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{{Renormalization and regularization}}
In mathematics, '''Hadamard regularization''' (also called '''Hadamard finite part''' or '''Hadamard's partie finie''') is a method of [[Regularization (physics)|regularizing]] divergent integrals by dropping some divergent terms and keeping the finite part, introduced by {{harvs|txt|authorlink=Jacques Hadamard|last=Hadamard|first=Jacques|year1=1923|loc1=book III, chapter I|year2=1932}}. {{harvs|txt|authorlink=Marcel Riesz|last=Riesz|first=Marcel|year1=1938|year2=1949}} showed that this can be interpreted as taking the [[meromorphic continuation]] of a convergent integral.

==Description==

If the [[Cauchy principal value]] integral 
<math display="block">\mathcal{C}\int_a^b \frac{f(t)}{t-x} \, dt \quad (\text{for } a<x<b)</math>
exists, then it may be differentiated with respect to {{mvar|x}} to obtain the Hadamard finite part integral as follows:
<math display="block">\frac{d}{dx} \left(\mathcal{C}\int_{a}^{b} \frac{f(t)}{t-x} \,dt\right)=\mathcal{H}\int_a^b \frac{f(t)}{(t-x)^2}\, dt \quad (\text{for } a<x<b).</math>

Note that the symbols <math>\mathcal{C}</math> and <math>\mathcal{H}</math> are used here to denote Cauchy principal value and Hadamard finite-part integrals respectively.  

The Hadamard finite part integral above (for {{math|''a'' &lt; ''x'' &lt; ''b''}}) may also be given by the following equivalent definitions:
<math display="block">\mathcal{H}\int_a^b \frac{f(t)}{(t-x)^2}\, dt = \lim_{\varepsilon \to 0^+} \left\{ \int_a^{x-\varepsilon} \frac{f(t)}{(t-x)^2} \, dt + \int_{x+\varepsilon}^b\frac{f(t)}{(t-x)^2}\,dt -\frac{f(x+\varepsilon)+f(x-\varepsilon)}{\varepsilon}\right\},</math>
<math display="block">\mathcal{H}\int_a^b \frac{f(t)}{(t-x)^2}\, dt = \lim_{\varepsilon \to 0^+} \left\{ \int_a^b \frac{(t-x)^{2}f(t)}
{((t-x)^2+\varepsilon^2)^2}\,dt -\frac{\pi f(x)}{2\varepsilon}- \frac{f(x)}{2} \left(\frac{1}{b-x}-\frac{1}{a-x}\right)\right\}. </math>   

The definitions above may be derived by assuming that the function {{math|''f''&thinsp;(''t'')}} is differentiable infinitely many times at {{math|''t'' {{=}} ''x'' for ''a'' &lt; ''x'' &lt; ''b''}}, that is, by assuming that {{math|''f''&thinsp;(''t'')}} can be represented by its Taylor series about {{math|''t'' {{=}} ''x''}}. For details, see {{harvs|txt|last=Ang|year1=2013}}. (Note that the term {{math|− {{sfrac|''f''&thinsp;(''x'')|2}}({{sfrac|1|''b'' − ''x''}} − {{sfrac|1|''a'' − ''x''}})}} in the second equivalent definition above is missing in {{harvs|txt|last=Ang|year1=2013}} but this is corrected in the errata sheet of the book.)

Integral equations containing Hadamard finite part integrals (with {{math|''f''&thinsp;(''t'')}} unknown) are termed hypersingular integral equations. Hypersingular integral equations arise in the formulation of many problems in mechanics, such as in fracture analysis.

==Example==

Consider the divergent integral
<math display="block">\int_{-1}^1 \frac{1}{t^2} \, dt = \left( \lim_{a \to 0^-} \int_{-1}^{a} \frac{1}{t^2} \, dt \right) + \left( \lim_{b \to 0^+} \int_{b}^{1} \frac{1}{t^2} \, dt \right) = \lim_{a\to 0^-} \left( -\frac{1}{a} - 1 \right) + \lim_{b\to 0^+} \left(-1 + \frac{1}{b}\right)= +\infty</math>
Its [[Cauchy principal value]] also diverges since
<math display="block">\mathcal{C} \int_{-1}^1 \frac{1}{t^2} \, dt = \lim_{\varepsilon \to 0^+} \left( \int_{-1}^{-\varepsilon} \frac{1}{t^2} \, dt + \int_{\varepsilon}^1 \frac{1}{t^2} \, dt \right) = \lim_{\varepsilon \to 0^+} \left( \frac{1}{\varepsilon} - 1 - 1 + \frac{1}{\varepsilon} \right) = +\infty </math>
To assign a finite value to this divergent integral, we may consider
<math display="block">\mathcal{H}\int_{-1}^1 \frac{1}{t^2} \, dt = \mathcal{H} \int_{-1}^1 \frac{1}{(t-x)^2} \, dt \Bigg|_{x=0} = \frac{d}{dx}\left( \mathcal{C} \int_{-1}^1 \frac{1}{t-x} \, dt \right) \Bigg|_{x=0}</math>
The inner Cauchy principal value is given by
<math display="block">\mathcal{C} \int_{-1}^1 \frac{1}{t-x} \, dt = \lim_{\varepsilon \to 0^+} \left( \int_{-1}^{-\varepsilon} \frac{1}{t-x} \, dt + \int_{\varepsilon}^1 \frac{1}{t-x} \, dt \right) = \lim_{\varepsilon \to 0^+} \left( \ln\left| \frac{\varepsilon +x}{1+x} \right| + \ln \left| \frac{1-x}{\varepsilon - x} \right| \right) = \ln\left| \frac{1-x}{1+x} \right|</math>
Therefore,
<math display="block">\mathcal{H}\int_{-1}^1 \frac{1}{t^2} \, dt = \frac{d}{dx}\left( \ln\left| \frac{1-x}{1+x} \right| \right) \Bigg|_{x=0} = \frac{2}{x^2-1}\Bigg|_{x=0} = -2</math>
Note that this value does not represent the area under the curve {{math|''y''(''t'') {{=}} 1/''t''<sup>2</sup>}}, which is clearly always positive. However, it can be seen where this comes from. Recall the Cauchy principal value of this integral, when evaluated at the endpoints, took the form
<math>\lim_{\varepsilon \to 0^+} \left( \frac{1}{\varepsilon} - 1 - 1 + \frac{1}{\varepsilon} \right) = +\infty</math>

If one removes the infinite components, the pair of <math>\frac{1}{\varepsilon}</math> terms, that which remains is
<math>\lim_{\varepsilon \to 0^+} \left(  - 1 - 1  \right) = -2</math>

which equals the value derived above.

==References==
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 | place =Oxford
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| archive-date=2016-03-05 
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*{{Citation
 |last1        = Riesz
 |first1       = Marcel
 |author1-link = Marcel Riesz
 |title        = Rectification au travail "Intégrales de Riemann-Liouville et potentiels"
 |language     = French
 |year         = 1938
 |url          = http://acta.fyx.hu/acta/showCustomerArticle.action?id=5667&dataObjectType=article
 |journal      = Acta Litt. AC Sient. Univ. Hung. Francisco-Josephinae, Sec. Sci. Math. (Szeged)
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[[Category:Integrals]]
[[Category:Summability methods]]