I'm trying to work through a couple improper integral problems to study for a final, but am somewhat stuck on showing that $\displaystyle \int^1_0 \frac{\sin x -x}{x^3}$ converges.
Edit: Can you use the weighted mean value theorem for integrals? If I can show that $\sin x -x \leq 1$ on $[0,1]$, then it seems like by that theorem, $\displaystyle \int^1_0 \frac{\sin x -x}{x^3} = \frac{1}{c^3} \int^1_0 \sin x -x$ for some $c \in [0,1]$. This should converge, if I can guarantee that $c \neq 0$.
On
$\sin(x)$ is an odd entire function. Since: $$ \sin(x) = \sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)!} \tag{1}$$ we have: $$ \frac{\sin(x)-x}{x^3} = \sum_{n\geq 1}\frac{(-1)^n x^{2n-2}}{(2n+1)!}\tag{2} $$ and: $$ \int_{0}^{1}\frac{\sin(x)-x}{x^3}\,dx=\sum_{n\geq 1}\frac{(-1)^n}{(2n-1)\cdot(2n+1)!}.\tag{3} $$ That fast convergent series can be written as $\frac{2-\sin(1)-\cos(1)-\text{Si}(1)}{2}$ by using the sine integral function.
We may also use integration by parts, leading to: $$ \int_{0}^{1}\frac{\sin(x)-x}{x^3}\,dx = \frac{2-\sin(1)-\cos(1)}{2}-\frac{1}{2}\int_{0}^{1}\frac{\sin(x)}{x}\,dx \tag{4}$$ and due to the concavity of the sine function over $[0,1]$, the very last integral is between $\sin(1)$ and $1$.
Observe that by l'Hopital's $$ \lim_{x\rightarrow 0^+}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow0^+}\frac{\cos(x)-1}{3x^2} =\lim_{x\rightarrow0^+}\frac{-\sin(x)}{6x}=-\frac{1}{6}. $$ Therefore, the function is continuous bounded on $(0,1]$, and, therefore integrable.