Integrability of $x^{-a}\log(x)$

Question

Take $b>a>1$ By considering $x^{-y}$ over $(1,\infty)\times (a,b)$, show that $$\int_{1}^{\infty}\frac{x^{-a}-x^{-b}}{\log(x)}dx$$ exists and find its value

I've assumed they want me to write the intagral as $$\int_{1}^{\infty}\int_{a}^{b}\frac{yx^{-y-1}}{\log(x)}dxdy$$ and use Tonelli's Theorem to justify reversing the order of integration and showing that the function is integrable that way. My problem is that I don't see how that would help, as the integrand still seems impossible to evaluate the other way around.

Answer 1

Let $x=e^u$ to get

$$I=\int_0^\infty\frac{e^{-(a-1)u}-e^{-(b-1)u}}u\ du$$

This happens to be Frullani's integral, and one easily finds that

$$I=\ln\left(\frac{b-1}{a-1}\right)$$

Answer 2

How about instead, $$\int_{1}^{\infty}\frac{x^{-a}-x^{-b}}{\log(x)}dx=\int_{1}^{\infty}\int_{a}^{b}x^{-y}dydx$$ EDIT: $\int x^{-y} dy = \int e^{-y\log(x)}= \frac{-1}{\log(x)}e^{-y\log(x)}= \frac{-1}{\log(x)}x^{-y}$.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{1}^{\infty}{x^{-a} - x^{-b} \over \ln\pars{x}}\,\dd x & = \int_{1}^{\infty}\pars{x^{-a} - x^{-b}} \int_{0}^{\infty}\pars{1 \over x}^{t}\,\dd t\,\dd x = \int_{0}^{\infty}\int_{1}^{\infty}\pars{x^{-a - t} - x^{-b - t}} \,\dd x\,\dd t \\[5mm] & = \int_{0}^{\infty}\pars{{1 \over t + a - 1} - {1 \over t + b - 1}}\,\dd t = \left.\ln\pars{t + a - 1 \over t + b - 1} \right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty} \\[5mm] & = \bbx{\ds{\ln\pars{b - 1 \over a - 1}}} \end{align}