= Solution
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. Set
$$
F_T(z)=\int_0^T f(t)e^{-tz}\,dt,\qquad |f(t)|\leq M
$$
almost 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 $-\delta\leq\operatorname{Re}z\leq0$, $|\operatorname{Im}z|\leq R$ 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> with
$$
K_T(z)=e^{Tz}\left(1+\frac{z^2}{R^2}\right)\frac1z.
$$
The <residue theorem> gives
$$
F(0)-F_T(0)=\frac1{2\pi i}
\left[\int_{C_+}(F-F_T)K_T\,dz
+\int_L FK_T\,dz-\int_L F_TK_T\,dz\right].
$$
On $C_+$, with $x=\operatorname{Re}z>0$,
$$
|e^{Tz}(F-F_T)(z)|\leq M/x,\qquad
\left|1+z^2/R^2\right|=2x/R.
$$
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 $x<0$,
$$
|e^{Tz}F_T(z)|
\leq M\int_0^T e^{(T-t)x}\,dt\leq M/|x|.
$$
The same circle factor gives another bound $M/R$. Consequently
$$
\limsup_{T\to\infty}|F_T(0)-F(0)|\leq\frac{2M}{R}.
$$
The radius $R$ is arbitrary, so
$$
\boxed{\int_0^\infty f(t)\,dt=\lim_{T\to\infty}F_T(0)=F(0).}
$$
This 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. \b[Analyticity only in the open right half-plane is insufficient.]
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