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?

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    $\begingroup$ $f'(x)=\frac{1}{1+x^{2a}}$ is not correct unless $a=1/2$. You have to use the chain rule. $\endgroup$ – Pink Panther Jun 16 '19 at 15:04
  • $\begingroup$ Hint: split it into 3x2 cases: $(a>0,a=0,a<0)\times(x\to0,x\to\infty)$. $\endgroup$ – Simply Beautiful Art Jun 16 '19 at 15:06
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    $\begingroup$ Alternatively you can split the integral to 0-1 and 1-infinity. In the first integral, you can use the first term of the taylor series of log(1+x). In the second integral the polynomial dominates. These two give you two conditions for a $\endgroup$ – Sandeep Silwal Jun 16 '19 at 15:08

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 $\frac{1}{5}<a<\frac{1}{3}$.

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  • $\begingroup$ I have a question about "as can easily be seen by comparison to $x^{\beta}$ for a suitable $\beta$": we know that integral $\frac{1}{x^{\beta}}$ is diverges for $\beta \le 1$ so $\int_0^{+\infty} \frac{1}{x^{5a}}$ is divergent for $a\le \frac{1}{5}$. But how to justify that also $\int_0^{+\infty}x^{-5a}\ln(1+x^{2a})\,dx$ is also divergent? My idea is that for $a\in [0,\frac{1}{5}]$ we have $\ln 2 \le \ln (1+x^{2a}) \le \ln (1+x^{\frac{2}{5}})$ so $\ln (1+x^{2a})\le 2$ and only $x^{-5a}$ have importance in this integral but I don't know if it is a good explanation. $\endgroup$ – MP3129 Jun 16 '19 at 21:27

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