 Solution

Assume $\varphi$ has period $2\pi$, $f>0$ in the regular bulk, and consider a unit proper length in $z$. The proper transverse area element is $dA=fdrd\varphi$, while the mass density is $T^{00}/c^2$. The axial boundary term in the mass of a cylindrical spacetime is thereforeChoose the usual deficit branch $0\le \lambda <1$, so $f_{\mathrm{out}}=(1-\lambda )r$. Continuity of the induced metric requires $f(R^{-})=(1-\lambda )R$. If there is no distributional thin shell at $R$, continuity of the extrinsic curvature also requires $f^{\prime }(R^{-})=1-\lambda$. A jump in $f^{\prime }$ would add a surface-layer stress and cannot simply be discarded.

Using the printed inner condition $f^{\prime }(0^{+})=0$, these assumptions yieldThis is the literal smooth-bulk answer. It is negative for the ordinary deficit range, so it must not be silently replaced by the usual positive cosmic-string mass formula.

For comparison, a regular axis with proper radial coordinate and angular period $2\pi$ requires $f(0)=0$ and $f^{\prime }(0)=1$, since the circumference-to-radius ratio must approach $2\pi$. Under that different regular-axis condition the same integral givesThe original PDF really prints the derivative limit zero. It describes either a nonregular axis or an inner geometry different from an ordinary smooth string core; any additional singular-axis matter is not included in the bulk integral above and would need its own stress specification. Standard regular-core matching can also require an offset in the exterior radial origin. Thus the regular-axis formula is a qualified physical interpretation, not a derivation from the printed zero. For a general exterior branch replace $1-\lambda$ by $|1-\lambda |$ in the boundary derivative.