i want to compute the product of convolution $1 * (\delta' * H)$ where $\delta$ is distribution of Dirac and $H$ is function of Heaviside.
first, we compute $\delta' * H.$ We have by definition that $\delta' * H (x) = \displaystyle\int_{-\infty}^x \delta' (y) dy$ how we can finish?
Thank's
In fact, this can easily be done if you already know some properties of distributions.
The distribution $\delta$ is the identity element of the convolution and it's easy to see that:
$\hspace{3cm}\delta^{(k)}*T=(\delta*T)^{(k)}=T^{(k)}$ for all $k\in\Bbb N$ and $T\in\cal{D}'(\Bbb R)$.
Now using the fact that $H'=\delta$, you get : $$1*(\delta'*H)=1*H'=1*\delta=1$$
If you don't know all this, you take $\varphi\in\cal{D}(\Bbb R)$, and start like you did, except you made an error :
$$\langle \delta'*H,\varphi\rangle=\langle\delta',H*\tilde{\varphi}\rangle$$
where $\tilde{\varphi}:x\mapsto\varphi(-x)$.
For the part where you were stuck, don't forget that $\langle T',\varphi\rangle=-\langle T,\varphi'\rangle$.