Show the integral $\int_{0}^{\infty} \frac{e^{-y} - e^{-ry}}{y} dy = ln(r)$ by applying Tonelli's Theorem on the set $[0,\infty) \times [1,r]$.
Well Im trying to apply Tonelli's theorem to prove this claim. However I have no idea how to set it up. I was trying to set it up like this: Fix an $t \in [1,r]$. Then I am defining $$g(x,y) = \begin{array}{cc} \{ & \begin{array}{cc} e^{-y} & (x,y) \in [t,r] \times [0,1] \\ \ e^{-xy} & (x,y) \in [t,r] \times [1,\infty) \end{array} \end{array}.$$ This is what I got so far in coming up with the iterated integral. Is this the way to go here. I need to use Tonelli's Theorem. Thank you for all of your help!
Supposing that $r \geq 1$ $$\int_0^{+\infty}\frac{e^{-y}-e^{-ry}}{y}dy=\int_0^{+\infty}(\int_1^re^{-xy}dx)dy=\int_{1}^r(\int_{0}^{+\infty}e^{-xy}dy)dx=\int_{1}^r\frac{1}{x}dx=\ln(r)$$
Supposing that $0<r<1,$
$\int_0^{+\infty}\frac{e^{-y}-e^{-ry}}{y}dy=-\int_0^{+\infty}(\int_r^1e^{-xy}dx)dy=-\int_{r}^1(\int_{0}^{+\infty}e^{-xy}dy)dx=-\int_{r}^1\frac{1}{x}dx=\ln(r)$
Which means $$\int_{0}^{+\infty}\frac{e^{-y}-e^{ry}}{y}dy=\ln(r).$$