study the convergence of this integral $\int_{0}^{1}x^a|\ln{x}|^bdx$ 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
 A: 
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 $\beta >1$

A: 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.
A: $\newcommand{\+}{^{\dagger}}
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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 .
