I am trying to evaluate the following definite double integral:

$$\int_0^\infty \int_0^\infty \frac{x^2 + y^2}{1 + (x^2 - y^2)^2} e^{-2xy} \:\mathrm{d}x \:\mathrm{d}y$$

I have tried the following substitution:

$$u(x, y) = x^2 - y^2 \\ v(x, y) = 2xy$$

This gives us the Jacobian:

$$\frac{\partial(x, y)}{\partial(u, v)} = \left(\frac{\partial(u, v)}{\partial(x, y)}\right)^{-1} = \begin{vmatrix}\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix}^{-1}$$

Evaluating this we get:

$$\mathrm{d}x\:\mathrm{d}y = \frac{\mathrm{d}u\:\mathrm{d}v}{4x^2 + 4y^2}$$

This allows us to transform the integral:

$$\int_0^\infty \int_0^\infty \frac{x^2 + y^2}{1 + (x^2 - y^2)^2}e^{-2xy}\:\mathrm{d}x\:\mathrm{d}y = \iint_\phi \frac{1}{4(1+ u^2)}e^{-v}\:\mathrm{d}u\:\mathrm{d}v$$

However I am unable to determine $\phi$ in terms of $u$, and $v$ and therefore cannot finish evaluating. I believe that $u$ and $v$ are related to parabolic coordinates, but I haven't been able to find anything useful so I'd appreciate any pointers from people.

  • 3
    $\begingroup$ why don't you use polar coordinates? Also $dxdy=rdrd\theta$. $\endgroup$
    – Fermat
    May 3, 2014 at 13:13
  • $\begingroup$ Writing $z = x + iy$ and $w = u + iv$, then your transform satisfies $w = z^{2}$. So the first quadrant is mapped to the upper half-plane $\Bbb{H} = \{(u, v) : v > 0 \}$. Of course, polar coordinates also work. $\endgroup$ May 3, 2014 at 13:21
  • $\begingroup$ I thought about $x=ve^u$, $y=ve^{-u}$. Haven't tried it but it might work. $\endgroup$
    – Kal S.
    May 3, 2014 at 13:39

1 Answer 1


Extracting $(x,y)$ from $(u,v)$ yields $$ x^2=\tfrac12(u+\sqrt{u^2+v^2}),\qquad y^2=\tfrac12(-u+\sqrt{u^2+v^2}), $$ hence every $(u,v)$ such that $v\gt0$ corresponds to some unique $(x,y)$ with $x\gt0$, $y\gt0$. Thus, the $(u,v)$-domain of integration is $u\in\mathbb R$, $v\gt0$.


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