Say you have the distribution function $\rho_i$ of a quark $i$ (depending on the momentum fraction of that quark), and a structure function $\hat{F}_i$ (depending on the "Bjorken variable" $x$ and the energy scale $q^2$) of such quark. i.e. a term describing its contribution to the hadronic tensor which describes some process, such as deep inelastic scattering of an electron off the hadron. Many sources, such as this Scholarpedia article by Guido Altarelli, say that in order to calculate the structure function of the whole hadron you have to combine the previous two function as the convolution \begin{equation} F(x,q^2)=\sum_i\int_x^1\rho_i(\xi)F_i\Bigl(\frac{x}{\xi},q^2\Bigr)\frac{1}{\xi}\,\mathrm{d}\xi. \end{equation} I can't find why it should have this form.
I tried to prove it as follows. If the quark/parton has a moment equal to $\xi p$ where $p$ is the momentum of the hadron, then its $x$ variable, defined as \begin{equation} x=-\frac{q^2}{2\xi p\cdot q}, \end{equation} is $1/\xi$ times the hadron's $x$, which has just $p$ instead of $\xi p$. (The $q$ vector, in deep inelasting scattering problems, is the momentum of the internal particle exchanged between the electron and the hadron. I don't know if there is a general definition of such $x$ outside of DIS problems.)
Therefore, if we have assigned a value of $x$ to the hadron, a parton with momentum fraction $\xi$ will be described by a structure function of the form $\hat{F}(x/\xi,q^2)$. In order to obtain the structure function of the hadron, I weigh this with the parton density distribution, and sum over all types of quarks, obtaining \begin{equation} F(x,q^2)=\sum_i\int_0^1\rho_i(\xi)\hat{F}_i\Bigl(\frac{x}{\xi},q^2\Bigr)\,\mathrm{d}\xi. \end{equation} My proof doesn't explain the factor $1/\xi$ in the integral, or the fact that the lower limit is $x$ and not $0$. Why is that so?