Let's say you have the following double integral:
$$\int_{-\infty}^0 \int_{a}^b -f(u-x) dudx, $$ where $F’(u-x) = f(u-x)$, $a \lt b$ and $F(\infty) = 0$.
Changing the order of integration, I would assume this solves to $\int_a^b-F(u) du$, but it seems this solves to $\int_a^b F(u) du$.
What am I missing here?
If $F'=f$, then $$ \frac{\partial}{\partial x} \Bigl( F(u-x) \Bigr) = F'(u-x) \cdot \frac{\partial}{\partial x} (u-x) = f(u-x) \cdot (-1) $$ by the chain rule. So the inner integral, after changing the order of integration, is $$ \int_{-\infty}^0 (-f(u-x)) dx = \Bigl[ F(u-x) \Bigr]_{x=-\infty}^0 = F(u) - \lim_{t \to \infty} F(t) = F(u) - 0 = F(u) . $$