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?
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.