the values of $R$ are those in the interval $[R_m,\infty)$. If we take $S<1$, the values of $R$ will be $(0,\infty)$. Nonetheless, unlike $R$, the domain of $r$ is always $ (r_0,\infty)$. As we shall see later, there is a throat at $r_m$ that “separates” the regions $(r_0,r_m)$ and $(r_m,\infty)$.

The wormhole in Wyman’s solution {#1032014b}
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Coordinate $w$ {#20022014a}
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In dealing with wormholes, it is sometimes interesting to work with a coordinate that does not posses coordinate singularities. Although $r$ is not singular in the interval $(r_0,\infty)$, let us define a coordinate $w$ analogous to $l$ through the integral $$w=\int_{r_m}^{r} W^{-S/2}dr, \label{18122013a}$$ where it is clear that $w=0$ corresponds to $r_m$ (the throat). We call $A$ the region with negative values of $w$ and $B$ the other region.

Since the analytic solution of (\[18122013a\]) for an arbitrary $S$ may not exist, consider the following expansion for the integrand. $$\begin{aligned}
(1-r_0/r)^{-S/2}= 1+\frac{Sr_0}{2}\frac{1}{r}+  \sum_{n=2}^{\infty} \frac{r_0^n}{2^n n!}r^{-n}\prod_{j=1}^{n}(S+2j-2). \label{15092013d}\end{aligned}$$ By substituting the expansion (\[15092013d\]) into the integral (\[18122013a\]), we obtain the expression $$\begin{aligned}
w= r-r_m+\frac{Sr_0}{2}\ln(r/r_m) +\sum_{n=2}^{\infty} \frac{r_0^n}{2^n n!(n-1)}\left(  r_m^{1-n}-r^{1-n} \right)   \prod_{j=1}^{n}(S+2j-2). \label{18122013b}\end{aligned}$$ From the ratio test, one can easily prove that the series above converges for $r>r_0$ and $S>1$, but Raabe’s test 