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I have a integral which I have to prove is finite.

$$\int_{-\pi }^{\pi } \left(\frac{x \cos x-\sin x}{x^2}\right)^2 dx $$

call the function inside $g(x)$, where $g(x) = (f'(x))^2$ and where $f(x) = \frac{\sin x}{x}$ more explicitly the function $f(x)$ is piece-wise that is:

$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(x) = \begin{cases} \frac{\sin x}{x} \ \ \text{for } x \neq 0 \\ \ \ 1 \ \ \ \ \ \text{for} \ x = 0 \end{cases} $

(I am not 100% sure this is how piecewise function works but correct me if I am wrong)

This implies

$$\rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ g(x) = \begin{cases} \left(\frac{x \cos x - \sin x}{x^2}\right)^2 \ \ \text{for} \ x \neq0 \\ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{for} \ x = 0 \end{cases} $$

Again correct me if the leap of logic from $f(x) \rightarrow g(x)$ does not make sense given that $g(x) = (f'(x))^2$.

Given that the above was done correctly it can be shown (using L'Hopital's ) that the function is continuous everywhere including $x = 0$. My logic is that since this function is well behaved (never goes to infinity) is continuous everywhere on the interval from $[-\pi,\pi]$ and the integral is over a finite domain. Therefore the integral must be finite. I have never heard of a theorem explicitly stating these conditions but let me know if this is true.

(The graph of $g(x)$ below)enter image description here

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    $\begingroup$ Yes, $g$ can be extended to a continuous function over $[-\pi,\pi]$. Then see math.stackexchange.com/questions/90939/… $\endgroup$ – Robert Z Nov 14 '18 at 17:51
  • $\begingroup$ Does the piecewise quality extend to $g(x)$ as a consequence of $f(x)$ being a piecewise function as well? $\endgroup$ – QuantumPanda Nov 14 '18 at 17:54
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    $\begingroup$ Sorry, I mean $g$... Anyway if $f$ is continuous then $f^2$ is continuous too. $\endgroup$ – Robert Z Nov 14 '18 at 17:56
  • $\begingroup$ Once you know the function is continuous on a closed interval, you know it has an upper and lower bound. In this particular case, the lower bound is $0$ and the integral will not exceed the area of the bounding rectangle. $\endgroup$ – John Wayland Bales Nov 18 '18 at 1:03

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