Changing the bounds in an improper double integral

Suppose we have a double integral of the form $$ \int_0^{\infty} \int_0^{\infty} f(xy) g(x,y) \, dx \, dy $$ Let's suppose the functions behave well so that the integrals converge, etc. I'm trying to perform a substitution to change this to an iterated integral. If we set $y=\frac{t}{x}$ so that $dy=\frac{dt}{x}$, would the bounds on the new variable $t$ also be zero and infinity, so we could write $$ \int_0^{\infty} \int_0^{\infty} f(xy) g(x,y) \, dx \, dy = \int_0^{\infty} \int_0^{\infty} f(t) g\left(x,\frac{t}{x}\right) \, dx \, \frac{dt}{x} = \int_0^{\infty} f(t) \int_0^{\infty} \frac{g \left(x,\frac{t}{x} \right)}{x} \, dx \, dt $$ or would the bounds on $t$ be something different? How should I think about this to determine what the correct bounds should be?

Solution 1:

I suggest you proceed with a bit more caution when you perform a multivariate substitution. Changing the bounds with a substitution on a multiple integral requires that you determine the image of your domain of integration under some appropriate bivariate transformation $(u,v)=T(x,y)$.

The substitution you're considering is $(u,v)=(x,xy)$ which has inverse $(x,y)=\left(u,\frac{v}{u}\right)$. You can check that this this transformations maps $\mathbb{Q}_1$ bijectively into $\mathbb{Q}_1$. This means your $u,v-$bounds don't change after the substitution. Therefore, $$\begin{eqnarray*}\iint_{\mathbb{Q}_1}f(xy)g(x,y)\mathrm{d}x\mathrm{d}y &=& \iint_{\mathbb{Q}_1}f\left(v\right)g\left(u,\frac{v}{u}\right)\Bigg|\frac{\partial(x,y)}{\partial(u,v)}\Bigg|\mathrm{d}u\mathrm{d}v \\ &=& \iint_{\mathbb{Q}_1}\frac{1}{u}f\left(v\right)g\left(u,\frac{v}{u}\right)\mathrm{d}u\mathrm{d}v\end{eqnarray*}$$