Careful thinking to update our minds

Let's start with $\displaystyle \int \dfrac{d x}{x} = \ln \color{#3D99F6}{| x |} + C$ while integrate across 0!

$\int_{-2}^{3} \dfrac{d x}{x} = \ln 3 - \ln 2 = \ln \frac32$ = $\int_{2}^{3} \dfrac{d x}{x}$

$\int_{2}^{-3} \dfrac{d x}{x} = \ln 3 - \ln 2 = \ln \frac32$ = $\int_{-2}^{-3} \dfrac{d x}{x}$

Both are correct! The second one can be realized from $- \int_{-3}^{2} \dfrac{d x}{x} = -(\ln 2 - \ln 3) = \ln \frac32.$ As thought as skipping area of $\int_{-n}^{n} \dfrac{d x}{x}$ of zero! They are reconnected after skipped. $\color{#3D99F6}{This~tells~that~ \int_{-1}^1 \dfrac{d x}{x}= 0.}$

$\int_{-1}^1 \dfrac{d x}{x}$ = \Large^\lim_{\epsilon \to 0}[\int_{-1}^{\epsilon^-} \dfrac{d x}{x} + \int_{\epsilon^+}^1 \dfrac{d x}{x}] = \Large^\lim_{\epsilon \to 0} (\ln \epsilon - \ln \epsilon) = 0.

Note that $\int_{-1}^{0^-} \dfrac{d x}{x} + \int_{0^-}^{1} \dfrac{d x}{x}$ = indeterminate but not preserving the integral of $\int_{-1}^1 \dfrac{d x}{x}.$

$\int \dfrac{d x}{x} = \ln x + C$ when $x \geq 0$ while $\int \dfrac{d (-x)}{-x} = \ln (-x) + C$ when $x < 0.$

To conclude with simplicity, we just need to write $\int \dfrac{d x}{x} = \ln | x | + C,$ where integrate crossing zero with zero on the line is inclusive! Hence, statement q is proven to be True! $$ \Large^\lim_{n \to \infty} \int_{-n}^n \dfrac{d x}{x} = $\Large^{\lim}_{n \to \infty}[\int_{-n}^{\epsilon^-} \dfrac{d x}{x} + \int_{\epsilon^+}^{n} \dfrac{d x}{x}]$ = $\Large^{\lim}_{n \to \infty} (\ln \frac{\epsilon^-}{-n} + \ln \frac{n}{\epsilon^+})$ = $\ln 1 = 0.$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^{\Large \epsilon \to 0}$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^{\Large \epsilon \to 0}$

Hence, statement p is also proven to be True!

$$

Statement p is conclusive which includes statement q rather than just for both infinities. To think about $\ln (-1),$ I think at least the difference from contour integral is there is a radius of another $\epsilon$ from singularity zero for the complex variable analysis. In the context of real integral like $\int_{-n}^n \dfrac{d x}{x},$ there is at least an on the line by the singularity zero! $\ln (-1)$ is defined by me as $\sqrt{-1} \beta \pi$ where $\beta$ = {..., -3, -1, 1, 3, 5, 7, ...}; but since an on the line is not synonym to passing by, an outcome of $\ln (-1)$ for $\int_{-n}^{n} \dfrac{d x}{x}$ is $not$ necessarily true. The trick is $\int \dfrac{d (-x)}{-x} = \int \dfrac{d x}{x}$ ought to be smart to make automatic selection with regard to singularity zero!

One interesting is $\ln x$ = $\displaystyle \frac{1 - x^{-\epsilon}}{\epsilon}$ where $\epsilon \to 0.$ You could substitute values to see.

In one round, we actually have angles of -180 $^\circ$ < $\theta \leq$ 180 $^\circ$ to have zero fault for a complex number to the power of another complex number, regardless of fault according to general number. Provided angles are highly restricted to the range mentioned above, having $2 \pi$ as angle which duplicated the angle 0 and etc. could be a fault of mistake. Therefore, I am doubted by Cauchy actually.

From point of view of seeing $\int_{-1}^1 \dfrac{d x}{x}$ = indeterminate $xor$ $\int_{-1}^1 \dfrac{d x}{x}$ = 0, with x = 0 occupying no width for area, but a Dirac's infinity, I would think that Dirac obtained his intuition from here but such infinity in the context of our discussion here does not have gender of positive or negative actually. For our case, we can still imagine that $it$ neutralized itself into zero as well! Therefore, 0 would be reasonable because we should not suddenly let it to produce something which cannot be certain or cannot be determined but anything like $\ln 2$ or etc. My instinct up to here tells me that some other things must only arise from our improper treatments.

Fair and lovely!

Answer: $\boxed{False~and~False}$ however as $d x$ cannot be 0 and therefore value of a $d A$ or $\frac10 d x$ does not exist.

$$

Generalizing a subject (allows banned expressions):

$n \times 0 = 0 \implies n = \frac00$ = indeterminate

$\frac10 \times 0 = 1$ while $\frac10 \times 0 \times 0$ = $0$ tells that the order of zeroes is decisive.

$\frac10 d x$ = 0 with high order of 0 for $d x.$ Generalization onto $\frac00$ and $\frac10$ in the past may not be sufficient, that had not enable people to handle them. But d x with multiple of width for others makes the dominated areas. The thing is equal extend or magnitude of different signs makes a cancellation into zero. I think indeterminate cannot be what is meant by false here. The only thing remained is invalid or not exist for false. Letting the singularity to contribute a value of 1 for the integral, the quest must fall onto extend of finest of $d x$ applied. Gender for $\infty$ here must only come from +1 which is not -1 where 0 is neutral.

Brilliant must show a convincing example which can prove these as $false$ before falsifying the solution written particularly for this question. I haven't agreed with the decision with reasons given.

The significance of my conclusion would affect the consequence for any \Large ^\lim_{m \to a}\int_{-m}^{m} \frac{d x}{x^n} = $\frac{a^{1 - n}}{1 - n}$ - $\frac{(-a)^{1 - n}}{1 - n}.$ Including those which converge at x = $0^-$ and x = $0^+$ , the consequence means that all integral for $\frac{1}{x^n}$ with x crossing zero would only be values not exist!

For \Large ^\lim_{\epsilon \to 0}(\int_{-1}^{-\epsilon} \frac{d x}{x^n}+ \int_{\epsilon}^{1} \frac{d x}{x^n}) however, all odd functions should produce integral of zero, which is called the principal value of an integral.

$\int_{0}^{\epsilon} \dfrac{d x}{x}$ = $\ln | \frac{\epsilon}{0} |$ hinted for integral of positive infinity when leaving 0 to positive side. But the value of $\frac10$ is a not exist.