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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
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1 Answer 1

up vote 2 down vote accepted

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.

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Thanks Davide. I'm unable to justify the change of the limit and variable of the second integral from $\int_0^x$ to $\int_0^1$. Is this change really necessary? –  Olumide Jan 2 '12 at 14:57
    
I don't know, but in any case the bound $\left|x^{\lambda}\int_0^xt\phi''(t)dt\right|\leq x^{\lambda+1}\sup_{t\in\mathbb R}|\phi''(t)|$ is not enough. –  Davide Giraudo Jan 2 '12 at 16:28
    
Sorry, in fact this bound is enough, since $x^{\lambda+1}$ is integrable. –  Davide Giraudo Jan 3 '12 at 15:56
    
Just to clarify. Are you saying that the first two steps (up till the integration by parts) are sufficient, and that the third and fourth steps (integration by substitution) is not required? –  Olumide Jan 4 '12 at 13:41
    
Yes, I will edit it. –  Davide Giraudo Jan 4 '12 at 18:45
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