For example in $R^2$ with area $D=(a,\infty)\times(b,\infty)$
Function $f(x,y)$ is defined on $(x,y)\in D$
And $x=a,y=b$ are branch points, which means $\forall y\in(a,\infty)\Rightarrow\lim_{x\to a}f(x,y)=\infty$ and $\forall x\in(a,\infty)\Rightarrow\lim_{y\to b}f(x,y)=\infty$
What would the official definition of improper integral on area $D$ converges??
To further illustrate the concept, I may list some more information here:
The formal definition I got for a 2-dimension improper integral with unlimited integration area are as this:
Suppose $D\in R^2$ is an unlimited area. There is a line $\Gamma\in D$ separate $D$ into a limited part $D_\Gamma$ and unlimited part. Say a point $(x,y)\in\Gamma$ we define $d(\Gamma)=\inf\left\{\sqrt{x^2+y^2}\big|(x,y)\in\Gamma\right\}$
So the definition is when $d(\Gamma)\to\infty$ disrespect of what kind of $\Gamma$ you choose, $\lim_{d(\Gamma)\to\infty}\int_{D_\Gamma}f(x,y)dxdy$ always converge, thus we have $\int_{D}f(x,y)dxdy=\lim_{d(\Gamma)\to\infty}\int_{D_\Gamma}f(x,y)dxdy$
Here the key point is, $\Gamma$ can be any shape.
So the formal definition of this would be $\forall\epsilon>0,\exists\delta$, for all $\Gamma$ that satisfy:
-
$D_\Gamma$ is limited
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$d(\Gamma)>\delta$
We have $\left|\int_{D_\Gamma}f(x,y)dxdy-I\right|<\epsilon$
Thus we can say $\int_{D}f(x,y)dxdy=I$
This also brings up the key thing that the convergence of $\int_{D}|f(x,y)|dxdy$ is a sufficient and necessary condition of $\int_{D}f(x,y)dxdy$ converges (in 1-detention, $\int_{D}f(x,y)dxdy$ converges does not support $\int_{D}|f(x,y)|dxdy$ converges).
Because you could choose any shape of $\Gamma$, thus for periodic function (Which do has limit say $|f(x,y)|<M$), I can select $\Gamma$ that contains all the positive area and with size of negative area as $\frac{1}{M}$.
Best Answer
Let $D \subset \mathbb{R}^2$ be an open set. If $D$ is unbounded and/or $f:D \to \mathbb{R}$ is unbounded at the boundary, then the improper (Riemann) integral has to be carefully defined and there can be problems as you suspect if we don't have convergence of the improper integral of $|f|$ under the definition.
First, we require that $f$ be Riemann integrable over any compact $A \subset D$ which is also rectifiable (measure of boundary is $0$).
We can define the improper integral over $D$ by selecting any sequence $(A_n)$ of compact rectifiable sets which are nested in the sense that $A_n \subset \text{int}(A_{n+1})$ and which exhaust $D$, meaning $D = \cup_{n=1}^\infty A_n$. Defining $f^+ = \max (f,0)$ and $f^- = \max(-f,0)$ the improper integral is
$$\int_D f = \lim_{n \to \infty} \int_{A_n} f^+ - \lim_{n \to \infty} \int_{A_n} f^- ,$$
when both limits exist. In this way $|f| = f^+ + f^-$ must also have a convergent improper integral. It can be shown that the improper integral is independent of the choice of $(A_n)$ provided the above conditions are met.
If we find only conditional convergence
$$I_{\mathcal{A}} = \lim_{n \to \infty} \int_{A_n} f^ ,$$
for a specific sequence $\mathcal{A} = (A_n)$ then there is no guarantee that we will get the same result for another choice of sets.
This is true even in the one-dimensional case as shown here.