4

let $a,b\in {\mathbb R}$, study the convergence of the following integrals: $$\int_{0}^{1}x^{a}\left\vert\,\ln\left(x\right)\,\right\vert^{b}\,{\rm d}x$$

My idea:

$$\int_{0}^{1}x^a|\ln{x}|^bdx=\int_{0}^{1/2}x^a|\ln{x}|^bdx+\int_{1/2}^{1}x^a|\ln{x}|^bdx$$ then I can't,Thank you

Felix Marin
  • 89,464

3 Answers3

3

Let $(\alpha,\beta)\in \mathbb{R}^2$, $$f:t\mapsto \frac{1}{t^\alpha|\ln{t}|^\beta}$$

$f$ defined on $(0,+\infty)$-$\{1\}$

  • Let $\alpha >1$ and $\gamma \in (1,\alpha)$ then, $$ f(t)=\frac{1}{t^\gamma t^{\alpha-\gamma}ln(t)^\beta}=\frac{1}{t^\gamma}g(t) $$ with $\lim_{t\to +\infty} g(t)=0$ Thus for $t$ sufficiently large we get $f(t)\leq \frac{1}{t^\gamma}$ and $$ \int_{e}^{+\infty}\frac{1}{t^\gamma}dt< +\infty $$
  • Let $\alpha <1$ and $\gamma \in (\alpha,1)$ then, $$ f(t)=\frac{t^{\gamma-\alpha}}{t^\gamma ln(t)^\beta}=\frac{1}{t^\gamma}h(t) $$ with $\lim_{t\to +\infty} g(t)=+\infty$ for all $\beta$

Thus for $t$ sufficiently large we get $f(t)\geq \frac{1}{t^\gamma}$ and $$ \int_{e}^{+\infty}\frac{1}{t^\gamma}dt= +\infty $$

  • For $\alpha=1$ , use the substitution $u=ln(t)$ and for $x>e$, $$ \int_{e}^{x}\frac{1}{tln(t)^\beta}dt=\int_{1}^{ln(x)}\frac{du}{u^\beta} $$

    Therefore the integral on $[e,+\infty)$ converge if and only if $\alpha > 1$ or $\alpha=1$ and $\beta >1$

Now, by substituting $u=\frac{1}{t}$ and for 0

Therefore the integral on $(0,\frac{1}e]$ converge if and only if $\alpha < 1$ or $\alpha=1$and $\beta >1$

0

Suppose $a+1>0$. Then, let $\log{x}=-\frac{u}{a+1}$. Clearly, $\frac{dx}{x} = -\frac{du}{a+1}$. Changing the limits of integration, we obtain $$ \int_0^1 x^a |\log{x}|^b\,dx = \int_0^\infty \left(\frac{u}{a+1}\right)^b \mathrm{e}^{-u}\,\frac{du}{a+1} = \frac{\Gamma(b+1)}{(a+1)^{(b+1)}}. $$

We will always have convergence when $b$ is not a negative integer.

As for $a \leq -1$, note $$ \int_0^1 x^a |\log{x}|^b\,dx \geq \int_0^1 x^{-1} |\log{x}|^b\,dx = \int_{-\infty}^0 |u|^b\,du = \int_0^\infty u^b\,du = \lim_{u\to\infty} \frac{u^{b+1}}{b+1}. $$ If $b+1>0$, then we must diverge. However, we still don't know the case $a\leq-1,b\leq-1$.

We treat this finally; let $u=\log{x}$. Then, $$ \int_0^1 x^a |\log{x}|^b\,dx > \int_0^1 \frac{dx}{x|\log{x}|} = \int_{-\infty}^0 \frac{du}{|u|} = \int_0^\infty \frac{du}{u} \to \infty.$$

To summarize, $a\leq-1$ implies divergence. For $a>1$, we have convergence if $b$ is not a negative integer.

Jason
  • 1,518
0

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{a, b\ \in\ {\mathbb R}}$: \begin{align} \int_{0}^{1}x^{a}\verts{\ln\pars{x}}^{b}\,\dd x&= \int_{\infty}^{1}x^{-a}\verts{\ln\pars{1 \over x}}^{b}\,\pars{-\,{\dd x \over x^{2}}} = \int_{1}^{\infty}x^{-a - 2}\ln^{b}\pars{x}\,\dd x \\[3mm]&=\int_{0}^{\infty}\expo{-\pars{a + 2}t}t^{b}\expo{t}\,\dd t \quad\mbox{with}\ \pars{~x \equiv \expo{t}\quad\iff\quad t = \ln\pars{x}~} \end{align}

\begin{align} \int_{0}^{1}x^{a}\verts{\ln\pars{x}}^{b}\,\dd x&= \int_{0}^{\infty}t^{b}\expo{-\pars{a + 1}t}\,\dd t \end{align} $\ds{\large\tt\mbox{This integral converges when}\ a > -1\ \mbox{and}\ b > -1}$.

\begin{align}\color{#00f}{\large% \int_{0}^{1}x^{a}\verts{\ln\pars{x}}^{b}\,\dd x}&= \pars{a + 1}^{-b - 1}\int_{0}^{\infty}t^{b}\expo{-t}\,\dd t =\color{#00f}{\large\pars{a + 1}^{-b - 1}\,\Gamma\pars{b + 1}} \end{align} where $\ds{\Gamma\pars{z}}$ is the Gamma Function .

Felix Marin
  • 89,464