I am trying to prove that $f(x)=\frac{x-\sin(x)}{x^3}$ is Riemman integrable in $[0,\infty)$. Here is my attempt of a solution, I think I made a valid solution, nevertheless I would like to find a more elegant (less hand-wavy) approach.
My approach: First note that this function's Taylor Series is $$f(x)=\frac{x-\sum_{n=0}^\infty (-1)^n x^{2n+1}/(2n+1)!}{x^3}=\sum_{n=1}^\infty (-1)^{n+1} \frac{x^{2(n-1)} }{(2n+1)!}=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n+3)!}$$ Now, we may calculate first $\int_0^1 f$, this is $$\int_0^1f=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+3)!} \int_0^1x^{2n}\; dx=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+3)!(2n+1)}$$ Which by convergence criteria converges, now it just remain to be shown that $\int_1^\infty f<\infty $, which is true since $$ \int_1^\infty f\leq \int_1^\infty \bigr\vert f\bigr\vert =\int_1^\infty \frac{\vert x-\sin(x) \vert}{\vert x^3 \vert}dx $$ On the other hand, we have $$\vert x - \sin(x)\vert=\vert x+\sin(-x)\vert\leq \vert x\vert + \vert \sin(-x) \vert \leq \vert x\vert +1 $$ Thus, $$\int_1^\infty f \leq \int_1^\infty \frac{ x +1}{x^3 } dx=\frac{3}{2}<\infty$$ And we are done. So, is this approach valid, and more importantly, is there a nicer way to bound $f$ and show its integrability without braking the integral into parts?
You got the good ideas : taylor expansion and check the integrability on infinity. But you don't need to use series. Define $f(x) = \frac{x-sin(x)} {x^3}$. f is continuous on $] 0, +\infty[$. In fact f is also continuous at $0$ with $ f(0) = \frac16$ (Taylor expansion). Consequently we just need to check the integrability at the infinity. Fortunately $|f(x)| \leq \frac{2}{x^2}$ which is integrable.