This is from a derivation in Electrodynamics, however I don't follow the math.
I don't understand why equation $(2)$ and $(3)$ are equivalent.
If $\tau =0$ in $(2)$ we only have $\chi(t,t')=\chi(t,t')$ and not $\chi(t-t',0)$ ...?
$$ \int_{-\infty}^{t-\tau} \bigg(\chi(t-\tau,t')-\chi(t,t'+\tau)\bigg)E(t')\, dt'=0 \tag 1 $$ Here $E$ is an arbitrary function so $$ \chi(t-\tau,t')=\chi(t,t'+\tau) \tag 2 $$ for all $t$, $\tau$ and $t'$. Or equivalent $$ \chi(t,t')=\chi(t-t',0) \tag 3 $$ Therefore we can write $$ \int_{-\infty}^{t} \chi(t-t') E(t')\, dt'=0 \tag 4 $$
The equality $\chi(t-\tau,t')=\chi(t,t'+\tau)$ holds for every value of all three variables.
Set the first argument of $\chi$ to $t-t'$ and the second to $0$, $\chi(t-t',0)$, now remove in the first argument whatever is subtracting ($t'$) and add it to whatever is in the second, $\chi(t,0+t')$. They are equal.
Other way, make the change in the equality $\tau\to t'$, $t'=0$ and no change for $t$. We get $\chi(t-t',0)=\chi(t,t')$