I am trying to get to grips with Altarelli-Parisi-type equations. In chapter 17 of Peskin/Schroeder, they first develop the equations for a similar problem in QED. Equation $(17.123)$ introduces the sum rule $$ \int_0^1 dx ( f_e(x,Q) - f_\bar{e}(x,Q) ) = 1 $$ where $f_e$ and $f_\bar{e}$ are the distribution functions of electrons and antielectrons 'inside' an electron. I'm trying to prove that this is independent of $Q$.
The evolution equations are ($(17.120)$ in Peskin/Schroeder) $$ \frac{d}{d\log Q} f_e(x,Q)= \frac{\alpha}{\pi}\int_x^1 \frac{dz}{z} \left( P_{e\leftarrow e}(z) f_e(\frac{x}{z},Q) + P_{e\leftarrow\gamma}(z)f_\gamma(\frac{x}{z},Q)\right) $$ $$ \frac{d}{d\log Q} f_\bar{e}(x,Q)= \frac{\alpha}{\pi}\int_x^1 \frac{dz}{z} \left( P_{e\leftarrow e}(z) f_\bar{e}(\frac{x}{z},Q) + P_{e\leftarrow\gamma}(z)f_\gamma(\frac{x}{z},Q)\right) $$ where the relevant splitting functions is given by (equation $(17.121)$ in Peskin/Schroeder) $$ P_{e\leftarrow e}(z) = \frac{1+z^2}{(1-z)_+}+\frac{3}{2}\delta(1-z) $$ Using $\frac{d}{d\log Q}$ on $(17.123)$ gives (the part involving $P_{e\leftarrow\gamma}(z)$ cancels): $$ \frac{\alpha}{\pi}\int_0^1 dx \int_x^1 \frac{dz}{z} P_{e\leftarrow e}(z) \left( f_{e}(\frac{x}{z},Q) - f_\bar{e}(\frac{x}{z},Q) \right) $$ Insterting $(17.121)$ and using that $$ \int_0^1 dx \frac{f(x)}{(1-x)_+} = \int_0^1 dx \frac{f(x)-f(1)}{(1-x)} $$ I get $$ \frac{\alpha}{\pi}\int_0^1 dx \int_x^1 \frac{dz}{z} \left(\frac{1+z^2}{(1-z)_+}+\frac{3}{2}\delta(1-z) \right) \left( f_{e}(\frac{x}{z},Q) - f_\bar{e}(\frac{x}{z},Q) \right) $$ $$ = \frac{\alpha}{\pi}\int_0^1 dx \left( \int_x^1 dz \left( \frac{1+z^2}{(z-z^2)}\Delta(\frac{x}{z},Q) -\frac{2}{1-z} \Delta(x,Q) \right) + \frac{3}{2} \Delta(x,Q) \right) $$ This expression is singular and it seems that the singularities in the first two terms should cancle. However, I'm at a loss what to do here. My idea was to extract the singularity in the first term, but that seems like i'm doing it backwards (and I haven't figured out how to do it). Any hint would be appreciated, I'm not looking for full solutions.
