I want to find the area of the space between $\frac{\sin{x}}{x}$ and its derivative, $\frac{x\cos(x)-\sin(x)}{x^2}$, that includes the origin. More specifically, between the sinc function and its derivative, so that the function is defined at $x = 0$ as $1$.
The difficulty seemed to arise in finding the intersection points of the two functions. Even if the area or the intersection points can't be defined in terms of algebraic numbers, can they be defined as some output of the sinc function or another trigonometric function?
If you need the interections, you need to solve for $x$ the equation $$(x+1) \sin (x)-x \cos (x)=0$$ (the trivial $x=0$ will be excluded from the discussion.
By inspection, you can notice that the $n^{\text{th}}$ root is located very close to $\left(\frac \pi 4+n\pi\right)$ (if $x$ is large, you almost need to solve $\sin(x)-\cos(x)=0$).
Using series expansion around the point and then power series reversion, an approximation write, with $\large\color{red}{t=(4n+1)\pi+2}$, $$\large\color{blue}{x_{(n)}=\frac{t}{4}-\frac{1}{2}-\frac{2}{t}-\frac{40}{3 t^3}-\frac{2656}{15 t^5}-\frac{2832}{t^7}-\frac{36288}{t^9}+O\left(\frac{1}{ t^{11}}\right)}$$ Just for illustration, for the first roots, the results are $$\left( \begin{array}{ccc} n & \text{approximation} & \text{solution} \\ 1 & \color{red}{3.811538}96245987 & 3.81153864777939 \\ 2 & \color{red}{7.002033}30100463 & 7.00203329571326 \\ 3 & \color{red}{10.163320735}5184 & 10.1633207350938 \\ 4 &\color{red}{13.315593576}9062 & 13.3155935768387 \\ 5 &\color{red}{16.4638956000}733 & 16.4638956000575 \\ 6 & \color{red}{19.61009607296}78 & 19.6100960729630 \\ 7 & \color{red}{22.75504934382}26 & 22.7550493438209 \\ 8 & \color{red}{25.89920168969}33 & 25.8992016896926 \\ 9 & \color{red}{29.04280906906}11 & 29.0428090690607 \\ 10 &\color{red}{32.186028836711}9 & 32.1860288367118 \\ \end{array} \right)$$
Edit
Writing $x_{(n)}= \left(n+\frac{1}{4}\right)\pi-\epsilon_n$, the equation reduces to $$\frac{2 \cos (\epsilon_n )-( (4 n+1)\pi+2-4 \epsilon_n ) \sin (\epsilon_n )}{2 \sqrt{2}}$$ Using the value of $\epsilon_n$ as given in the "blue" formula, this is smaller than $\frac{671}{53760 \sqrt{2} \pi ^6 n^6}$ which is smaller than $\frac {10^{-5} } {n^6}$ (this is not $0$ but quite close to it).
Similarly $$x_{(n+1)}- x_{(n)} <\pi +\frac{1}{2 \pi n^2}$$