I understand that you can't find the area under a function thats not bounded using Riemann sums, but why can't the function have an antiderivative that is undefined at that point?
Instead of functions not being integrable, if the antiderivative has a discontinuity/is not bounded at the same point, the area still can't be found at that point, but the antiderivative would still be able to be used at other points.
For instance, see The domain of $\ln x$?
The domain of $\ln x$ has to be restricted because otherwise it would contain an unbounded value. Why can't it just have an asymptote like $1/x$?
Answer
As it happens also unbounded function can be "integrable", but I guess you meant the basic definition of Riemann integral (right?), so:
if the function isn't bounded in say $\;[a,b]\;$ then
$$\forall\,M\in\Bbb N\;\;\exists\,x_M\in[a,b]\;\;s.t.\;\;|f(x_M)|>M$$
Now prove that you can construct from the above a sequence so that the values of $\;f\;$ on it diverge to $\;\infty\;$ and check what happens with a partition of $\;[a,b]\;$ and its refinements that'll include points from that sequence...
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