How to determine if $ \int_0^1 \frac{(\ln x)^5}{\sqrt{ x}}\mathrm dx$ diverges or converges without calculating the integral itself

Question

I'm new here and I hope I follow the rules of asking a question correctly.

This integral $$ \int_0^1 \frac{(\ln x)^5}{\sqrt x} \mathrm dx$$ is giving me a hard time proving divergence or convergence.

Can you guys help me? Thanks in advance! <3

Answer 1

Make the substitution $y=-\ln x$. Then the integral equals $$-\int_0^\infty y^5e^{-y/2}\,dy$$ whose convergence/divergence seems much more obvious...

Answer 2

Consider the integrand $-\dfrac{(\ln x)^{5}}{\sqrt{x}}=\dfrac{(\ln x^{-1})^{5}}{\sqrt{x}}$, we know that for large $u>0$, $\ln u<u^{1/20}$, so for small $x>0$, we have $(\ln x^{-1})^{5}<x^{-1/4}$, then $\dfrac{(\ln x^{-1})^{5}}{\sqrt{x}}<\dfrac{1}{x^{3/4}}$ and we know that $\displaystyle\int_{0}^{\eta}\dfrac{1}{x^{3/4}}dx<\infty$.

Answer 3

The trick is use limit comparison test with

$$\frac{1}{x^{\frac{1+\frac12}{2}}}=\frac{1}{x^{\frac{3}{4}}}$$

Indeed $\int_0^1 \frac{1}{x^\frac34} \mathrm dx$ converges and since $\forall a>0$ and n we have that

$$x^a\log^n x \to 0$$

in this case we have

$$\frac{\frac{(\ln x)^5}{\sqrt x}}{\frac1{x^\frac34}}=x^\frac14\ln^5x\to0$$

and thus $\int_0^1 \frac{(\ln x)^5}{\sqrt x} \mathrm dx$ concerges too.