TITLE: Showing that $\int_0^1 x^{\lambda} [ \: \phi(x) - \phi(0)\: ] dx$ is convergent for $\lambda > -2$
QUESTION [6 upvotes]: 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.

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

REPLY [2 votes]: 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.