This is question 10.20c from Apostol's Mathematical Analysis. Basically, I am trying to show that $$ f(x)=\frac{1}{x^4 \sin^2(x)+1} \in L^1([0,\infty)) $$
I know that for some value $k$, I can create $A=\{x| \frac{1}{x^4 \sin^2(x)+1}<\frac{1}{x^k}\}$ which is measurable and also $$ \int_A \frac{1}{x^4 \sin^2(x)+1} dx $$ is finite.
I am stuck trying to show that $\int_{A^c} f(x)$ is finite. I have been trying to do this by constructing a set $B=\cup_n Bn$ where each $B_n$ is disjoint and is centered around some multiple of $\pi$. I am stuck, however, in trying to construct these sets such that the boundaries of each $B_n$ are in $A$ and so that $\mu(B) = \sum_n \mu(B_n)$ is finite. Any help would be greatly appreciated
Let $A(\alpha)=\{x\mid x^4\sin^2(x)\geqslant x^\alpha\}$ and $B(\alpha)=\{x\mid x^4\sin^2(x)\leqslant x^\alpha\}$ for some $\alpha\gt0$, then $$ \int_{A(\alpha)}f\leqslant\int_1^\infty\frac{\mathrm dx}{x^\alpha}, $$ which converges if $\alpha\gt1$.
On the other hand, $B_n(\alpha)=B(\alpha)\cap[n\pi-\frac\pi2,n\pi+\frac\pi2]$ is roughly the set of real numbers $n\pi+z$ with $|z|\leqslant\frac\pi2$ such that $(n\pi)^4\sin^2(z)\leqslant(n\pi)^\alpha$, that is $|\sin z|\leqslant(n\pi)^{\alpha/2-2}$. Thus, if $\alpha\lt4$, $$ \int_{B_n(\alpha)}f\leqslant|B_n(\alpha)|\sim2(n\pi)^{\alpha/2-2}. $$ Note that the series $$ \sum_nn^{\alpha/2-2}, $$ converges for every $\alpha\lt2$. Finally the decomposition into $A(\alpha)$ and $B(\alpha)=\bigcup\limits_nB_n(\alpha)$ shows that $f$ is integrable for every $\alpha$ in $(1,2)$.