1

I would like to solve the following double integral:

$$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} (x y)^{-\theta - 1} dx \hspace{1mm} dy$$

Using Wolfram Alpha for symbolic algebra, it gives a solution of $$=c_1 x + c_2 + \frac{(x y)^{-\theta}}{\theta^2}$$

but what are the $c$'s and where did they come from? are they complex numbers, or constants of integration? am I allowed to set them to $0$?

What solution would a derivation by hand result in?

develarist
  • 1,514
  • what is your definition of $(xy)^{-\theta-1}$ when $xy<0$ and $-\theta-1$ is not an integer? – Calvin Khor Sep 23 '20 at 11:39
  • the double integral shown is actually a component of a larger integration problem I'm struggling with, which actually restricts $x$ and $y$ to be $[0,1]$, and $\theta \in [-1,\infty)$. https://math.stackexchange.com/questions/3829062/analytical-closed-form-solution-for-the-entropy-of-a-parametric-copula – develarist Sep 23 '20 at 12:04
  • ok. It doesn't seem related or relevant to consider the integral on $x,y<0$ IMO. – Calvin Khor Sep 23 '20 at 12:08

2 Answers2

2

They are constants of integration. You get them because you don't haven't given WolframAlpha the limits of your integral.

md2perpe
  • 26,770
1

They are in fact constants of integration as mentioned by md2perpe.

When you substitute the limits, the integral diverges (as can be seen from WA).

To compute the indefinite integral we have:

$$\int \int (xy)^{-\theta-1} dx dy=\int \int (xy)^{-\theta-1} dy dx$$ $$=\int \int x^{-\theta-1} y^{-\theta-1} dydx$$ $$=\int [-x^{-\theta-1}\frac{y^{-\theta}}{\theta}+c_{1}]dx$$ $$=\frac{x^{-\theta}y^{-\theta}}{\theta^2}+c_{1}x+c_{2}=c_{1}x+c_{2}+\frac{(xy)^{-\theta}}{\theta^2}.$$

Also note that you can have $$\int \int (xy)^{-\theta-1} dx dy=c_{1}y+c_{2}+\frac{(xy)^{-\theta}}{\theta^2}$$ due to symmetry.

Alessio K
  • 10,599
  • i wasn't sure how to express this with double integral signs, but the question is actually a second mixed partial derivative, first w.r.t. $x$, and then w.r.t. $y$. Does this change the solution and divergence property? – develarist Sep 23 '20 at 12:01
  • 1
    If you had partial derivatives $\partial x \partial y$, then after the first integration, instead of constants we would have functions of $x$ and $y$. Since clearly $\frac{(xy)^{-\theta}}{\theta^2}$ disappears when you put the limits, it will depend on the functions of $x$ and $y$ if the integral diverges. In the case above, the integral diverges since the term $c_{1}y$ or $c_{1}x$ diverges. – Alessio K Sep 23 '20 at 12:18
  • thanks. how could I have indicated that the double integral was in fact a second mixed partial derivative? are there partial $\int \int$ signs? replace $dx dy$ with $\partial x \partial y$ in the same formula? – develarist Sep 23 '20 at 12:22
  • 1
    Yeah you would write $\int \int \partial x \partial y$. – Alessio K Sep 23 '20 at 12:26