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:


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


Your suspicion is correct, $GL_n$ has two components, and $\det$ may be used to show there are at least two of them. The other direction is slightly more involved and requires linear algebra rather than topology. Here is a sketch of how to do this:

i) If $b$ is any vector, let $R_b$ denote the reflection through the hyperplane perpendicular to $b$. These are all reflections. Any two reflections $R_a, R_b$ with $a, b$ linear independent can be joined by a path consisting of reflections, namely $R_{ta+ (1-t)b}, t\in[0,1]$.  

ii) Any $X\in O^+(n)$ (orthogonal matrices with positive determinant) is the product of an even number of reflections. Since matrix multiplication is continuous $O(n)\times O(n) \rightarrow O(n)$ and by i) you can join any product $R_a R_b$ with $R_a R_a = Id$ it follows that $O^+(n)$ is connected.

iii) $\det$ shows $O(n)$ is not connected.

iv) $O^-(n) = R O^+ (n)$ for any reflection $R$. Hence $O^-(n)$ is connected.

v) Any $ X\in GL_n$ is the product $AO$ of a positive matrix $A$ and $O \in O(n)$ (polar decomposition). Now you only need to show that the positive matrices are connected, which can be shown again using convex combination with $Id$. This proves the claim.
