Is there a reason to prefer the improper integral for some cases over the Cauchy principal value? Suppose one integrates
$$ \int^1_0 \frac{1}{4y - 1} \, \mathrm{d} y $$
Evaluating one obtains:
$$-\infty - 0 + \infty + \frac{\ln 3}{4}  $$
which is indeterminate.
However,
$$\lim_{y \rightarrow \frac{1}{4}} \frac{\ln (4y - 1)}{4} - 0 - \frac{\ln (4y - 1)}{4} + \frac{\ln 3}{4} =\frac{\ln 3}{4} $$
I think this is the Cauchy principal value.
Are there cases where it is useful to prefer the improper integral answer of undefined over the answer of the Cauchy principal value?
Why shouldn't I just always use the Cauchy principal value over the improper integral? Are there cases that using the Cauchy principal value produces the wrong result?
 A: Notice that $\frac{1}{4\cdot y+1} = \frac{1}{4}\cdot\frac{1}{y-1/4}$, so $\int_0^1 \frac{dy}{4\cdot y+1} = \frac{1}{4}\cdot\int_0^1 \frac{dy}{y-1/4} = \frac{1}{4}\cdot\int_{-1/4}^{3/4} \frac{dx}{x} = \frac{1}{4}\cdot\int_{-1/4}^{1/4} \frac{dx}{x} + \frac{1}{4}\cdot\int_{1/4}^{3/4} \frac{dx}{x} = \frac{1}{4}\cdot\int_{-1/4}^{1/4} \frac{dx}{x} + \frac{1}{4}\cdot\ln(\frac{3/4}{1/4}) = \frac{1}{4}\cdot\int_{-1/4}^{1/4} \frac{dx}{x} + \frac{\ln 3}{4}$. As such, it can be more easily seen where the problem is coming from: it comes from $\frac{1}{4}\cdot\int_{-1/4}^{1/4} \frac{dx}{x}$. In fact, $\frac{1}{4}\cdot\int_{-1/4}^{1/4} \frac{dx}{x} = \int{-1}^{1} \frac{dx}{x}$, and this is a well-known and controversial integral symbol that is known to be ill-defined. Rather than linking another stack exchange post explaining why it is ill-defined, though, I will just give you my explanation here.
Consider the function $f : [-1,1] \rightarrow $ where $f(x) = 1/x$ for every nonzero $x$ and $f(0)=0$. Remember that $\int_{-1}^1 f dx$ is defined as the limit of the Riemann sums of $f$, if it exists. However, this limit does not exist, in this case. This can be shown, but is tedious. The reason the Riemann sums do not converge is because $f$ is unbounded, which means, intuitively, that it blows up. This happens at $x = 0$. $f$ is defined, but discontinuous, at $0$, and the limit as $x \rightarrow 0$ of $f$ does not exist, because $f$ blows up near $0$. Since the Riemann sums do not converge, $\int_{-1}^1 f dx$ simply does not exist.
Since the discontinuity at $0$ is what is causing the problem, it is reasonable to restrict $f$ to $[-1,\alpha]\cup[\beta, 1]$, evaluate the Riemann integral of $f$, and let $\alpha \rightarrow 0^-, \beta \rightarrow 0^+$. However, this also does not converge, again because $f$ is unbounded near $0$. So the integral you want does not exist, even as an improper integral.
The Cauchy principal value is what happens when you let $\alpha = -\beta$ in the limit. What it does is pick out a specific path, and then evaluate the limit, rather than do the limit independently of the paths. This is also equivalent to simply taking the average of the Riemann integral of $f$ over every possible path, and this average can be parametrized as varying with respect to the radius $\epsilon$. Once you let $\epsilon \rightarrow 0$, you get the Cauchy principal value.
Now it should be clear why the Cauchy principal value is not the "correct" answer. Intuitively, you can think of the Cauchy principal value as an averaging operation that "smoothes out" the singularity of $1/x$ at $0$, and thus, the Cauchy principal value answers the question, "what would the integral be if the function was well-behaved?" In many applications, this is a useful tool. Nonetheless, it also acknowledges that the function is not well-behaved to begin with, and the singularity at 0 is there, regardless of how much it inconveniences us. So the integral in question simply cannot exist.
