# integral with dirac delta of a periodic argument

Is my solution of the following integral correct: $$I(\varphi,\varphi_0)=\int_{-\infty}^\infty x e^{-jx\cos(\varphi-\varphi_0)}\mathrm dx \qquad ?$$ I know that: $$\displaystyle \int_{-\infty}^{\infty} x^n e^{-i\nu x}\, \mathrm dx=\displaystyle 2\pi i^n\delta^{(n)} (\nu)\,$$ That means that in my case (lets call $\psi=\varphi-\varphi_0$): $$I(\psi)=\displaystyle 2\pi i\delta^{'} (\cos(\psi))$$ Using functional transformation of Dirac delta-function I get $$\delta^{'} (g(x))=\left[ \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}\right]^{'}=\sum_i \frac{\delta^{'}(x-x_i)}{|g'(x_i)|}-\sum_i\frac{\delta(x-x_i)|g'(x_i)|'}{|g'(x_i)|^2}$$ where $x_i$ are the roots of $g(x)$. I have $g=\cos(\psi)$ and $\psi_i=\frac{\pi}{2}+i\pi, \ i\in\mathbb{Z}$.
Since $|g''(x_i)|=|g(x_i)|=0$ the second term goes to zero.
Then $$\delta^{'} (\cos(\psi))= \sum_i \frac{\delta^{'}(\psi-\frac{\pi}{2}-i\pi)}{|\sin(\frac{\pi}{2}+i\pi)|}=\sum_i \delta^{'}\left(\psi-\frac{\pi}{2}-i\pi\right)$$ But how to deal with the derivative of the Dirac delta? Can I move further? Or it is dead end?
For instane if after that I'd like to compute someting like:

$$\int_{0}^{\pi}\delta^{'} (\cos(\psi))f(\psi)\mathrm d\psi$$ Will it be correct to do it like: $$\begin{eqnarray} \int_{0}^{\pi}\delta^{'} (\cos(\psi))f(\psi)\mathrm d\psi&=&\int_{0}^{\pi}\sum_i \delta^{'}\left(\psi-\frac{\pi}{2}-i\pi\right)f(\psi)\mathrm d\psi=\\&=&-\sum_i\int_{0}^{\pi}\delta\left(\psi-\frac{\pi}{2}-i\pi\right)f^{'}(\psi)\mathrm d\psi=\\&=&-f^{'}(\pi/2) \end{eqnarray}$$