Does $\int_0^\infty g(y)dy$ converge and $\int_0^\infty f(x)dx=\int_0^\infty g(y)dy$?

Question

Suppose $F(x,y)$ is continuous on $(0,\infty)\times (0,\infty)$. $\forall x\in (0,\infty)$, $f(x)=\int_0^\infty F(x,y)dy$ converges, and $f(x)$ is continuous on $(0,\infty)$. $\forall y\in (0,\infty)$, $g(y)=\int_0^\infty F(x,y)dx$ converges, and $g(y)$ is continuous on $(0,\infty)$. If $\int_0^\infty f(x)dx$ converges, does $\int_0^\infty g(y)dy$ converge and $\int_0^\infty f(x)dx=\int_0^\infty g(y)dy$?

Answer 1

As a counterexample take

$$F(x,y) = \frac{x-y}{(2+x+y)^3}$$

Then,

$$g(y) = \int_0^\infty F(x,y) \, dx = - \left.\frac{1+x}{(2+x+y)^2} \right|_{x = 0}^{x = \infty}= \frac{1}{(2+y)^2},$$

which is continuous on $(0,\infty)$ with a convergent improper integral $$\int_0^\infty \frac{dy}{(2+y)^2} = - \left.\frac{1}{2+y}\right|_0^\infty = \frac{1}{2}$$

Similarly,

$$f(x) = \int_0^\infty F(x,y) \, dy = -\frac{1}{(2+x)^2},$$

is continuous, but with an improper integral converging to $-1/2$.

Thus,

$$\int_0^\infty f(x) \, dx \neq \int_0^\infty g(y) \, dy$$