= Solution
<Fubini's theorem> states that an absolutely integrable measurable function on a product of two $\sigma$-finite <measure spaces> has integrable sections almost everywhere, the corresponding iterated <integrals> exist, and both orders equal its product <integral>. In particular, for $f\in L^1(\mathbb R^2)$,
$$
\iint f(x,y)\,dx\,dy=\int\left(\int f(x,y)\,dx\right)dy=\int\left(\int f(x,y)\,dy\right)dx.
$$
To establish integrability of the nonnegative function here, first use <Tonelli's theorem>: for a nonnegative measurable function the product <integral> equals either iterated <integral> with infinity allowed. This avoids assuming the conclusion needed for <Fubini's theorem>.
If $\alpha\geq-1$, $\int_0^\infty(1+x+y)^\alpha dx=\infty$ for every fixed $y$. If $\alpha<-1$, it is
$$
\int_0^\infty(1+x+y)^\alpha dx=\frac{(1+y)^{\alpha+1}}{-\alpha-1}.
$$
The outer <integral> is finite exactly when $\alpha+1<-1$, that is, $\alpha<-2$. In that case
$$
\boxed{\iint_{(0,\infty)^2}(1+x+y)^\alpha dx\,dy=\frac1{(-\alpha-1)(-\alpha-2)},\qquad f\in L^1\iff\alpha<-2.}
$$
For $-2\leq\alpha<-1$ the outer <integral> diverges, including logarithmic divergence at $\alpha=-2$. Nonnegativity means finiteness of this <integral> is exactly absolute integrability; after proving it, <Fubini's theorem> allows either order.
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