$u(x)$ is the step function and $T_u(x)$ be the distribution defined by
$$\forall \varphi \in D, \langle T_u , \varphi \rangle = \lim_{\epsilon \to 0} \left (\varphi(0) \ln(\epsilon) + \int^{+ \infty } _ {\epsilon} \frac{\varphi(x)}{x} dx \right )$$
In the sense of distributions, Prove $\forall \varphi \in D$
$$ \langle (u(x) Ln(x))', \varphi \rangle = \lim_{\epsilon \to 0} \left ( \varphi(\epsilon)\ln (\epsilon ) + \int^{+ \infty}_{\epsilon} \frac{\varphi(x)}{x} dx \right ) $$
Try 1
$$\begin{aligned} \langle (u(t) Ln(t) ) ' , \varphi\rangle &= \langle u'(t) ln(x)+ ln'(x) u(t) , \varphi \rangle \\&= \langle u'(t) ln(x) , \varphi \rangle +\langle ln'(x) u(t) , \varphi \rangle \\&= \langle \delta(t) ln(x) , \varphi\rangle +\langle u(t)/x , \varphi \rangle \\&= ln(0)\varphi(0) + \int^{+\infty}_{0} \varphi(x) /x dx \end{aligned} $$
and I just want to say that
$$ln(0)\varphi(0) + \int^{+\infty}_{0} \varphi(x) /x dx= \lim_{\epsilon \to 0} \left ( \varphi(\epsilon)\ln (\epsilon ) + \int^{+ \infty}_{\epsilon} \frac{\varphi(x)}{x} dx \right ) $$
Anything wrong,thanks!
Notice that differentiating distributions in this manner generates things like $\delta(x) \ln x$, and how do you define the action of that object on an arbitrary test function? Instead, going strictly by definition, $$((u(x) \ln x)', \phi) = - (u(x) \ln x, \phi') = -\int_0^\infty \phi'(x) \ln x dx = \lim_{\epsilon \to 0} \Big( -\phi(x) \ln x \Big\rvert_{x = \epsilon}^\infty + \underbrace {\int_\epsilon^\infty \frac {\phi(x)} x dx}_{= I_\epsilon} \Big) = \lim_{\epsilon \to 0} ( \phi(\epsilon) \ln \epsilon + I_\epsilon ) = \\ \lim_{\epsilon \to 0} ( (\phi(0) + O(\epsilon)) \ln \epsilon + I_\epsilon ) = \lim_{\epsilon \to 0} ( \phi(0) \ln \epsilon + I_\epsilon ).$$