Taylor expansion and Distribution Let $u(x)$ be the step function and $p_u(x)$ be the distribution defined by 
$$\forall \varphi \in D, 
\langle p_u , \varphi  \rangle
= \lim_{\epsilon \to  0} \left (\varphi(0) \ln(\epsilon)  + \int^{+ \infty } _ {\epsilon} \frac{\varphi(x)}{x} dx  \right )$$ 
Using a Taylor expansion of $\varphi(\epsilon)$, Show that  (in the sense of distributions) $$(u(x) ln(x)))'=P_u $$  

From previous post 
part(a) (asked in another post) Show that, In the sense of distributions, we have $\forall \varphi \in D$ 
$$ \langle (u(x) Ln(x))', \varphi \rangle
= \lim_{\epsilon \to 0} \left  (  \varphi(\epsilon)\ln (\epsilon ) + \int^{+ \infty}_{\epsilon} \frac{\varphi(x)}{x} dx  \right )  $$

Attempt
Taylor Expansion  of $\varphi(\epsilon)$
$$\varphi(\epsilon) = \sum^{n}_{k=0} \frac{\varphi^{(n)}(\epsilon)}{n!} (\epsilon- \epsilon_0)^n  $$
Diffirentiating $u(x) ln(x)$ in parts
$$ (u(x) ln(x))'= u'(x)ln(x)+u(x) \frac{1}{x}= \delta(x)ln(x)+ u(x)/x$$
I am guessing that $P_u =\langle u,\varphi  \rangle$
$$\begin{aligned}
<u ,\varphi> = \int^{+\infty}_{0} \varphi(x)dx
             = \int^{\infty}_{0}\sum^{n}_{k=0} \frac{\varphi^{n}(\epsilon)}{n!} (\epsilon- \epsilon_0)^n  d\epsilon
\end{aligned}$$

Kind of lost at this point can't thread the needle appreciate a nudge towards the right direction
 A: Since you already have the "part (a)", this is much simpler than what you are trying to do. In short, we want to "replace" in the limit the $\varphi(\epsilon)$ by $\varphi(0)$. To do this, we only need to prove that $\lim_{\epsilon\to 0}(\varphi(0)\ln(\epsilon) - \varphi(\epsilon)\ln(\epsilon)) = 0$.
In more details, we have $\varphi(\epsilon) = \varphi(0) + \mathcal O(\epsilon)$, so $\varphi(0)\ln(\epsilon) - \varphi(\epsilon)\ln(\epsilon) = \mathcal O(\epsilon)\ln(\epsilon) = o(1)$, which proves the limit claimed just before. Then, according to part (a),
\begin{align*}
\langle(u\ln)',\varphi\rangle
&=\lim_{\epsilon\to 0}\left( \varphi(\epsilon)\ln(\epsilon) +\int_\epsilon^{+\infty} \frac{\varphi(x)}x \,\mathrm d x\right)\\
&=\lim_{\epsilon\to 0}\left( \varphi(0)\ln(\epsilon) +\int_\epsilon^{+\infty} \frac{\varphi(x)}x \,\mathrm d x + (\varphi(\epsilon)\ln(\epsilon)-\varphi(0)\ln(\epsilon))\right)
\end{align*}
and according to the previous claim, this is equal to
\begin{align*}
\langle(u\ln)',\varphi\rangle
&=\lim_{\epsilon\to 0}\left( \varphi(0)\ln(\epsilon) +\int_\epsilon^{+\infty} \frac{\varphi(x)}x \,\mathrm d x\right)+0
\end{align*}
which is by definition $\langle p_u,\varphi\rangle$.
