distributional derivative I want to calculate the first and second distributional derivate of the $2\pi$-periodic function $f(t) = \frac{\pi}{4} |t|$, it is
$$
 \langle f', \phi \rangle = - \langle f, \phi' \rangle = -\int_{-\infty}^{\infty} f \phi' \mathrm{d}x 
$$
and
$$
 \langle f'', \phi \rangle = \langle f, \phi'' \rangle = \int_{-\infty}^{\infty} f \phi'' \mathrm{d}x 
$$
so have to evaluate those integrals, but other than $\phi$ is smooth and has compact support i know almost nothing about $\phi$, so it is enough to integrate over finite bound, but how does i use the definition of $f$? So how could i calculate those integrals?

Let $G(t)=\max(t,0)$. The $G'=H$ where $H$ is the Heaviside function ($H(t)=1$ if $t>0$ and $0$ otherwise). Indeed, for any test function $\phi$ we have 
$$\langle G, \phi'\rangle = \int_0^{\infty} t\phi'(t)\,dt = - \int_0^{\infty} \phi(t)\,dt = - \langle H, \phi\rangle$$
in agreement with the definition of the distributional derivative. Also, $H'=\delta_0$ by a similar argument:
$$\langle H, \phi'\rangle = \int_1^{\infty} \phi'(t)\,dt = -\phi(0) = -\langle \delta_0, \phi\rangle$$
Any piecewise linear function can be written in terms of the translates of $G$. To do this for your function, you can begin with $f_0(t)=\frac{\pi}{4}G(t)-\frac{\pi}{2}G(t-\pi)+\frac{\pi}{4}G(t-2\pi)$ which agrees with $f$ on $[0,2\pi]$ and is zero elsewhere. Note that $f(t)=\sum_{n\in\mathbb Z}f_0(t+2\pi n )$, where the sum converges in a very strong sense: for any finite interval $I$, there exists $N$ such that $f(t)=\sum_{|n|\le N}f_0(t+2\pi n )$ for $t\in I$. This kind of convergence implies convergence of all distributional derivatives, since test functions have compact support. 
