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Motivated by this wiki page, I put my question here:

How to prove $$\lim_{\varepsilon\rightarrow 0^+} \int_a^b \frac{x^2}{x^2+\varepsilon^2} \, \frac{f(x)}{x}dx=p.v.\int_a^b \frac{f(x)}{x}dx$$ where $a<0<b$?

The proof is partially done in words in that article. What I didn't understand is that how to elaborate it mathematical symbols.

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Let $c > 0$ be such that $a < -c < c < b$. Observe that $$\lim_{\epsilon \rightarrow 0} \bigg(\int_a^{-c}{f(x) \over x} {x^2 \over x^2 + \epsilon^2}\,dx + \int_{c}^b {f(x) \over x} {x^2 \over x^2 + \epsilon^2}\,dx\bigg) = \int_a^{-c}{f(x) \over x} \,dx + \int_{c}^b {f(x) \over x}\,dx $$ (We can just plug in $\epsilon = 0$ when we take limits here since the denominators are never zero.) So it suffices to replace $a$ by $-c$ and $b$ by $c$; in other words it suffices to show that $$\lim_{\epsilon \rightarrow 0} \int_{\epsilon < |x| < c} {f(x) \over x} {x^2 \over x^2 + \epsilon^2}\,dx = p.v. \int_{-c}^c {f(x) \over x}$$ Since ${\displaystyle {f(0) \over x}{x^2 \over x^2 + \epsilon^2}}$ is an odd function and the domain of integration is symmetric about $x = 0$, you can subtract it from the integrand on the left-hand side without changing the limit. So this limit is equal to $$\lim_{\epsilon \rightarrow 0} \int_{\epsilon < |x| < c} {f(x) - f(0) \over x} {x^2 \over x^2 + \epsilon^2}\,dx$$ Note that $|{f(x) - f(0) \over x}| < \sup_x|f'(x)|$, and that $|{x^2 \over x^2 + \epsilon^2}| < 1$. So the integrand above is uniformly bounded and one can just take the limit as $\epsilon$ goes to zero, using the dominated convergence theorem for example. The result is $$\int_{|x| < c} {f(x) - f(0) \over x} \,dx$$ $$= \lim_{\epsilon \rightarrow 0} \int_{\epsilon < |x| < c} {f(x) - f(0) \over x} \,dx$$ We can now go in the reverse direction, and add the odd function ${f(0) \over x}$ to the integrand, and the above is equal to $$= \lim_{\epsilon \rightarrow 0} \int_{\epsilon < |x| < c} {f(x) \over x} \,dx$$ $$= p.v. \int_{-c}^c {f(x) \over x}\,dx$$

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Is it generally required that $f\in C^1$ for this equation? I thought not... also, what happenned to the other answer? –  Glen Wheeler May 1 '11 at 15:18

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