Is this the idea behind Cauchy principal value? I came across this improper integral here
Improper integrals - exercises
$$\int_1^{4} \frac{1}{(x+3)(x-2)} \tag{*}$$
The problematic point here is $x=2$
I rewrote it as:
$$\int_1^{4} \frac{dx}{(x+3)(x-2)} = \frac{1}{5}\left[\int_1^{2} \frac{dx}{(x-2)} + \int_2^{4} \frac{dx}{(x-2)} - \int_1^{4} \frac{dx}{(x+3)}\right]$$
The last integral here is proper and I compute it as: $$(-1/5)\cdot \ln\frac{7}{4}$$
The first integral (after some rework) is:
$$(1/5)\cdot \lim_{a \to 0+} \ln{a}$$
And the second integral is:
$$(1/5)\cdot \ln{2} - (1/5)\cdot \lim_{b \to 0+} \ln{b}$$
So the first and second integrals diverge.
So I guess the conclusion is that the integral $(*)$ also diverges. Is that so?
But... if we take the sum of the first and second integrals, the two limit terms cancel out, and so in some sense the sum of these two divergent improper integrals exists (is finite) even though they are both divergent.
So in a way... the overall value of the integral $(*)$ turns out to be $$(1/5) \cdot \ln{2} - (1/5) \cdot \ln \frac{7}{4} \tag{**}$$
After doing this computations, I then came across this concept Cauchy principal value
So... is this what Cauchy principal value is about?
WA gives me this same value as Cauchy principal value $(1/5) \cdot \ln \frac{8}{7}$ which coincides with my number $(**)$. I just need to know if that indeed is the idea behind the Cauchy principal value i.e. to handle the situation where we have two improper integrals which both diverge but their sum converges i.e. is finite.
So in that case I guess we say the integral is divergent but it still has a finite Cauchy principal value, is that the correct wording?

At $x = 2$, denoting $f$ your integrande, you get
$$
f(x) \sim \frac{1}{5(x-2)}
$$
which is not integrable in the Riemann or Lebesgue sens. Your computation is not valid in this theory because the integral is not additive when your functions are not positive or integrable. So you cannot write that
$$
\int_1^4 \frac{1}{5(x-2)} - \frac{1}{5(x+3)}dx = \int_1^4 \frac{1}{5(x-2)} dx - \int_1^4 \frac{1}{5(x+3)}dx.
$$
But as your illegal and formal computation shows, it is possible to remove the singularity of the integral taking the principal value. This process allows to cumpute integrals which are not defined in a classical way, when the divergence is compensated by the contribution at both sides of the singularity. In your case, up to a translation, you discovered that the Cauchy Principal Value for an odd function is null. Your formulation is totally correct.
