$\def\R{\mathbf{R}}$ The Riemann-integral is defined only for bounded functions. So functions such as $f:\R-\{0\}$ defined by $f(x):=1/x$ are not Riemann-integrable over any interval that contains $0$ as a limit point. However, at the core of the motivation behind the rigorous definition of the Riemann-integral is the innate desire to prescribe a value that best represents the "area underneath the curve." With such a compelling motive, there is an overwhelming temptation to assign the value $0$ to an expression like $\int_{-1}^{1}{f}$. Likewise, the expression $\int_{-1}^{3}{f}$ could be assigned the value of $\int_{-1}^{1}{f}+\int_{1}^{3}{f}=0+\int_{1}^{3}{f}=\int_{1}^{3}{f}$ (which is also consistent with integral-laws!).
Unfortunately, even armed with a zealous belief in geometry is not enough to make sense of an expression like $\int_0^3{f}$. But I'd personally be okay with letting this be undefined as it already is. The crux of the issue here is why something like $\int_{-2}^{2}{f}$ must be undefined when it really begs to "equal" $0$?
Putting my question another way, is there a way to meaningfully extend the definition of the Riemann integral that remains consistent with integral laws and theorems, such that it can integrate a certain class of unbounded functions when the interval of integration is carefully chosen? Or is this whimsical enterprise a pointless endeavor that is doomed to fail?
My attempt at a primitive extension would be something like:
We could define \begin{align}\int_{-1}^{1}{f}:&=\lim_{r\to0, r\neq0}\int_{-1}^{-r}{f}+\int_{r}^{1}{f}\\&=0\end{align} Which can be generalized for functions with a finite number of limit points where the function is unbounded.
