Hadamard finite part

Remarks

The general rule is the same. Expand the integral over the region at distance more than $\varepsilon$ from the singularity in powers of $\varepsilon$ and $\log \varepsilon$. Then drop the terms that diverge as $\varepsilon \to 0$ and keep the limit of the rest. For example, $\int_\varepsilon^1 t^{-3/2}\, dt = 2\varepsilon^{-1/2} - 2$, so $\mathrm{f.p.}\!\int_0^1 t^{-3/2}\, dt = -2$.

Example: $\varphi = 1$ on $[-1, 1]$ and $x = 0$. Then $\int_{\varepsilon < \abs{t} \le 1} t^{-2}\, dt = \frac{2}{\varepsilon} - 2$, so the finite part is $-2$.

Non-example: for $\varphi(t) = \abs{t}^{1/2}$ and $x = 0$, the derivative condition fails at $0$. The integral over $\varepsilon < \abs{t} \le 1$ equals $4\varepsilon^{-1/2} - 4$, and subtracting $2\varphi(0)/\varepsilon = 0$ leaves a divergent quantity, so this limit does not exist.

For $\varphi$ as above, the finite part is the derivative of a principal value: $$\mathrm{f.p.}\!\int_a^b \frac{\varphi(t)}{(t - x)^2}\, dt = \frac{d}{dx}\, \mathrm{PV}\!\int_a^b \frac{\varphi(t)}{t - x}\, dt,$$ where $\mathrm{PV}$ is the Cauchy principal value.