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Does the following integral converge for $x < 0$

$$\int _x^0\:\cfrac{\ln^2 \ (|t |)}{ \sqrt[3]{t} }dt$$

I tried splitting it into two separate integrals under the assumption that both are convergent:

$$\int _x^{-1}\:\cfrac{\ln^2 \ (|t |)}{ \sqrt[3]{t} }dt + \int _{-1}^{0}\:\cfrac{\ln^2 \ (|t |)}{ \sqrt[3]{t} }dt$$

By comparison test:

$$\bigg| \int _x^{-1}\:\cfrac{\ln^2 \ (|t |)}{ \sqrt[3]{t} }dt \bigg| \leq \int _x^{-1}\:{\ln^2 \ (|t|)} dt $$

this part is convergent.

The problem is , I cannot find a way to show it for the other part:

$$\bigg| \int _{-1}^{0}\:\cfrac{ln^2 \ (|t |)}{ \sqrt[3]{t} }dt \bigg| \leq \int _{-1}^{0}\textit{something} $$

Is this even the right approach?

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  • $\begingroup$ How do you define $t^{\frac{1}{3}}$ for $t<0$? $\endgroup$
    – Alex
    Jun 25, 2020 at 10:59
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    $\begingroup$ It is a monotonically increasing function - its domain and range are all real numbers $\endgroup$
    – vlad arete
    Jun 25, 2020 at 11:10
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    $\begingroup$ $t\mapsto t^3$ is a bijection from $\mathbb{R}$ to $\mathbb{R}$ so it is legitimate to define $t\mapsto t^{\frac{1}{3}}$ on $\mathbb{R}$ $\endgroup$
    – charlus
    Jun 25, 2020 at 11:11

2 Answers 2

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The function $t\mapsto \frac{\ln^2\left(\lvert t\rvert\right)}{t^{1/3}}$ is continuous over $(-\infty, 0)$ so the only possible issue with your integral is at 0. If we show that, for example $$\int _{-1}^0\:\cfrac{\ln^2 \ (|t |)}{ t^{1/3} }dt$$ converges then your problem is solved. To make things more practical, let's go back to positive numbers. Let $\varepsilon\in(0,1)$ and write the change of variable $u=-t$ $$\int _{-1}^{-\varepsilon}\cfrac{\ln^2 \ (|t |)}{ t^{1/3} }dt = \int _{-1}^{-\varepsilon}\cfrac{\ln^2 \ (-t)}{ t^{1/3} }dt = -\int_{\varepsilon}^1\frac{\ln^2u}{u^{1/3}}du$$ Letting $\varepsilon\rightarrow 0$, your problem is thus equivalent to showing the convergence of $$\int_{0}^1\frac{\ln^2u}{u^{1/3}}du$$ Write one more change of variable $v=\ln(u)$ $$\int_{\varepsilon}^1\frac{\ln^2u}{u^{1/3}}du = \int_{\ln \varepsilon}^0e^{\frac{2v}{3}}v^2dv$$ Letting $\varepsilon\rightarrow 0$ again, we find $$\int_{0}^1\frac{\ln^2u}{u^{1/3}}du = \int_{-\infty}^0e^{\frac{2v}{3}}v^2dv=\frac{27}{4}$$ This method also allows you to compute your integral modulo some work on the integration boundaries.

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  • $\begingroup$ The $O(u^{-1/6})$ might be a bit not easy to show. How about changing the 2nd last equation to: $$0 \leq \frac{(ln(u))^2}{u^{1/3}} \leq \frac{(1/u)^2}{u^{1/3}} = u^{-7/3} \text{ }?$$ $\endgroup$
    – Yuki.F
    Jun 25, 2020 at 11:49
  • $\begingroup$ Thank you very much for your answer. Could you please explain the part with ...= o(u^-1/6) $\endgroup$
    – vlad arete
    Jun 25, 2020 at 11:52
  • $\begingroup$ Indeed it might not be easy to show because it is false... Let me edit my answer $\endgroup$
    – charlus
    Jun 25, 2020 at 12:03
  • $\begingroup$ I have edited my answer. The upper bound $u^{-7/3}$ cannot be used since $\int_0^1 u^{-7/3}du = \left[\frac{u^{-4/3}}{-4/3}\right]_0^1=+\infty$ $\endgroup$
    – charlus
    Jun 25, 2020 at 12:11
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Assume $t>0$ and let $t=u^3$ to make $$\int _x^0\:\cfrac{\ln^2 \ (|t |)}{ \sqrt[3]{t} }dt=27\int_{\sqrt[3]x}^0 u \log^2(u)\,du=-\frac{3x^{2/3}}{4} \, (2 (\log (x)-3) \log (x)+9)$$

Assume $t<0$ and let $t=-u^3$

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  • $\begingroup$ $\log(u^3)=3\log(u)$ $\endgroup$ Jun 25, 2020 at 12:12

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