= Differentiating a principal-value integral across a fixed simple pole
{title2=$\frac d{dx}\operatorname{PV}\int_{a(x)}^{b(x)}q(t)dt=q(b)b'-q(a)a'$}
Suppose $q(t)=A/(t-c)+r(t)$, where $r$ is continuous, the pole $c$ is fixed, and differentiable endpoints obey $a(x)<c<b(x)$ locally. Then
$$
\operatorname{PV}\int_{a(x)}^{b(x)}q(t)dt=A\ln\frac{|b(x)-c|}{|a(x)-c|}+\int_{a(x)}^{b(x)}r(t)dt.
$$
Apply the <chain rule> and the <fundamental theorem of calculus> to get
$$
\boxed{\frac d{dx}\operatorname{PV}\int_{a(x)}^{b(x)}q(t)dt=q(b(x))b'(x)-q(a(x))a'(x).}
$$
The same boundary expression as in the <Leibniz integral rule> thus applies to this explicitly specified <Cauchy principal value>. If $A\ne0$, the two ordinary one-sided <improper integrals> diverge, so their sum is not an ordinary <integral>. A finite boundary derivative alone does not make that divergent integral meaningful.
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