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Id' appreciate help understanding why the integral

$ \int_0^1 x^{\lambda} [ \: \phi(x) - \phi(0)\: ] dx $

is convergent provided $\lambda > -2$, where $\phi \in \mathcal{D}(\mathbb{R})$.

To provide some context: this integral arises in the regularization of the (divergent) integral of $x^{\lambda}_+$ ; i.e.

$ \langle x^{\lambda}_+ , \phi \rangle = \int_0^\infty x^{\lambda} \phi \: dx $

By analytic continuation, this integral can be expressed as

$ \int_0^1 x^{\lambda} [ \: \phi(x) - \phi(0)\: ] dx + \int_1^\infty x^{\lambda} \: \phi(x) \: dx \: + \: \frac{\phi(0)}{\lambda + 1} $

The following texts all state that the first integral is convergent provided $\lambda > -2$, but its not obvious to me how it does.

  1. Generalized Functions, Volume 1 by Gelfand and Shilov (1964) -- page 47 & 48
  2. Theory of Distributions by M. A. Al-Gwaiz (1992), page 64
  3. Asymptotic approximation of integrals by R. Wong (2001) -- page 258

1 Answers 1

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We have for $0\leq x\leq 1$: \begin{align*} \left|x^{\lambda}(\phi(x)-\phi(0))\right|&=\left|x^{\lambda}\int_0^x\phi'(t)dt\right|\\ &=\left|x^{\lambda}\left(x\phi'(x)-\int_0^xt\phi''(t)dt\right)\right|\\ &\leq x^{\lambda+1}|\phi'(x)|+x^{\lambda}\int_0^xt\phi''(t)dt\\ &\leq x^{\lambda+1}\sup_{t\in\mathbb R}|\phi''(t)|+x^{\lambda+1}\int_0^x|\phi''(t)|dt \\ &\leq x^{\lambda+1}\sup_{t\in\mathbb R}|\phi''(t)|+x^{\lambda+2}\sup_{t\in\mathbb R}|\phi'(t)| \\ &\leq x^{\lambda+1}\left(\sup_{t\in\mathbb R}|\phi''(t)|+\sup_{t\in\mathbb R}|\phi'(t)|\right) \end{align*} Since $\lambda+1>-1$ and \phi' and \phi'' are bounded, the integral is (absolutely) convergent.

  • 0
    BTW, I've just posted a follow up to this question at: http://math.stackexchan$g$e.com/questions/100979/simplifyin$g$-the-$g$eneralized-function-x-lambda-in-the-strip-n-1-mbo2012-01-21