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The displayed kernel has residue one at zero and vanishes at the endpoints of a radius-$R$ imaginary-axis segment. On either radius-$R$ semicircle, $|1+z^2/R^2|=2|\mathrm{Re}z|/R$, cancelling the inverse real-part bound on a bounded function's Laplace transform tail. Use the full transform only on a right semicircle and a thin analytic leftward path; deform the finite-interval entire transform to the large left semicircle. This proves the Newman Tauberian theorem without assuming analytic continuation across an entire left half-disk.

Let $M=\sup |f|$ and $F_T(z)={\int }_0^{T}f(t)e^{-tz}dt$. Each $F_T$ is an entire function: on any compact set of $z$, differentiation under this finite-interval integral is justified by a bounded integrable majorant. We prove $\boxed{F_T(0)\to F(0)}$, which gives ordinary convergence of the improper integral, not just an Abel limit.

Fix $R>0$. Since $\Omega$ contains the imaginary segment from $-iR$ to $iR$ and is open, choose $0<\delta <R$ so small that its closed left strip of width $\delta$ belongs to $\Omega$. Let $C_{+}$ be the right semicircle $|z|=R$ oriented from $-iR$ to $iR$, and let $\Gamma$ be the left half-ellipse $z=\delta \cos \theta +iR\sin \theta$, $\pi /2\le \theta \le 3\pi /2$, oriented back to $-iR$. The enclosed right half-disk and thin left region lie in $\Omega$. The residue theorem, applied to the analytic difference with its simple pole at zero, givesFor $x=\mathrm{Re}z>0$,On the right circle, $|1+z^2/R^2|=2x/R$ and $|z|=R$. The exponential cancels the displayed damping, so the absolute integrand per unit arc length is at most $2M/R^2$. The right-arc contribution is therefore at most $M/R$ after the factor $1/(2\pi )$.

For the $F_T$ term on $\Gamma$, deform that left path to the left semicircle $C_{-}$ of radius $R$. The Cauchy integral theorem allows this because $F_T$ is an entire function and the region between the paths excludes the pole at zero. On $C_{-}$, $x<0$ andThe same factor $|1+z^2/R^2|=2|x|/R$ bounds this contribution by $M/R$. The arc endpoints can be omitted from the estimates, since they have zero arc length and the integrands are continuous there.

Finally, the $F$ term on the original $\Gamma$ tends to zero as $T\to \infty$. This compact path stays away from zero, so $F(z)(1+z^2/R^2)/z$ is bounded there. Its interior has $\mathrm{Re}z<0$, hence $e^{Tz}\to 0$, and $|e^{Tz}|\le 1$. Dominated convergence on the finite-length path gives the claim. Combining the three pieces yieldsSince $R$ is arbitrary, the difference tends to zero. This proves the Newman Tauberian theorem in the required setting; the proof crucially uses analytic continuation across every finite segment of the imaginary axis.

The printed differential in the transform integral is $dz$; it must be $dt$, since $z$ is the transform parameter. We prove the Newman Tauberian theorem in this corrected interpretation. Setalmost everywhere. The finite-interval transform $F_T$ is entire.

Fix $R>0$. Because the analytic domain contains the whole imaginary axis, compactness supplies a $\delta >0$ such that the thin rectangle $-\delta \le \mathrm{Re}z\le 0$, $|\mathrm{Im}z|\le R$ lies in the domain. Let $C_{+}$ be the right semicircle of radius $R$, oriented from $-iR$ to $iR$. Join its endpoints by a leftward path $L$ along the other three sides of that rectangle. This is a closed positively oriented contour.

Use contour damping for bounded Laplace transforms withThe residue theorem givesOn $C_{+}$, with $x=\mathrm{Re}z>0$,Thus the integrand has modulus at most $2M/R^2$, and this half-circle contributes at most $M/R$ after division by $2\pi$.

On $L$, the $F$ term tends to zero as $T\to \infty$: every interior point of the path has negative real part, $F$ is bounded on this fixed compact path, and the remaining kernel factor is bounded because the path avoids zero. Dominated convergence applies, with the two endpoints irrelevant to the path integral.

For the $F_T$ term, deform $L$ to the left semicircle $C_{-}$ of radius $R$. This deformation uses only the entire function $F_T$; it does not demand that $F$ extend across a large left half-disk. Both paths lie to the left of zero and their enclosed deformation region avoids the kernel pole. On $C_{-}$, for $x<0$,The same circle factor gives another bound $M/R$. ConsequentlyThe radius $R$ is arbitrary, soThis proves convergence of the ordinary improper integral, not merely a damped limit.

For the weakened domain hypothesis, take $f(t)=1$. Its Laplace transform is $F(z)=1/z$, analytic on the open right half-plane, but ${\int }_0^{T}f(t)dt=T$ diverges. Analyticity only in the open right half-plane is insufficient.