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Is there a way to define the integral of say $f(x)=-1/x^2$ between 0 and y so that it is safely valued as $F(y)=1/y$?

Put another way, usually the Cauchy integral $F(x)$ is defined symmetrically around the pathological point, say from $-y$ to $y$ of $f(x)$, and is equal to 0, on the basis that the integration point at 0 is somewhat discarded. Loosely, $F(x)$ is made of two halves $F(x)=1/x$ and $F(-x)=-1/x$ that cancel each other.

Is there an alternative way to define either function $f(x)$ or $F(x)$ in question on the positive half-line only, using for example Dirac or unit step functions, in order to obtain a finite $F(x)$ value and to arrive at $F(y)=\int_0^y f(x)dx = 1/y$ for $y>0$?

My inclination would be to define $F(x)$ for example like $F(x)=\mathbb{1}(x>0)/x$. However, if I derive such $F(x)$, I get $f(x)=\delta(x)/x -\mathbb{1}(x>0)/x^2$. The first part integrates back to 0 (it seems) but what is the integral of the second part?

Additional question, in this example, can I in turn calculate the integral of $-f(x)/F(x)$ from 0 to $y$ to get $\ln(y)$?

Douar Gwenn
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1 Answers1

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Yes there is a way to do this with Hadamard regularization. Using the definition $$ {\displaystyle {\mathcal {H}}\int _{a}^{b}{\frac {f(t)}{(t-x)^{2}}}\,dt=\lim _{\varepsilon \to 0^{+}}\left\{\int _{a}^{x-\varepsilon }{\frac {f(t)}{(t-x)^{2}}}\,dt+\int _{x+\varepsilon }^{b}{\frac {f(t)}{(t-x)^{2}}}\,dt-{\frac {f(x+\varepsilon )+f(x-\varepsilon )}{\varepsilon }}\right\},} $$ we write $$ \begin{aligned} \mathcal H\int_0^y(-1)\frac{\mathrm dx}{x^2} &=-\lim_{\epsilon\to0^+}\left(\int_\epsilon^y\frac{\mathrm dx}{x^2}-\frac{1}{\epsilon}\right)\\ &=-\lim_{\epsilon\to0^+}\left(-\frac{1}{x}\bigg|_{x=\epsilon}^y-\frac{1}{\epsilon}\right)\\ &=-\lim_{\epsilon\to0^+}\left(-\frac{1}{y}+\frac{1}{\epsilon}-\frac{1}{\epsilon}\right)\\ &=\frac{1}{y}. \end{aligned} $$

Aaron Hendrickson
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