t}{e^{ts}} e^t dt \\ &= \int_{\ln a}^{\ln y} t e^{-t(s-1)} dt \\ &= \left[ -\frac{t e^{-t(s-1)}}{s-1} \right]_{\ln a}^{\ln y} + \int_{\ln a}^{\ln y} \frac{e^{-t(s-1)}}{s-1} dt \\ &= \frac{(\ln a) a^{-(s-1)}}{s-1} + \frac{a^{-(s-1)}}{(s-1)^2} - \underbrace{\left( \frac{(\ln y) y^{-(s-1)}}{s-1} + \frac{y^{-(s-1)}}{(s-1)^2} \right)}_{\to 0 \text{ as } y \to \infty} \end{align}

Hence $$\int_a^\infty \frac{\ln(x)}{x^s} dx = \frac{(\ln a) a^{-(s-1)}}{s-1} + \frac{a^{-(s-1)}}{(s-1)^2}.$$

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