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As its told on the title I want to check the convergence/divergence of the improper integral when $a\in \mathbb{R}$: \begin{equation} \int_{0}^{1}{\frac{\mid{\log(x)}\mid^a}{\sqrt{1-x^2}}}dx \end{equation} So, it's improper at $x=0$ and $x=1$, so I split the integral in : \begin{equation} \int_{0}^{\frac{1}{2}}{\frac{\mid{\log(x)}\mid^a}{\sqrt{1-x^2}}}dx + \int_{\frac{1}{2}}^{1}{\frac{\mid{\log(x)}\mid^a}{\sqrt{1-x^2}}}dx \end{equation} I see, that on the first one the integrals it's like $\int\mid{\log(x)}\mid^a dx $ by the comparison limit test, but I don't know how to prove that $\int\mid{\log(x)}\mid^a dx $ converges.

Hopefully you can help me. Much thanks!

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Note that

$$\int_{0}^{\frac{1}{2}}{\frac{\mid{\log(x)}\mid^a}{\sqrt{1-x^2}}}dx$$

converges $\forall a$ by comparison test with $\frac1{\sqrt x}$.

For the second part

$$\int_{\frac{1}{2}}^{1}{\frac{\mid{\log(x)}\mid^a}{\sqrt{1-x^2}}}dx $$

let $1-x^2=y^2$

$$\int_{\frac{\sqrt 3}{2}}^{0}{\frac{\mid{\log(\sqrt{1-y^2}}\mid^a}{y}}\cdot \left(\frac{-y}{\sqrt{1-y^2}}\right)dy = \int_{0}^{\frac{\sqrt 3}{2}}{\frac{\mid{\log(\sqrt{1-y^2}}\mid^a}{\sqrt{1-y^2}}}dy $$

and note that for $y\to 0$

$$\mid\log\left(\sqrt{1-y^2}\right)\mid\sim \frac{y^2}{2}$$

thus the integral converges for $-2a<1$ that is $a>-\frac12$ and diverges for $-2a\ge1$ that is $a\le-\frac12$ by comparison with $y^{2a}$.

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For $\frac12\le x\le1$, $$ 1-x\le|\log(x)|\le2\log(2)(1-x) $$ and $$ \sqrt{\tfrac32}\sqrt{1-x}\le\sqrt{1-x^2}\le\sqrt2\sqrt{1-x} $$ Therefore, $$ \sqrt{\tfrac12}\int_{1/2}^1(1-x)^{a-\frac12}\,\mathrm{d}x \le\int_{1/2}^1\frac{|\log(x)|^a}{\sqrt{1-x^2}}\,\mathrm{d}x \le(2\log(2))^a\sqrt{\tfrac23}\int_{1/2}^1(1-x)^{a-\frac12}\,\mathrm{d}x $$ which converges for $a\gt-\frac12$ and diverges for $a\le-\frac12$.


For $0\le x\le\frac12$, $$ \frac1{\sqrt{1-x^2}}\le\frac2{\sqrt3} $$ Therefore, $$ \int_0^{1/2}\frac{|\log(x)|^a}{\sqrt{1-x^2}}\,\mathrm{d}x \le\frac2{\sqrt3}\int_{\log(2)}^\infty x^ae^{-x}\,\mathrm{d}x $$ which converges for all $a$.


Therefore, $$ \int_0^1\frac{|\log(x)|^a}{\sqrt{1-x^2}}\,\mathrm{d}x $$ converges for $a\gt-\frac12$ and diverges for $a\le-\frac12$.

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  • $\begingroup$ thus I've been correct too! thanks $\endgroup$ – gimusi Jan 30 '18 at 23:05
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    $\begingroup$ @gimusi: indeed, and been upvoted, too. $\endgroup$ – robjohn Jan 30 '18 at 23:08
  • $\begingroup$ mee too, it is a very nice approch by inequalities and thanks again for the check! $\endgroup$ – gimusi Jan 30 '18 at 23:10
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$$\int_{0}^{1}\frac{x^\beta}{\sqrt{1-x^2}}\,dx =\frac{\sqrt{\pi}\,\Gamma\left(\frac{\beta+1}{2}\right)}{2\,\Gamma\left(\frac{\beta+2}{2}\right)}$$ by Euler's Beta function. The RHS is a $C^\infty$ function in a neighbourhood of the origin, hence by applying $\left.\frac{d^a}{d\beta^a}\left(\ldots\right)\right|_{\beta=0}$ to both sides we have that the given integral is finite for any $a\in\mathbb{N}$. By the Cauchy-Schwarz inequality the function $$ f(a)=\int_{0}^{1}\frac{\left|\log(x)\right|^a}{\sqrt{1-x^2}}\,dx\stackrel{x\mapsto e^{-t}}{=}\int_{0}^{+\infty}\underbrace{\frac{t^a}{\sqrt{e^{2t}-1}}}_{g(t)}\,dt $$ is log-convex on its maximal domain. $g(t)$ behaves like $C t^{a-1/2}$ in a right neighbourhood of the origin and like $t^a e^{-t}$ in a left neighbourhood of $+\infty$, hence integrability is ensured by $a>-\frac{1}{2}$, which is also a necessary condition.

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  • $\begingroup$ Okey, that was intense. We haven't studied the Euler's beta function, so must be a way to prove it without using it. But thanks anyway Jack, good to know ways to prove it like this. $\endgroup$ – Ginger Jan 30 '18 at 21:59
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    $\begingroup$ @Jack Sorry I've obtained that the integral converges for any $a >-\frac12$, could you please indicate where I went wrong? Thanks! $\endgroup$ – gimusi Jan 30 '18 at 23:01
  • $\begingroup$ @gimusi: I concur. I've just posted my answer. $\endgroup$ – robjohn Jan 30 '18 at 23:03
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    $\begingroup$ @gimusi: you are right, indeed my last inequality did not hold, now fixed. $\endgroup$ – Jack D'Aurizio Jan 30 '18 at 23:37
  • $\begingroup$ @JackD'Aurizio Thanks, unfortunately I can't follow your work in detail at the moment, I have to upvote for faith! $\endgroup$ – gimusi Jan 30 '18 at 23:50

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