Integral involving the derivative of the Dirac's Delta "function"

Question

I'm dealing with the integral \begin{equation} \int\limits_{-1}^{+1}\delta'(\cos(\pi x))(x^2+1)\ \mathrm{d}x \label{1}\tag{1} \end{equation} This makes me in me lot of confusion about the derivative of the Dirac's Delta, as I only know how to deal with the derivative of the Dirac Delta Distribution, that is the usual derivative of a distribution $$ \delta_0^{(n)}(f) = (-1)^n\delta_0(f') $$ Also, I've found some exercises where it is said that it is true the derivation rule for function composition as \begin{equation} \frac{\mathrm{d}}{\mathrm{d}x}\delta(f(x)) = \delta'(f(x))f'(x) \label{2}\tag{2} \end{equation} so that i can make a swap in \eqref{1} between the derivative of the distribution and the delta of the argument function of the Delta: \begin{gather} \delta'_{\frac{2k+1}{2}}(x^2+1) \equiv \langle\delta'_{\frac{2k+1}{2}}, x^2+1 \rangle = \int\limits_{-1}^{+1}\delta'(\cos(\pi x))(-\pi\sin(\pi x))\frac{(x^2+1)}{(-\pi\sin(\pi x))}\ \mathrm{d}x = \\ \int\limits_{-1}^{+1}\frac{\mathrm{d}\delta(\cos(\pi x))}{\mathrm{d}x}\frac{(x^2+1)}{(-\pi\sin(\pi x))}\ \mathrm{d}x = -\langle\delta_{\frac{2k+1}{2}}, \left(\frac{x^2+1}{-\pi\sin(\pi x)}\right)' \rangle =\\ \frac{1}{\pi}\int\limits_{-1}^{+1}\delta(\cos(\pi x))\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^2+1}{\sin(\pi x)}\right)\ \mathrm{d}x \end{gather} from the last relation we can simplify the delta by finding the zeros of the function inside the delta \begin{align} \delta(\cos(\pi x)) = \frac{1}{\pi}\sum_{k\in\mathbb{Z}}\delta_{\frac{2k+1}{2}} \end{align} Inside the interval of integration there are just two zeros, for k = 0, -1, so \begin{gather} \frac{1}{\pi}\int\limits_{-1}^{+1}\frac{1}{\pi}\left(\delta\left(x - \frac{1}{2}\right) + \delta\left(x + \frac{1}{2}\right)\right)\frac{2x\sin(\pi x) - (x^2+1)\pi\cos(\pi x)}{\sin^2(\pi x)}\ \mathrm{d}x\\ = \frac{2}{\pi^2} \end{gather} result which is not right, as I checked on Mathematica and with the result of the exercise. Summing up all my doubts:

1. How do I solve cases like this with the derivative of the delta function?
2. Can I apply \eqref{2} all the times I deal with the function? Under which circumstance can I use it?
3. Does anybody know some good literature very rigorous and helpful to solve this kind of doubt, maybe with solved exercises? Thanks.

Answer 1

You look for all points in the domain of integration where $\cos(\pi x)=0$. You evaluate the negative of the derivative of the test function at those points, which in this case are $1/2$ and $-1/2$. Then you divide by the absolute value of the derivative of $\cos(\pi x)$ at those points and sum. So overall you have

$$\frac{1}{\pi} \left ( -\left. \frac{d}{dx} \frac{1}{x^2+1} \right |_{x=1/2} - \left. \frac{d}{dx} \frac{1}{x^2+1} \right |_{x=-1/2} \right ).$$

The main idea of this construction is to locally change variables around the zeros of $\cos(\pi x)$. That is you split your domain of integration into $[0,1]$ and $[-1,0]$ and take $u=\cos(\pi x)$ on each of those. Then you use the definition of the Dirac delta and the distributional derivative, which tell you that $\langle \delta',f \rangle = -f'(0)$.