两个1/x类的广义函数

【转载请注明出处】http://www.cnblogs.com/mashiqi

2017/04/15

1、$ ext{p.v.}\,frac{1}{x}$

因为$(x ln x - x)' = ln x$， 所以$int_0^a ln x mathrm{\,d}x = lim_{epsilon o 0^+} int_epsilon^a ln x mathrm{\,d}x = lim_{epsilon o 0^+}(x ln x - x)ig|_epsilon^a = a ln a - a$，即是说对任意的$varphi in C_c^infty(mathbb{R}^1)$， $langle ln|x|, varphi angle$有意义，且连续性也显而易见。所以$ln|x| in mathcal{D}'(mathbb{R}^1)$。所以$(ln|x|)' in mathcal{D}'(mathbb{R}^1)$。将$(ln|x|)'$定义为$ ext{p.v.}\,frac{1}{x}$：$$oxed{ ext{p.v.}\,frac{1}{x} riangleq (ln|x|)'. quad ext{Then}quad ext{p.v.}\,frac{1}{x} in mathcal{D}'(mathbb{R}^1).}$$

2、$frac{1}{x pm i0}$

设${f_n} subset mathcal{D}'(mathbb{R}^1)$，则${f_n'} subset mathcal{D}'(mathbb{R}^1)$。若存在$f inmathcal{D}'(mathbb{R}^1)$使得$f_n o f ext{ in }mathcal{D}'(mathbb{R}^1)$，则$lim_{n o +infty} f_n'$存在且$f_n' o f' ext{ in }mathcal{D}'(mathbb{R}^1)$.

当$epsilon > 0$时，$ln(x + iepsilon) = ln|x + iepsilon| + i cdot arg(x + iepsilon)$。所以 $frac{1}{x + iepsilon} = ig( ln(x + iepsilon) ig)' in mathcal{D}'(mathbb{R}_x^1), \,forall epsilon > 0$。因为$lim_{epsilon o 0^+} ln(x + iepsilon) = ln|x| - ipi cdot (H(x)-1) in mathcal{D}'(mathbb{R}_x^1)$，所以在$mathcal{D}'(mathbb{R}_x^1)$上$lim_{epsilon o 0^+} frac{1}{x + iepsilon}$存在且

egin{align*}
lim_{epsilon o 0^+} frac{1}{x + iepsilon} & = lim_{epsilon o 0^+} ig( ln(x + iepsilon) ig)'
= ig( lim_{epsilon o 0^+} ln(x + iepsilon) ig)' \
& = ig( ln|x| - ipi cdot (H(x)-1) ig)' \
& = ext{p.v.}\,frac{1}{x} - ipidelta(x).
end{align*}

现在将$frac{1}{x + i0}$定义为$lim_{epsilon o 0^+} frac{1}{x + iepsilon}$：$$oxed{frac{1}{x + i0} riangleq lim_{epsilon o 0^+} frac{1}{x + iepsilon}. quad ext{Then}quad frac{1}{x + i0} = ext{p.v.}\,frac{1}{x} - ipidelta(x) in mathcal{D}'(mathbb{R}^1).}$$ 同理，对正负的$i0$，我们有$$frac{1}{x pm i0} = ext{p.v.}\,frac{1}{x} mp ipidelta(x) in mathcal{D}'(mathbb{R}^1).$$