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If $ P(1/x) $ means the principal value of $1/x$ and $ H(x) $ is the Heaviside step function is this then correct (regularization)

$$\operatorname{P.V.}\int_{-\infty}^{\infty}\frac{H(x)f(x)}{x}\mathrm dx=\int_{0}^{\infty}\frac{H(x)f(x)-H(0)f(0)}{x}\mathrm dx.$$

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Sorry, but what is P.V? – Tomás Jun 26 '13 at 13:50
Cauchy's principal value – Jose Garcia Jun 26 '13 at 13:52
Sorry to bother you again, but I can't understand the symbol $P.V\int_{-\infty}^{\infty}\frac{H(x)f(x)}{x}$. In you notation what this mean? – Tomás Jun 26 '13 at 13:56
Multiplication by $H(x)$ amounts to restricting the integration to half-line. This renders P.V. pointless: you just have the ordinary integral $\int_0^\infty \frac{f(x)}{x}\,dx$ (which may or may not converge). – ˈjuː.zɚ79365 Jun 26 '13 at 15:28

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