TITLE: For which $a\in \mathbb R$ the integral is convergent?
QUESTION [5 upvotes]: For which $a\in \mathbb R$ the integral is convergent?
  $$ \int_0^{+\infty} x^{-5a} \ln(1+x^{2a})dx$$

Firstly I tried to use: $$f(x)=\ln (1+x^{2a}), f'(x)=\frac{1}{1+x^{2a}}$$
$$g(x)=\begin{cases}{\frac{1}{1-5a} x^{1-5a} , a\neq \frac{1}{5}\\\ln x, a= \frac{1}{5}}\end {cases} , \text{              }g'(x)=x^{-5a} $$
Then for $a\neq \frac{1}{5}$:
$$\int_0^{+\infty} x^{-5a} \ln(1+x^{2a})dx= \lim_{M \rightarrow +\infty} [\frac{1}{5-a} x^{1-5a} \ln (1+x^{2a})]^M_0 -\frac{1}{1-5a}\cdot \int_0^{+\infty} x^{1-5a} \frac{1}{1+x^{2a}}dx$$However I know only that $\lim_{M \rightarrow +\infty} [\frac{1}{5-a} x^{1-5a} \ln (1+x^{2a})]^M_0$ is convergent for $a>\frac{1}{5}$ and again I create an integral ($\int_0^{+\infty} x^{1-5a} \frac{1}{1+x^{2a}}dx$) that I can't easily calculate so this method is not effective.
Secondly I tried to use Direct comparison test but then I have:
$$0\le x^{-5a} \ln(1+x^{2a}) \le x^{-5a}(1+x^{2a})$$and also don't know what I can do with it in this moment.Have you got any better ideas?

REPLY [2 votes]: If $a\leq 0$, then clearly the integral diverges.
If $a>0$, then for fixed $\delta>0$ an the integral of the form $\int_0^{\delta}x^{-5a}\ln(1+x^{2a})\,dx$ converges if and only if $a<\frac{1}{3}$, 
because near zero $x^{-5a}\ln(1+x^{2a})\approx x^{-3a}$. On the other hand, the integral $\int_{\delta}^{\infty}x^{-5a}\ln(1+x^{2a})\,dx$ converges if and only if $a>\frac{1}{5}$, as can easily be seen by comparison to $x^{\beta}$ for a suitable $\beta$. In summary, the integral converges if and only if $\frac{1}{5}<a<\frac{1}{3}$.