Integrate $e^{xy}$ over a hyperbolic "rectangle"

Question

While reviewing multivariate calculus with a student, I encountered this exercise:

Evaluate the integral $\displaystyle\iint_We^{xy}\,\mathrm dx\,\mathrm dy$, where $W$ is the region bounded by $1\le xy\le4$ and $1\le\dfrac xy\le2$.

I made the change of coordinates

$$\begin{cases}u(x,y)=xy\\[1ex]v(x,y)=\dfrac xy\end{cases}$$

for which the Jacobian and its determinant are

$$\mathbf J=\begin{bmatrix}y&x\\[1ex]\dfrac1y&-\dfrac x{y^2}\end{bmatrix},\quad\det\mathbf J=-\dfrac{2x}y=-2v$$

So the integral is

$$\iint_We^{xy}\,\mathrm dx\,\mathrm dy=2\int_{v=1}^{v=2}\int_{u=1}^{u=4}ve^u\,\mathrm du\,\mathrm dv=3e^4-3e$$

I check my answer with Mathematica to find out the answer instead appears to be

Integrate[
    E^(x y) Boole[1 <= x y <= 4 && 1 <= x/y <= 2],
    {x, 0, Infinity}, {y, 0, Infinity}
]

which returns a different answer of $\dfrac{(e^4-e)\log2}2$.

Is there a mistake in my work? Or is Mathematica correct? Or is the integral I'm computing with Integrate not equivalent to the one I'm supposed to compute?

Answer 1

Yes indeed, as you noted, we have

$$du\,dv=|\det\mathbf J|dx\,dy=-\dfrac{2x}y=-2v\implies dx\,dy =-\frac1{2v}du\,dv$$

and it should be

$$\color{red}{\frac12}\int_{v=1}^{v=2}\int_{u=1}^{u=4}\frac{e^u}v\,\mathrm du\,\mathrm dv$$

Answer 2

I've realized I'm using the incorrect Jacobian. Using $\dfrac1{\det\mathbf J}$ instead, I get the correct answer:

$$\frac12\int_{v=1}^{v=2}\int_{u=1}^{u=4}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\log2}2$$