What are $c$'s in antiderivative of double integral in Wolfram Alpha

Question

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?

Answer 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.

Answer 2

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