Is the answer to this improper integral ∞ or -∞? The following integral is discontinuous at x = 0, $$\int_0^{1}\frac{1}{x}\:dx$$
Of course the proper answer to this integral is that it diverges, but I'm just curious whether the prior answer before concluding that this integral diverges is $\infty$ or $-\infty$?

1) When just computing the integral with standard integration rules, it becomes the following:
$${ln|x|_0^1}\: = ln(1) - ln(0) = 0 -∞ = -∞ $$ 
2) When applying improper integral techniques, it becomes the following:
$$\lim_{t \to 0^+} (ln|x|_t^1) = lim_{t \to 0^+}-ln|t|=-(-∞) = ∞$$
Noting that both diverge regardless of being $-\infty$ or $\infty$, I just want to know which prior answer before the divergence  conclusion is correct, $-\infty$ or $\infty$?  Note sure if my math is wrong but I think there is some ambiguity here.
 A: It seems to me that you think 
$$\int_0^1 \, \frac{1}{x} \, dx = \ln|x| \,\bigg|_0^1 = \ln(1) - \ln(0)$$
is a valid move, but it's not. You can't simply plug-in $0$ into the $\ln(x)$ because $\ln$ is undefined at $x = 0$. In fact, the domain of $\ln$ is $(0,\infty)$. Even if we generalize $\ln(x)$ to $\ln|x|$ the enlarged domain of $\ln|x|$ is $(-\infty,0) \cup (0,\infty)$. 
Therefore, you must take the right-hand limit. 
For a clearer example, consider the integral
$$\int_0^1 \frac{1}{x-1} dx.$$
If we evaluate this integral without limits, we get
$$\int_0^1 \frac{1}{x-1} dx = \ln|x-1|\,\bigg|_0^1 = -1 - \frac{1}{0},$$
which is clearly ridiculous. Again, this is because the number $1$ does not belong to the domain of the function $f(x) = \frac{1}{x-1}$, similar to how $0$ does not belong to the domain of the natural logarithm.
Therefore, the correct way to evaluate this integral is to take a left-hand limit:
$$\int_0^1\frac{1}{x-1} dx = \lim_{t\to 1^-}\int_0^t \frac{1}{x-1} dx = \lim_{t \to 1^-} \, \ln|x-1|\,\bigg|_0^t = -\lim_{t \to 1^-} \ln|x-1| = -(-\infty) = \infty$$
Edit
If you ever see notation like $\ln(0)$, or I've even seen $\ln(\infty)$, understand that this is shorthand for some limit.
