Should the eigenkets be weighted in $|P\rangle = \sum\limits_{r}|\xi^r\rangle$? Page 37 of Dirac's book The Principles of Quantum Mechanics, states

The condition for the eigenstates of $\xi$ to form a complete set must thus be formulated, that any ket $|P\rangle$ can be expressed as an integral plus a sum of eigenkets of $\xi$, i.e.
$$|P\rangle = \int |\xi'c\rangle\; d\xi' + \sum\limits_{r}|\xi^rd\rangle$$
where the $|\xi'c\rangle$, $|\xi^rd\rangle$ are all eigenkets of $\xi$, the labels c and d being inserted to distinguish them when the eigenvalues $\xi'$ and $\xi^r$ are equal, and where the integral is taken over the whole range of eigenvalues and the sum is taken over any selection of them.

I find it inconsistent that the eigenkets under the integral are weighted by what appears to be a differential eigenvalue $d\xi'$, yet all the eigenkets under the summation are weighted by the value 1. Is this correct?
 A: If eigenkets are defined up to arbitrary constants, it is possible to write the sum without any coefficients.
A: The situation can be described more precisely by means of the resolution of
the identity for a self-adjoint operator. Thus let $X=X^{\ast }$ (Dirac's $%
\xi $) be a self-adjoint operator acting in a separable Hilbert space $%
\mathcal{H}$ with empty singular continuous spectrum. Then
\begin{equation*}
X=\int_{\Lambda }\lambda E(d\lambda ),
\end{equation*}
where $\{E(..)\}$ is the resolution of the identity for $X$ and $\Lambda
\subset \mathbb{R}$ its spectrum. In particular $E(\mathbb{R})=I$, the
identity operator. We can decompose into discrete and continuous spectrum
\begin{equation*}
X=\sum_{n}\sum_{m}\lambda _{n}|\varphi _{nm}><\varphi _{nm}|+\int \lambda
E_{cont}(d\lambda ),
\end{equation*}
where the $\varphi _{nm}$ are the orthonormal discrete eigenstates.The
subindex $m$ labels the possible degeneracy of the discrete states (think of
angular momentum in a problem with a spherical potential such as the Coulomb
potential). With the continuous spectrum no square integrable eigenstates
can be associated but in many cases an eigenfunction expansion exists
\begin{equation*}
E_{cont}(d\lambda )=|\psi _{ls}><\psi _{ls}|d\lambda .
\end{equation*}
Think of the plane wave eigenfunctions of the momentum operator
\begin{equation*}
p=\int dkk|\psi _{k}><\psi _{k}|,\;\psi _{k}(x)=<x|\psi _{k}>=(2\pi
)^{-1/2}\exp [ikx],
\end{equation*}
where the $\psi _{k}$'s are $\delta $-function normalised,
\begin{equation*}
\int dx\psi _{k}(x)\overline{\psi _{l}(x)}=\delta (k-l).
\end{equation*}
Note that there can be discrete eigenvalues embedded in the continuous
spectrum. In more complicated cases there can also degeneracies be present
here.
Now consider Dirac's $|P>$.
\begin{eqnarray*}
|P &>&=\int_{\Lambda }E(d\lambda )|P>=\sum_{n}\sum_{m}|\varphi
_{nm}><\varphi _{nm}|P>+\sum_{s}\int_{\Lambda ^{cont}}d\lambda |\psi
_{\lambda s}><\psi _{\lambda s}|P> \\
&\leftrightarrow &\sum_{r}|\xi ^{r}d>+\int d\xi ^{\prime }|\xi ^{\prime }c>.
\end{eqnarray*}
Note that the  $\xi ^{r}d$ $\leftrightarrow <\varphi _{nm}|P>$ are in
general not normalised to unity.
In fact Dirac's approach, although intuitively appealing, is
somewhat imprecise.
