As specified in the title, I desire to determine if the integral $$ \int_0^{+\infty} \frac{1}{x} \, dx $$ either converges or diverges. If it diverges, I would also like to be able to say where to (if the limit goes to $+\infty, -\infty$ or if it simply doesn't exist.
My attempt. This is an integral over an infinite interval that also contains a discontinuous integrand. To do this integral we’ll need to split it up into two integrals so each integral contains only one point of discontinuity. With this in mind, we do the following split:
$$ \int_0^{+\infty} \frac{1}{x} \, dx = \int_0^1 \frac{1}{x} \, dx + \int_1^{+\infty} \frac{1}{x} \, dx $$
In order for the integral above to be convergent we will need BOTH of these to be convergent. If one or both are divergent then the whole integral will also be divergent. Let us compute, for example, the second integral on the RHS:
$$ \int_1^{+\infty} \frac{1}{x} \, dx = \lim_{t \to +\infty} \int_1^t \frac{1}{x} \, dx = \lim_{t \to +\infty} \ln(x)\Big|_1^t = \lim_{t \to +\infty} \ln(t) = +\infty. $$ With this, it is confirmed that the integral of interest is divergent.
My doubt comes now: How can I find where it diverges to?
Thanks for any help in advance.