I have integral $$ \int_A\frac{x+y}{(1+(x+y)^2)^2}dxdy $$ $A=\{(x,y):x>0,y\in\mathbb{R}\}$
First: $$\int_{-\infty}^{\infty}\int_{0}^{\infty}\frac{x+y}{(1+(x+y)^2)^2}dxdy={1\over 2}\int_{-\infty}^{\infty}\frac{dy}{1+y^2}={\pi\over2}$$
Second try: $$\int_{0}^{\infty}\int_{-\infty}^{\infty}\frac{x+y}{(1+(x+y)^2)^2}dydx=\int_{0}^{\infty}0dx=0$$
So should I conclude that Lebesgue integral does not exist?
On
You are right. Assuming that $f(x,y)=\frac{(x+y)}{(1+(x+y)^2)^2}\in L^1(A)$, $$ \int_{-\infty}^{+\infty}\int_{0}^{+\infty}f(x,y)\,dx\,dy=\int_{0}^{+\infty}\int_{-\infty}^{+\infty}f(x,y)\,dy\,dx \tag{1}$$ has to hold, but the LHS of $(1)$ is zero while the RHS is $\frac{\pi}{2}$, hence $f\not\in L^1(A)$.
Some of the comments indicate that there is still a bit of confusion about what is going on, so I'm just going to add a bit more detail to the previous answer.
We can calculate the two iterated integrals as follows. For the first integral, the substitution $u=1+(x+y)^2$ gives $du=2(x+y)dx$ and $$\int^{\infty}_{-\infty}\int_{0}^{\infty}\frac{(x+y)}{(1+(x+y)^2)^2}du dy=\frac{1}{2}\int^{\infty}_{-\infty}\int_{1+y^2}^{\infty}\frac{1}{u^2}du dy=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{1+y^2}dy=\int_{0}^{\infty}\frac{1}{1+y^2}\;.$$ The last integral is evaluated as $\lim_{x\to \infty} tan^{-1}(x)=\pi/2$. The other integral is slightly more nuanced. For each $x$, the inner integral of the iterated integral can be evaluated via the same $u$-substitution. This will lead to the improper Riemann integral $$\lim_{M\to -\infty}\lim_{N\to \infty}\int_{1+(x+M)^2}^{1+(x+N)^2}\frac{1}{u^2}=0\;.$$ Thus, the iterated integral must be zero. The issue here of course is that Fubini's theorem does not apply. In fact, Fubini's theorem for $\sigma$-finite measure spaces implies that since the two iterated integrals do not agree, the function $f(x,y)=\frac{(x+y)}{(1+(x+y)^2)^2}$ cannot be in $L^1(A)$ (whenever the improper Riemann integral is finite it agrees with the Lesbesgue integral).