---
tag: R519NKG3V6
kind: definition
name: Hadamard finite part
msc: ["45E99", 26A42]
defines: G7VE41JHHH
uses: [0SYR10HD4R, 012PBA0FMD]
sources:
  - url: https://en.wikipedia.org/wiki/Hadamard_regularization
---

## Definition

Let $a < x < b$ and let $\varphi \colon [a, b] \to \R$ be integrable, differentiable near $x$, with $\varphi'$ [[Hölder continuous|012PBA0FMD]] near $x$.
The *Hadamard finite part* integral of $\varphi(t)/(t - x)^2$ is
$$\mathrm{f.p.}\!\int_a^b \frac{\varphi(t)}{(t - x)^2}\, dt = \lim_{\varepsilon \to 0^+} \left( \int_{\set{t \in [a, b] : \abs{t - x} > \varepsilon}} \frac{\varphi(t)}{(t - x)^2}\, dt - \frac{2\varphi(x)}{\varepsilon} \right).$$
Under these hypotheses the limit exists.

## 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|0SYR10HD4R]].