Simplifying the generalized function $x^{\lambda}_+$ in the strip $-n - 1 < \mbox{Re}\lambda < -n$

Question

Note: this post is a follow up to an earlier question.

The (divergent) integral of $x^{\lambda}_+$ can be analytically continued into the region Re $\lambda > -n - 1$, $\lambda \ne -1, -2 , \ldots , -n$ by the expression

$$ \begin{align*} \int_0^\infty x^{\lambda} \phi \: dx &= \int_0^1 x^{\lambda} \left[ \phi(x) - \phi(0) - x \phi^\prime(0) - \cdots -\frac{x^{n - 1}}{(n - 1)!}\phi^{(n - 1)}(0) \right] dx \\\ &+ \int_1^\infty x^\lambda \phi(x) dx + \sum_{k = 1}^n \frac{\phi^{(k - 1)}(0)}{(k - 1)! (\lambda + k)}. \end{align*} $$

This is achieved by a repeatedly applying the first $n$ terms of the Taylor expansion of $\phi$ at $x = 0$, by a generalization of the method discussed in my earlier question.

According to the texts that I'm studying, in any strip of the form $-n - 1 < \mbox{Re}\lambda < -n$ the above equation can be written in the simple form:

$$ \langle x^{\lambda}_+ , \phi \rangle = \int_0^\infty x^{\lambda} \left[ \phi(x) - \phi(0) - x \phi^\prime(0) - \cdots -\frac{x^{n - 1}}{(n - 1)!}\phi^{(n - 1)}(0) \right] dx. $$

It seems like a trivial step, as none of the texts that I'm studying bother to explain it; However, I've spent exactly one week trying to figure out this step but I'm nowhere near doing so.

Reference texts:

1. Generalized Functions, Volume 1 by Gelfand and Shilov (1964) -- page 47 & 48
2. Asymptotic approximation of integrals by R. Wong (2001) -- page 258
3. Generalized Functions: theory and technique by Ram P. kanwal (1998) -- page 86

Answer 1

Because $$ \int_1^{\infty} x^{\lambda}dx \; = \; -\frac{1}{\lambda + k} $$ It follows that $$ \sum_{k = 1}^{n} \frac{\phi^{(k - 1)}(0)}{(k - 1)! (\lambda + k)} \; = \; \sum_{k = 1}^{n} \frac{-\phi^{(k - 1)}(0)}{(k - 1)!} \int_1^{\infty} x^{\lambda + k - 1} dx $$ Therefore $$ \int_1^\infty x^\lambda \phi(x) dx \; + \; \sum_{k = 1}^n \frac{\phi^{(k - 1)}(0)}{(k - 1)! (\lambda + k)} \; = \; \int_1^{\infty} x^{\lambda} \left[ \phi(x) - \sum_{k = 1}^{n} \frac{\phi^{(k - 1)}(0) \, x^{k - 1}}{(k - 1)!} \right] dx $$ Adding this integral to the integral from $0$ to $1$ completes the proof.

Answer 2

We have \begin{align*} \int_0^{+\infty}x^{\lambda}\left(\phi(x)-\sum_{j=0}^{n-1}\frac{\phi^{(j)}(0)}{j!}x^j\right)dx &=\int_0^1x^{\lambda}\left(\phi(x)-\sum_{j=0}^{n-1}\frac{\phi^{(j)}(0)}{j!}x^j\right)dx \\ &+\int_1^{+\infty}x^{\lambda}\phi(x)dx-\sum_{j=0}^{n-1}\frac{\phi^{(j)}(0)}{j!}\int_1^{+\infty}x^{\lambda+j}dx\\ &=\int_0^1x^{\lambda}\left(\phi(x)-\sum_{j=0}^{n-1}\frac{\phi^{(j)}(0)}{j!}x^j\right)dx\\ &+\int_1^{+\infty}x^{\lambda}\phi(x)dx-\sum_{j=0}^{n-1}\frac{\phi^{(j)}(0)}{j!}\frac 1{\lambda+j+1}\\ &=\int_0^1x^{\lambda}\left(\phi(x)-\sum_{j=0}^{n-1}\frac{\phi^{(j)}(0)}{j!}x^j\right)dx\\ &+\int_1^{+\infty}x^{\lambda}\phi(x)dx-\sum_{k=1}^n\frac{\phi^{(k-1)}(0)}{(k-1)!}\frac 1{\lambda+k}.\\ \end{align*} The integrals $\int_1^{+\infty}x^{\lambda+j}dx$ are absolutely convergent since $|x^{\lambda+j}|=x^{\operatorname{Re}\lambda+j}\leq x^{\operatorname{Re}\lambda+n-1}$ and the integral $\int_1^{+\infty}x^{\operatorname{Re}\lambda+n-1}dx$ is convergent since $\operatorname{Re}\lambda+n-1<-1$.