# Complicated Improper Integral convergence/divergence

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?

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

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.
• The $O(u^{-1/6})$ might be a bit not easy to show. How about changing the 2nd last equation to: $$0 \leq \frac{(ln(u))^2}{u^{1/3}} \leq \frac{(1/u)^2}{u^{1/3}} = u^{-7/3} \text{ }?$$ Jun 25, 2020 at 11:49
• I have edited my answer. The upper bound $u^{-7/3}$ cannot be used since $\int_0^1 u^{-7/3}du = \left[\frac{u^{-4/3}}{-4/3}\right]_0^1=+\infty$ Jun 25, 2020 at 12:11
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 $$t=-u^3$$
• $\log(u^3)=3\log(u)$ Jun 25, 2020 at 12:12