I asked a previous question on using a Heaviside step function to change the order of integration. Now I need some help in understanding how the limits are determined, for example in this double integral:
\begin{eqnarray*} \int_{0}^{1}\int_{1}^{e^y}f\left(x,y\right)\,{\rm d}x\,{\rm d}y & = & \int_{0}^{1}\left\lbrack\int_{1}^{?}\Theta\left(e^y - x \right) {\rm f}\left(x, y\right)\,{\rm d}x\right\rbrack{\rm d}y \end{eqnarray*}
And what would the upper limit of the inner integral be? Is it $e$, since $0 \leq y \leq 1$, so that $\Theta(e^y - x) = 0$ from the interval $e$ to $\infty$?
My understanding: \begin{eqnarray*} \int_{0}^{1}\int_{1}^{e^y}f\left(x,y\right)\,{\rm d}x\,{\rm d}y & = & \int_{0}^{1}\left\lbrack\int_{1}^{e}\Theta\left(e^y - x \right) {\rm f}\left(x, y\right)\,{\rm d}x\right\rbrack{\rm d}y \end{eqnarray*}
The bounds of the inner integral in the first step is $\{1 \leq x \leq e\}$, such that $e^y - x \geq 0$ on the interval $e^0$ to $e^1$. The interval is determined by the bounds of $y$.
\begin{eqnarray*} \int_{0}^{1}\left\lbrack\int_{1}^{e}\Theta\left(e^y - x \right) {\rm f}\left(x, y\right)\,{\rm d}x\right\rbrack{\rm d}y & = & \int_{1}^{e}\left\lbrack\int_{}^{}\Theta\left(e^y - x \right) {\rm f}\left(x, y\right)\,{\rm d}y\right\rbrack{\rm d}x \end{eqnarray*}
Fubini's Theorem.
\begin{eqnarray*} \int_{1}^{e}\left\lbrack\int_{}^{}\Theta\left(e^y - x \right) {\rm f}\left(x, y\right)\,{\rm d}y\right\rbrack{\rm d}x & = & \int_{1}^{e}\left\lbrack\int_{\log{x}}^{1}\Theta\left(e^y - x \right) {\rm f}\left(x, y\right)\,{\rm d}y\right\rbrack{\rm d}x \end{eqnarray*}
The bounds for the inner integral is $\{\log{x} \leq y \leq 1\}$. Since $y$ is being varied, $y \geq \log{x}$, $0 \leq x \leq 1$. The upper bound for $y$ is 1 and the lower bound is $\log{x}$.
\begin{eqnarray*} \int_{1}^{e}\left\lbrack\int_{\log{x}}^{1}\Theta\left(e^y - x \right) {\rm f}\left(x, y\right)\,{\rm d}y\right\rbrack{\rm d}x & = & \int_{1}^{e}\int_{\log{x}}^{1}{\rm f}\left(x, y\right)\,{\rm d}y{\rm d}x \end{eqnarray*}
Remove the Heaviside function.
