# Show $\int^1_0 \frac{\sin x -x}{x^3}$ converges.

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 $c \neq 0$.

• Are you allowed to use Taylor series? Expanding Sin and cancelling the first term would be very easy. – Jeremy Upsal Dec 15 '15 at 18:47
• Take the absolute value, $|\sin(x)-x| \le |1-x|$ then decompose the fraction and use the linearity. – Hexacoordinate-C Dec 15 '15 at 18:47
• If you know that the limit of $(\sin x - x)/x^3$ as $x \to 0$ exists and is finite, then you're just integrating a continuous function over a finite integral, so there's nothing "improper" to worry about. But you can find the limit, e.g. using Taylor series or l'Hospital's rule. – Ravi Fernando Dec 15 '15 at 18:50
• Thanks! Yeah, Taylor series is fine. – Jay Dec 15 '15 at 18:51

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.
$\sin(x)$ 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 $1$.