Convergence of logarithm/polynomial improper integrals My instructor has a fondness for asking questions regarding the convergence of such integrals:
$$ \int_{0}^{1} \frac{\ln(x)}{x^{1/2}}\,\mathrm dx $$
$$ \int_{0}^{1} \frac{\ln(x)}{x^{3/2}}\,\mathrm dx $$
What is the best way to determine the convergence of such integrals? Comparison tests require that $ f(x), g(x) \ge 0 $ so let's consider $-\ln x$.
How can I proceed from here?
 A: Hint: For the second one, use this fact that $$\lim_{x\to 0^+}(x-0)^\frac{3}{2}\frac{\ln x}{x^{\frac{3}{2}}}$$ is infinite.
A: More or less  André Nicolas' answer in a slightly different form:


*

*We have $f(x) = |\ln x|\,x^{-1/2}= -(\ln x) \,x^{-1/2}$ which we claim to be smaller than $g(x)= \alpha\,x^{3/4}$ for all $x\in[0,1]$ for an appropriate $\alpha>0$. To prove that we introduce $h(x)= g(x)/f(x)$ and seek for its extrema $h'(x^*)=0$ with the only solution $x^*=e^{-4}$. It is easy to show that $h''(x^*) >0$ such that the extremum is in fact a minimum. With $g(x^*)= e \alpha/4$ we see that $g(x) > f(x)$, e.g., for $\alpha=2$. In conclusion, we have
$$\int_0^1\frac{|\ln x|}{x^{1/2}}  dx
\leq \int_0^1 \frac{2}{x^{3/4}}dx  =8.$$

*In the second case, we have $f(x) =|\ln x|\,x^{-3/2}= -(\ln x) \,x^{-3/2}$. We choose $g(x)= (1-x)x^{-3/2}$ and have $g(x)\leq f(x)$ because $(1-x) \leq -\ln x$. Thus, $$\int_0^1\frac{|\ln x|}{x^{3/2}}  dx
\geq \int_0^1 \frac{1-x}{x^{3/2}}dx \to \infty.$$
With similar arguments, you can show that
$$\int_0^1 \frac{|\ln x|}{x^p} dx$$
converges for $p<1$. In fact, these integral can be easily evaluated explicitly via integration by parts. We have 
$$\int_0^1 \frac{|\ln x|}{x^p} dx =
-\int_0^1 \frac{\ln x}{x^p} dx =\underbrace{\lim_{x\downarrow 0} \frac{\ln x}{(1-p) x^{p-1}}}_{0}+ \underbrace{\int_0^1 \frac{1}{(1-p)x^p} dx}_{(1-p)^{-2}} .$$
The results below the underbrace hold for $p<1$ otherwise both terms diverge.
A: Look at the leading order behavior at the singularity. In both cases you gave, the singularity is at $x=0$. The logarithm integrates to something finite near $x=0$, and $x^a$ integrates to something finite near $x=0$ as long as $a>-1$.
Then in your specific cases, $x^{-1/2}\ln x$ will have a finite integral on $(0,1)$ while $x^{-3/2}\ln x$ will diverge.
These comments are more heuristic than anything, but it sounds like your teacher wants you to be able to make an assessment by eye.
A: The second one is automatic: Already $\displaystyle\int_0^1\dfrac{dx}{x^{3/2}}$ blows up, and the $\log$ makes things worse.
For $\dfrac{\log x}{x^{1/2}}$, I would rewrite it as, say, $\dfrac{x^{1/4}\log x}{x^{3/4}}$.
By L'Hospital's Rule, or otherwise, we can show that $\lim_{x\to 0+}x^{1/4}\log x=0$. But $\displaystyle\int_0^1 \dfrac{dx}{x^{3/4}}$ converges, so our integral does. 
