Hadamard regularization

Template:Renormalization and regularization 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 Template:Harvs. Template:Harvs showed that this can be interpreted as taking the meromorphic continuation of a convergent integral.

If the Cauchy principal value integral $$𝒞\underset{a}{\overset{b}{\int }}\frac{f(t)}{t-x}dt(\text{for }a<x<b)$$ exists, then it may be differentiated with respect to Template:Mvar to obtain the Hadamard finite part integral as follows: $$\frac{d}{dx}\left(𝒞\underset{a}{\overset{b}{\int }}\frac{f(t)}{t-x}dt\right)=\mathscr{H}\underset{a}{\overset{b}{\int }}\frac{f(t)}{(t-x{)}^2}dt(\text{for }a<x<b).$$

Note that the symbols $𝒞$ and $\mathscr{H}$ are used here to denote Cauchy principal value and Hadamard finite-part integrals respectively.

The Hadamard finite part integral above (for Template:Math) may also be given by the following equivalent definitions: $$\mathscr{H}\underset{a}{\overset{b}{\int }}\frac{f(t)}{(t-x{)}^2}dt=\underset{\epsilon \to 0^{+}}{\lim }\{\underset{a}{\overset{x-\epsilon }{\int }}\frac{f(t)}{(t-x{)}^2}dt+\underset{x+\epsilon }{\overset{b}{\int }}\frac{f(t)}{(t-x{)}^2}dt-\frac{f(x+\epsilon )+f(x-\epsilon )}{\epsilon }\},$$ $$\mathscr{H}\underset{a}{\overset{b}{\int }}\frac{f(t)}{(t-x{)}^2}dt=\underset{\epsilon \to 0^{+}}{\lim }\{\underset{a}{\overset{b}{\int }}\frac{(t-x{)}^{2}f(t)}{((t-x{)}^2+{\epsilon }^2{)}^2}dt-\frac{\pi f(x)}{2\epsilon }-\frac{f(x)}{2}(\frac{1}{b-x}-\frac{1}{a-x})\}.$$

The definitions above may be derived by assuming that the function Template:Math is differentiable infinitely many times at Template:Math, that is, by assuming that Template:Math can be represented by its Taylor series about Template:Math. For details, see Template:Harvs. (Note that the term Template:Math in the second equivalent definition above is missing in Template:Harvs but this is corrected in the errata sheet of the book.)

Integral equations containing Hadamard finite part integrals (with Template:Math 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 $$\underset{-1}{\overset{1}{\int }}\frac{1}{t^2}dt=\left(\underset{a\to 0^{-}}{\lim }\underset{-1}{\overset{a}{\int }}\frac{1}{t^2}dt\right)+\left(\underset{b\to 0^{+}}{\lim }\underset{b}{\overset{1}{\int }}\frac{1}{t^2}dt\right)=\underset{a\to 0^{-}}{\lim }(-\frac{1}{a}-1)+\underset{b\to 0^{+}}{\lim }(-1+\frac{1}{b})=+\mathrm{\infty }$$ Its Cauchy principal value also diverges since $$𝒞\underset{-1}{\overset{1}{\int }}\frac{1}{t^2}dt=\underset{\epsilon \to 0^{+}}{\lim }(\underset{-1}{\overset{-\epsilon }{\int }}\frac{1}{t^2}dt+\underset{\epsilon }{\overset{1}{\int }}\frac{1}{t^2}dt)=\underset{\epsilon \to 0^{+}}{\lim }(\frac{1}{\epsilon }-1-1+\frac{1}{\epsilon })=+\mathrm{\infty }$$ To assign a finite value to this divergent integral, we may consider $$\mathscr{H}\underset{-1}{\overset{1}{\int }}\frac{1}{t^2}dt=\mathscr{H}\underset{-1}{\overset{1}{\int }}\frac{1}{(t-x{)}^2}dt{|}_{x=0}=\frac{d}{dx}\left(𝒞\underset{-1}{\overset{1}{\int }}\frac{1}{t-x}dt\right){|}_{x=0}$$ The inner Cauchy principal value is given by $$𝒞\underset{-1}{\overset{1}{\int }}\frac{1}{t-x}dt=\underset{\epsilon \to 0^{+}}{\lim }(\underset{-1}{\overset{-\epsilon }{\int }}\frac{1}{t-x}dt+\underset{\epsilon }{\overset{1}{\int }}\frac{1}{t-x}dt)=\underset{\epsilon \to 0^{+}}{\lim }(\ln \left|\frac{\epsilon +x}{1+x}\right|+\ln \left|\frac{1-x}{\epsilon -x}\right|)=\ln \left|\frac{1-x}{1+x}\right|$$ Therefore, $$\mathscr{H}\underset{-1}{\overset{1}{\int }}\frac{1}{t^2}dt=\frac{d}{dx}(\ln \left|\frac{1-x}{1+x}\right|){|}_{x=0}=\frac{2}{x^2-1}{|}_{x=0}=-2$$ Note that this value does not represent the area under the curve Template:Math, 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 $\underset{\epsilon \to 0^{+}}{\lim }(\frac{1}{\epsilon }-1-1+\frac{1}{\epsilon })=+\mathrm{\infty }$

If one removes the infinite components, the pair of $\frac{1}{\epsilon }$ terms, that which remains is $\underset{\epsilon \to 0^{+}}{\lim }(-1-1)=-2$

which equals the value derived above.

References

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