# Showing that $\int_0^1 x^{\lambda} [ \: \phi(x) - \phi(0)\: ] dx$ is convergent for $\lambda > -2$

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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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.
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