Let $f(x) = \log(1+\frac{1}{x^2})$. I am trying to show $f \in L^1(\mathbb{R}^+)$ where $\mathbb{R}^+ = (0,\infty)$.
What I have done so far
Since $f$ is strictly positive, I just have to show that $\int_{\mathbb{R}^+}f < \infty$. I already proved that this is true on $(\varepsilon,\infty)$ for any $\varepsilon>0$: since $\log(1+1/x^2) \leq 1/x^2$ for all $x > 0$, we have $$ \int_{(\varepsilon,\infty)}\log\left(1+\frac{1}{x^2}\right)dx \leq \int_{(\varepsilon,\infty)}\frac{1}{x^2}dx = \frac{1}{\varepsilon}, $$ so $\int_{(\varepsilon,\infty)}f < \infty$, where I used the Riemann integrability of $1/x^2$.
But here I run into trouble where I can't figure out how to bring the lower bound to zero, as $1/\varepsilon$ blows up as $\varepsilon \to 0$.
I attempted to use the fact that for any $\alpha >0$, $$ \lim_{x \to \infty}\frac{\log(1+x)}{x^{\alpha}} = 0, $$ i.e. there must exist an $M_{\alpha}>0$ such that $x>M_{\alpha}$ implies that $\log(1+x)\leq x^{\alpha}$. Letting $z = 1/x^2$, we see that for $x<1/\sqrt{M_{\alpha}}$ we have $$ \log\left(1+\frac{1}{x^2}\right)\leq \frac{1}{x^{2\alpha}}. $$ Then we set $\alpha = 1/3$ or something like that, so for $x < 1/\sqrt{M_{1/3}} = \varepsilon$ we have $f(x)\leq 1/x^{2/3}$. I'm thinking then to use some sort of Monotone Convergence Theorem argument, i.e. let $f_n(x) = f(x)\chi_{[1/n,\varepsilon]}$, making sure to start at large enough $n$ such that $1/n < \varepsilon$, then $$ \int_{(0,\varepsilon)}f_n(x)dx = \int_{(0,\varepsilon)}\log\left(1 + \frac{1}{x^2}\right)\chi_{[1/n,\varepsilon]}dx \leq \int_{1/n}^{\varepsilon}\frac{1}{x^{2/3}}dx = 3\left(\sqrt[3]{\varepsilon} - \frac{1}{\sqrt[3]{n}}\right), $$ where I again used Riemann integrability, which implies Lebesgue integrability. Sending $n \to \infty$ and using monotone convergence theorem we have that $\int_{(0,\varepsilon)}f(x)dx \leq 3 \sqrt[3]{\varepsilon}< \infty$.
My main question
I have shown then that $f$ is Lebesgue integrable on $(0,\varepsilon)$ for sufficiently small $\varepsilon$, and also that it is Lebesgue integrable on $(\varepsilon,\infty)$ for any $\varepsilon>0$. Is it possible to "glue" these results together to get that $f\in L^1(\mathbb{R}^+)$? This solution seems kind of sloppy and disconnected to me, but it's the closest I could come to a solution and I'd really appreciate any help someone could provide.
Please let me know if something is unclear, thank you!
The Lebesgue integral is additive for all nonnegative measurable functions, even those not necessarily integrable, so for any $ ε > 0$,
$$ ∫_{ℝ_+} f \ \mathrm dλ= ∫_{ℝ_+} f ( 1_{(0,ε)} + 1_{[ε,∞)}) \ \text dλ = ∫_{(0,ε)} f \ \text dλ + ∫_{[ε,∞)} f \ \text dλ, $$ which justifies the gluing procedure.