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} $
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)
Best Answer
It's easy to show that there are a finite number of points that are unbounded. In fact, the only point that is undefined is $x=0$. Note that $g(x)\in[0,1]$ for all $x\in[-\pi,\pi]\backslash 0$. (There are, of course, better bounds, but these are fine.)
Since the function is bounded on a compact set except for a finite set of points, the integral is bounded.