How Changing the order of integration(Elementary proof of the prime number theorem)?

Question

I'm studying the exchange of integration order, I need help, any hint? For every real number $\rho \geq 0$, write $V(\rho)=e^{-\rho}R(e^{\rho})=e^{-\rho}\psi(e^{\rho})-1$ where $\psi(x)$ is the chebyshev's function, write $x=e^{\xi}$, with the substitution $x/t=e^{\eta}$, we have $$\displaystyle{\int_{1}^{x}\left|R\left(\frac{x}{t}\right)\right|\log t\ dt=x\int_{0}^{\xi}|V(\eta)|(\xi-\eta)d\eta=x\int_0^{\xi}\int_{\eta}^{\xi}|V(\eta)|d\tau\ d\eta}$$ and the last integal is equal to $\displaystyle{x\int_0^{\xi}\left(\int_0^{\tau}|V(n)|d\eta\right)d\tau}$ This last integral , I do not know how to get to get it , help me .

Answer 1

The integrand is nonnegative, so by Tonelli's theorem the value of the double integral is insensitive to the order (i.e. switching the order of integration is justified).

Just consider the region of integration as a two-dimensional set: the set of points $(\tau, \eta)$ that comprises the domain of integration for $\int_0^\xi \int_\eta^\xi d\tau d\eta$ is

$$\{ (\tau,\eta) : 0 \le \eta \le \xi, \eta \le \tau \le \xi \} = \{ (\tau,\eta) : 0 \le \eta \le \tau \le \xi \} = \{ (\tau,\eta) : 0 \le \tau \le \xi, 0 \le \eta \le \tau \},$$

Which corresponds exactly to the double integral $\int_0^\xi \int_0^\tau d\eta d\tau$. It boils down to just showing that the conditions $0 \le \eta \le \xi, \eta \le \tau \le \xi$ are logically equivalent to $ 0 \le \tau \le \xi, 0 \le \eta \le \tau $ (and it's easier to see this when you note that either one is equivalent to $0 \le \eta \le \tau \le \xi$).