Is the following always true? (i.e. if both converges, limits are equal; if one diverges, the other must diverge; EXCLUDE the case where the limit keeps "jumping")
$$ \int_{0}^{\infty}f(x)\;dx=\sum_{n=0}^{\infty}\int_{n}^{n+1}f(x)\;dx $$
When $f(x)$ is continuous on $(0,\infty)$, how to argue that the above equality holds?
In fact they are not the same. But it is just nitpicking on definition of integral.
Take for example $f(x) = \sin(2\pi x)$. Than $$ F(x) = \int_0^x \sin(2\pi t) dt = \frac{-1}{2 \pi}\cos(2\pi x) $$ and limit $\lim_{x\rightarrow \infty} F(x)$ does not exists.
On other hand $$ \int^{n+1}_n \sin(2\pi t) dt = 0 $$ for every $n$.
So the sum on right side exists but the integral on left does not.