I know that in order to prove that $ \intop_{-\infty}^{\infty}f\left(x\right) $ is convergent, I have to show that both $ \intop_{-\infty}^{a}f\left(x\right),\intop_{a}^{\infty}f\left(x\right) $ for some $ a \in \mathbb{R} $.
Is it true also for :
$ \intop_{-\infty}^{0}f\left(x\right) $ for $ f(x) $ that isn't bounded around $ 0 $ ?
I mean, in order to show that $ \intop_{-\infty}^{0}f\left(x\right) $ is convergent, I have to show that $ \intop_{-\infty}^{a}f\left(x\right),\intop_{a}^{0}f\left(x\right) $ both convergent for some $ a<0 $ ? and even if just one of them divergent, it means that the whole integral $ \intop_{-\infty}^{0}f\left(x\right) $ is divergent ?
I can't understand it, because for instance, if $ \intop_{-\infty}^{a}f\left(x\right)=-\infty $ and $ \intop_{a}^{0}f\left(x\right)=\infty $, there is a chance that $ \intop_{a}^{0}f\left(x\right)+\intop_{-\infty}^{a}f\left(x\right) $ will be a real number.
A clarification of this subject would be really helpful for me. Thanks in advance.