Hadamard regularization explained
In mathematics, Hadamard regularization (also called Hadamard finite part or Hadamard's partie finie) is a method of regularizing divergent integrals by dropping some divergent terms and keeping the finite part, introduced by . showed that this can be interpreted as taking the meromorphic continuation of a convergent integral.
If the Cauchy principal value integral
Note that the symbols
and
are used here to denote Cauchy principal value and Hadamard finite-part integrals respectively.
The Hadamard finite part integral above (for) may also be given by the following equivalent definitions:\mathcal\int_a^b \frac\, dt = \lim_ \left\,\mathcal\int_a^b \frac\, dt = \lim_ \left\.
The definitions above may be derived by assuming that the function is differentiable infinitely many times at, that is, by assuming that can be represented by its Taylor series about . For details, see . (Note that the term in the second equivalent definition above is missing in but this is corrected in the errata sheet of the book.)
Integral equations containing Hadamard finite part integrals (with 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\int_^1 \frac \, dt = \left(\lim_ \int_^ \frac \, dt \right) + \left(\lim_ \int_^ \frac \, dt \right) = \lim_ \left(-\frac - 1 \right) + \lim_ \left(-1 + \frac\right)= +\inftyIts Cauchy principal value also diverges since\mathcal \int_^1 \frac \, dt = \lim_ \left(\int_^ \frac \, dt + \int_^1 \frac \, dt \right) = \lim_ \left(\frac - 1 - 1 + \frac \right) = +\infty To assign a finite value to this divergent integral, we may consider\mathcal\int_^1 \frac \, dt = \mathcal \int_^1 \frac \, dt \Bigg|_ = \frac\left(\mathcal \int_^1 \frac \, dt \right) \Bigg|_The inner Cauchy principal value is given by\mathcal \int_^1 \frac \, dt = \lim_ \left(\int_^ \frac \, dt + \int_^1 \frac \, dt \right) = \lim_ \left(\ln\left| \frac \right| + \ln \left| \frac \right| \right) = \ln\left| \frac \right|Therefore,\mathcal\int_^1 \frac \, dt = \frac\left(\ln\left| \frac \right| \right) \Bigg|_ = \frac\Bigg|_ = -2Note that this value does not represent the area under the curve, which is clearly always positive.
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