0
$\begingroup$

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 $c \neq 0$.

$\endgroup$
4
  • 1
    $\begingroup$ Are you allowed to use Taylor series? Expanding Sin and cancelling the first term would be very easy. $\endgroup$ – Jeremy Upsal Dec 15 '15 at 18:47
  • $\begingroup$ Take the absolute value, $|\sin(x)-x| \le |1-x|$ then decompose the fraction and use the linearity. $\endgroup$ – ParaH2 Dec 15 '15 at 18:47
  • 1
    $\begingroup$ 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. $\endgroup$ – Ravi Fernando Dec 15 '15 at 18:50
  • $\begingroup$ Thanks! Yeah, Taylor series is fine. $\endgroup$ – Jay Dec 15 '15 at 18:51
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$\sin(x)$ 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 $1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.