I’ve been studying Levinson’s elementary proof of the Prime Number Theorem lately, but I have encountered some writings that make me confused.
First, define $S(y)=\int_{2}^y \frac{\psi(x)-x}{x}dx$ when $x\geq2$, and $S(y)=0$ when $x<2$, where $\psi(x)$ is Chebyshev's Psi Function. It is known that $S(y)$ is $O(y)$ and Lipchitz. Then, further define $W(x) = e^{-x}|S(e^{x})|$.
Now, there is an inequality
$|W(x)|\leq\frac{2}{x^{2}}\int_{0}^{x}(x-v)|W(v)|dv + \frac{K_{4}}{x}$,
where $K_{4}$ is a constant. The author then rewrites the inequality as
$|W(x)|\leq \frac{2}{x^{2}}\int_{0}^{x}u\ du(\frac{1}{u}\int_{0}^{u}|W(v)|\ dv) = \frac{2}{x^{2}} \cdot \frac{x^{2}}{2}\cdot\frac{1}{u}\int_{0}^{u}|W(v)|\ dv$
and he mentioned that it "can be verified by inverting the order of integration".
Here's my question: I completely understand how inverting the order of integration works, but the author splits the two integrals after inverting the order. Why is this valid since he is not integrating over a square?
We can look at the integral as being of the form $\int_0^x \int_0^u f(u, v)\ dv\ du$. Then the region of integration is $\{(u, v) : 0 \leq u \leq x, 0 \leq v \leq u\}$ which is a triangle bounded by the $u$ and $v$ axes and the line $u + v = x$. So we can equally define the region as $\{(u, v): 0 \leq v \leq x, 0 \leq u \leq v\}$ and that allows us to swap the order of integration.