[proofplan]
We approximate the sharp cutoff $\mathbb{1}_{[0,T]}$ by Newman's contour kernel, whose Fourier-side factor vanishes at the two endpoints of a finite vertical segment. The vertical side of the contour recovers the smoothed truncated integral, while the part of the contour lying just to the left of the imaginary axis tends to zero as $T \to \infty$ because $e^{Tz}$ decays there. A uniform kernel estimate shows that the smoothed truncated integral differs from the ordinary partial integral by at most a constant multiple of $R^{-1}$, with a constant depending only on $\|A\|_{L^\infty}$. Letting first $T \to \infty$ and then $R \to \infty$ gives the asserted convergence and identifies the limit with $F(0)$.
[/proofplan]
[step:Fix the extension and the contour scale]
Let
\begin{align*}
M := \|A\|_{L^\infty([0,\infty))}.
\end{align*}
Since $A \in L^\infty([0,\infty))$, this number is finite and $|A(t)| \leq M$ for $\mathcal{L}^1$-a.e. $t \geq 0$.
Fix $R>0$. Define the holomorphy domain
\begin{align*}
\Omega := \{s \in \mathbb{C}:\operatorname{Re}(s)>0\}\cup U.
\end{align*}
Because $U$ is open and contains the compact set
\begin{align*}
S_R := \{iy : -R \leq y \leq R\},
\end{align*}
there exists $\eta_R>0$ such that
\begin{align*}
\{z \in \mathbb{C} : |\operatorname{Re}(z)| < \eta_R,\ |\operatorname{Im}(z)| \leq R\} \subset U.
\end{align*}
In estimates below, $C_R$ denotes a positive finite constant depending only on $R$; its value may increase from one displayed estimate to the next.
Let $\Gamma_R:[0,1]\to U$ be the three-sided path consisting of the top horizontal segment from $iR$ to $-\eta_R/2+iR$, the vertical segment from $-\eta_R/2+iR$ to $-\eta_R/2-iR$, and the bottom horizontal segment from $-\eta_R/2-iR$ to $-iR$. Thus
\begin{align*}
\Gamma_R(0)=iR,
\qquad
\Gamma_R(1)=-iR,
\end{align*}
and $\operatorname{Re}(\Gamma_R(\theta))<0$ except at the two endpoints. For every $0<\delta<\eta_R$, the left part of the strip below lies in $U$ by the preceding strip inclusion, while the part with positive real part lies in $\{s:\operatorname{Re}(s)>0\}$. Hence the whole region enclosed by the contour below is contained in $\Omega$:
\begin{align*}
\{z\in\mathbb C:-\eta_R/2\leq\operatorname{Re}(z)\leq\delta,\ |\operatorname{Im}(z)|\leq R\}.
\end{align*}
For $\delta>0$ small enough that $\delta<\eta_R$, define the vertical segment
\begin{align*}
V_{R,\delta} : [-R,R] &\to \mathbb{C} \\
y &\mapsto \delta+iy.
\end{align*}
Let $C_{R,\delta}$ be the positively oriented contour consisting of $V_{R,\delta}$ traversed upward from $\delta-iR$ to $\delta+iR$, a short connector to $iR$, then $\Gamma_R$, and then a short connector from $-iR$ back to $\delta-iR$. The two short connectors lie in $\Omega$ and their lengths tend to $0$ as $\delta\downarrow0$.
For $T>0$, define the [meromorphic function](/page/Meromorphic%20Function)
\begin{align*}
H_{T,R}: \Omega \setminus \{0\} &\to \mathbb{C} \\
z &\mapsto F(z)e^{Tz}\frac{1+z^2/R^2}{z}.
\end{align*}
The point $0$ lies in $\Omega$, but it is removed from the domain of $H_{T,R}$ by the factor $z^{-1}$. Thus the only singularity of $H_{T,R}$ inside $C_{R,\delta}$ is the simple pole at $z=0$, and its residue is $F(0)$ because
\begin{align*}
\lim_{z\to0}e^{Tz}\left(1+\frac{z^2}{R^2}\right)=1.
\end{align*}
Hence the residue theorem gives
\begin{align*}
F(0)
=
\frac{1}{2\pi i}\int_{C_{R,\delta}}F(z)e^{Tz}\frac{1+z^2/R^2}{z}\,dz.
\end{align*}
Letting $\delta \downarrow 0$ removes the connectors. Indeed, the connectors lie in small discs about $iR$ and $-iR$, hence for all sufficiently small $\delta$ they stay a positive distance from $0$. On those discs the quotient by $z$ is bounded, $F$ is continuous, and the factor $1+z^2/R^2$ vanishes at the endpoint on $S_R$; therefore the connector integrands are uniformly bounded while their lengths tend to $0$. Thus
\begin{align*}
F(0)
=
\lim_{\delta\downarrow 0}
\frac{1}{2\pi i}\int_{-R}^{R}
F(\delta+iy)e^{T(\delta+iy)}
\frac{1+(\delta+iy)^2/R^2}{\delta+iy}
\,i\,d\mathcal{L}^1(y)
+
E_R(T),
\end{align*}
where
\begin{align*}
E_R(T)
:=
\frac{1}{2\pi i}\int_{\Gamma_R}
F(z)e^{Tz}\frac{1+z^2/R^2}{z}\,dz.
\end{align*}
[/step]
[step:Show that the left contour contribution vanishes as $T \to \infty$]
The map
\begin{align*}
z \mapsto F(z)\frac{1+z^2/R^2}{z}
\end{align*}
is continuous on $\Gamma_R([0,1])$: the only possible singularity is at $z=0$, but $\Gamma_R$ lies strictly in the left half-plane except possibly at its endpoints $\pm iR$, and the endpoint factor $1+z^2/R^2$ vanishes there. Therefore there exists a finite constant
\begin{align*}
B_R := \sup_{z \in \Gamma_R([0,1])}\left|F(z)\frac{1+z^2/R^2}{z}\right|.
\end{align*}
Let $L_R$ denote the length of $\Gamma_R$. Since $\operatorname{Re}(\Gamma_R(\theta))<0$ for $0<\theta<1$, we have
\begin{align*}
e^{\,T\operatorname{Re}(\Gamma_R(\theta))} \to 0
\end{align*}
for every $0<\theta<1$ as $T \to \infty$. Also
\begin{align*}
\left|F(\Gamma_R(\theta))e^{\,T\Gamma_R(\theta)}
\frac{1+\Gamma_R(\theta)^2/R^2}{\Gamma_R(\theta)}
\Gamma_R'(\theta)\right|
\leq
B_R|\Gamma_R'(\theta)|
\end{align*}
for $\mathcal{L}^1$-a.e. $\theta \in [0,1]$, and $B_R|\Gamma_R'|$ is integrable on $[0,1]$. By dominated convergence,
\begin{align*}
\lim_{T\to\infty}E_R(T)=0.
\end{align*}
[/step]
[step:Identify the vertical integral with a smoothed truncation]
For $R>0$, define Newman's kernel
\begin{align*}
K_R:\mathbb{R} &\to \mathbb{R} \\
u &\mapsto \frac{1}{2}+\frac{1}{\pi}\int_0^R
\frac{\sin(yu)}{y}\left(1-\frac{y^2}{R^2}\right)\,d\mathcal{L}^1(y).
\end{align*}
The value of $K_R(0)$ is irrelevant for the integrals below, since single points have $\mathcal{L}^1$-measure zero.
For $\delta>0$, substitute the Laplace representation of $F(\delta+iy)$ into the vertical integral. Since $|A(t)|e^{-\delta t}$ is integrable on $[0,\infty)$ and the $y$-interval is compact, [Fubini's theorem](/theorems/2961) applies and gives
\begin{align*}
&\frac{1}{2\pi i}\int_{-R}^{R}
F(\delta+iy)e^{T(\delta+iy)}
\frac{1+(\delta+iy)^2/R^2}{\delta+iy}
\,i\,d\mathcal{L}^1(y) \\
&\qquad =
\int_0^\infty A(t)
\left[
\frac{1}{2\pi i}\int_{-R}^{R}
e^{(\delta+iy)(T-t)}
\frac{1+(\delta+iy)^2/R^2}{\delta+iy}
\,i\,d\mathcal{L}^1(y)
\right]d\mathcal{L}^1(t).
\end{align*}
For $u \in \mathbb{R}$ and $\delta>0$, define the Abel-regularized Newman kernel as the map
\begin{align*}
Q_{R,\delta}:\mathbb{R} &\to \mathbb{C} \\
u &\mapsto
\frac{1}{2\pi i}\int_{-R}^{R}
e^{(\delta+iy)u}
\frac{1+(\delta+iy)^2/R^2}{\delta+iy}
\,i\,d\mathcal{L}^1(y).
\end{align*}
Then the preceding identity says that the vertical integral equals
\begin{align*}
\int_0^\infty A(t)Q_{R,\delta}(T-t)\,d\mathcal{L}^1(t).
\end{align*}
We use the following Abel limit estimate instead of pointwise dominated convergence in the singular $y$-integral.
[claim:The Abel-regularized Newman kernels converge after integration against bounded functions]
For every $R>0$, every $T>0$, and every bounded locally integrable map $A:[0,\infty)\to\mathbb{C}$,
\begin{align*}
\lim_{\delta\downarrow 0}
\int_0^\infty A(t)Q_{R,\delta}(T-t)\,d\mathcal{L}^1(t)
=
\int_0^\infty A(t)K_R(T-t)\,d\mathcal{L}^1(t).
\end{align*}
[/claim]
[proof]
Define
\begin{align*}
w_\delta:[-R,R]&\to\mathbb C\\
y&\mapsto 1+\frac{(\delta+iy)^2}{R^2},
\end{align*}
and put
\begin{align*}
u:=T-t.
\end{align*}
For a complex number $a$ with $\operatorname{Re}(a)>0$ one has the two elementary Abel identities
\begin{align*}
\frac{e^{au}}{a}
&=
\int_{-\infty}^{u}e^{av}\,d\mathcal{L}^1(v) && (u<0),\\
\frac{e^{au}}{a}
&=
\frac{1}{a}+\int_{0}^{u}e^{av}\,d\mathcal{L}^1(v) && (u>0).
\end{align*}
Applying these with $a=\delta+iy$ separates the singular factor. The residue contribution in the case $u>0$ is
\begin{align*}
C_{R,\delta}
:=
\frac{1}{2\pi}\int_{-R}^{R}\frac{w_\delta(y)}{\delta+iy}\,d\mathcal{L}^1(y).
\end{align*}
Since
\begin{align*}
\frac{1}{\delta+iy}
=
\frac{\delta}{\delta^2+y^2}
-
i\frac{y}{\delta^2+y^2},
\end{align*}
the odd imaginary part has integral $0$ on the symmetric interval, while
\begin{align*}
\frac{1}{2\pi}\int_{-R}^{R}\frac{\delta}{\delta^2+y^2}\,d\mathcal{L}^1(y)\to\frac12.
\end{align*}
The remaining contribution is
\begin{align*}
\frac{1}{2\pi R^2}\int_{-R}^{R}(\delta+iy)\,d\mathcal{L}^1(y)
=
\frac{\delta}{\pi R},
\end{align*}
which tends to $0$. Hence the half-residue that survives in the limit is exactly the constant $1/2$ in $K_R$.
For $u\neq0$, the preceding identities give the pointwise boundary value
\begin{align*}
\lim_{\delta\downarrow0}Q_{R,\delta}(u)
&=
\frac12+\frac{1}{2\pi}
\lim_{\varepsilon\downarrow0}
\int_{\varepsilon\leq |y|\leq R}
e^{iyu}\left(\frac{1}{iy}+\frac{iy}{R^2}\right)
\,d\mathcal{L}^1(y)\\
&=
\frac12+\frac1\pi\int_0^R
\frac{\sin(yu)}{y}\left(1-\frac{y^2}{R^2}\right)
\,d\mathcal{L}^1(y)\\
&=K_R(u).
\end{align*}
This derives explicitly both the half-residue and the factor $1-y^2/R^2$.
It remains to justify passage through the $t$-integral. Split
\begin{align*}
w_\delta=w_0+r_\delta,
\end{align*}
where
\begin{align*}
w_0:[-R,R]&\to\mathbb C&
y&\mapsto 1-\frac{y^2}{R^2},
\end{align*}
and
\begin{align*}
r_\delta:[-R,R]&\to\mathbb C&
y&\mapsto \frac{2i\delta y}{R^2}+\frac{\delta^2}{R^2}.
\end{align*}
The main term $w_0$ vanishes at $y=\pm R$. For $u< -1$, the first Abel identity gives
\begin{align*}
Q_{R,\delta}(u)
=
\frac{1}{2\pi}
\int_{-\infty}^{u}e^{\delta v}
\left(\int_{-R}^{R}e^{iyv}w_\delta(y)\,d\mathcal{L}^1(y)\right)
d\mathcal{L}^1(v).
\end{align*}
For the $w_0$ part we use the oscillation at the endpoints, not merely the absolute bound for its [Fourier transform](/page/Fourier%20Transform). A direct [integration by parts](/theorems/2098) calculation gives, for $v\neq0$,
\begin{align*}
\int_{-R}^{R}e^{iyv}w_0(y)\,d\mathcal{L}^1(y)
&=
\frac{4(\sin(Rv)-Rv\cos(Rv))}{R^2v^3}.
\end{align*}
Indeed, this formula follows by evaluating the even integral
\begin{align*}
\int_{-R}^{R}e^{iyv}\left(1-\frac{y^2}{R^2}\right)\,d\mathcal{L}^1(y)
=
2\int_0^R\cos(yv)\left(1-\frac{y^2}{R^2}\right)\,d\mathcal{L}^1(y).
\end{align*}
Put $x=t-T$. If $x>1$, then $u=T-t=-x$ and the $w_0$ contribution is bounded by a constant multiple of
\begin{align*}
\left|
\int_x^\infty e^{-\delta w}
\left(
\frac{\sin(Rw)}{R^2w^3}
-
\frac{\cos(Rw)}{Rw^2}
\right)d\mathcal{L}^1(w)
\right|.
\end{align*}
The sine term is bounded absolutely by $C_Rx^{-2}$. For the cosine term, [integration by parts](/theorems/210) gives
\begin{align*}
\left|
\int_x^\infty e^{-\delta w}\frac{\cos(Rw)}{w^2}\,d\mathcal{L}^1(w)
\right|
&\leq
\frac{1}{Rx^2}
+
\frac{1}{R}\int_x^\infty
\left(\frac{\delta e^{-\delta w}}{w^2}+\frac{2e^{-\delta w}}{w^3}\right)d\mathcal{L}^1(w)\\
&\leq
\frac{C_R}{x^2},
\end{align*}
because
\begin{align*}
\int_x^\infty\frac{\delta e^{-\delta w}}{w^2}\,d\mathcal{L}^1(w)
\leq
\frac{1}{x^2}
\int_x^\infty\delta e^{-\delta w}\,d\mathcal{L}^1(w)
\leq
\frac{1}{x^2}.
\end{align*}
Thus the $w_0$ part satisfies the uniform tail estimate
\begin{align*}
\left|
\frac{1}{2\pi}
\int_{-\infty}^{\,T-t}e^{\delta v}
\int_{-R}^{R}e^{iyv}w_0(y)\,d\mathcal{L}^1(y)
d\mathcal{L}^1(v)
\right|
\leq
\frac{C_R}{(t-T)^2}
\end{align*}
for $t>T+1$ and all $\delta>0$.
It remains to handle the small remainder $r_\delta$. For $v\neq0$, direct integration gives
\begin{align*}
\int_{-R}^{R}e^{iyv}r_\delta(y)\,d\mathcal{L}^1(y)
&=
\frac{4\delta}{R^2}\frac{Rv\cos(Rv)-\sin(Rv)}{v^2}
+
\frac{2\delta^2}{R^2}\frac{\sin(Rv)}{v}.
\end{align*}
For $x>1$, put
\begin{align*}
J_\delta(x)
:=
\int_x^\infty e^{-\delta w}
\left(
\frac{\delta\cos(Rw)}{w}
+
\frac{\delta\sin(Rw)}{w^2}
+
\frac{\delta^2\sin(Rw)}{w}
\right)d\mathcal{L}^1(w).
\end{align*}
The $r_\delta$ contribution for $t>T+1$ is bounded by $C_R|J_\delta(t-T)|$. For the first term in $J_\delta$, integration by parts gives
\begin{align*}
\left|
\int_x^\infty e^{-\delta w}\frac{\delta\cos(Rw)}{w}\,d\mathcal{L}^1(w)
\right|
&\leq
C_R\frac{\delta e^{-\delta x}}{x}
+
C_R\int_x^\infty
\left(
\frac{\delta^2e^{-\delta w}}{w}
+
\frac{\delta e^{-\delta w}}{w^2}
\right)d\mathcal{L}^1(w).
\end{align*}
The other two terms in $J_\delta$ are bounded absolutely by
\begin{align*}
C_R\int_x^\infty
\left(
\frac{\delta e^{-\delta w}}{w^2}
+
\frac{\delta^2e^{-\delta w}}{w}
\right)d\mathcal{L}^1(w).
\end{align*}
Integrating these bounds in $x\in[1,\infty)$ and reversing the order of integration gives
\begin{align*}
\int_1^\infty |J_\delta(x)|\,d\mathcal{L}^1(x)
\leq
C_R\delta\left(1+\log\frac{1}{\delta}\right),
\end{align*}
for $0<\delta<1/2$. This tends to $0$ as $\delta\downarrow0$. Hence the remainder contributes nothing to the limit in $L^1([T+1,\infty),\mathcal{L}^1)$, while the $w_0$ part has the uniform tail majorant just proved. To control the bounded interval $0\leq t\leq T+1$, note first that
\begin{align*}
\left|\frac{r_\delta(y)}{\delta+iy}\right|
&=
\left|\frac{\delta(\delta+2iy)}{R^2(\delta+iy)}\right|
\leq
\frac{2\delta}{R^2}
\end{align*}
for every $y\in[-R,R]$. Hence the $r_\delta$ part of $Q_{R,\delta}(u)$ is bounded by $C_{R,T}\delta$ whenever $-1\leq u\leq T$. The $w_0$ part is also bounded on this same interval: for $u>0$ it is the bounded residue term plus an integral over $0\leq v\leq u$, and for $-1\leq u<0$ it is the tail integral over $-\infty<v\leq u$ whose integrand is bounded near finite $v$ and is bounded by $C_Rv^{-2}$ for $|v|\geq1$. Therefore, after increasing $C_1$, we have
\begin{align*}
|Q_{R,\delta}(T-t)|
&\leq C_1 &&\text{for }0\leq t\leq T+1.
\end{align*}
Hence
\begin{align*}
t\mapsto M C_1\mathbb{1}_{[0,T+1]}(t)+\frac{M C_R}{(t-T)^2}\mathbb{1}_{(T+1,\infty)}(t)
\end{align*}
belongs to $L^1([0,\infty),\mathcal{L}^1)$ and dominates the $A(t)$ times the $w_0$ part of $Q_{R,\delta}(T-t)$ for all sufficiently small $\delta$, while the $r_\delta$ part has $L^1$ norm tending to $0$. For each fixed $u\neq 0$, the Abel identity gives the one-sided boundary value
\begin{align*}
\lim_{\delta\downarrow 0}Q_{R,\delta}(u)=K_R(u),
\end{align*}
where the half-residue at $u=0$ is immaterial for the $t$-integral because $\{T\}$ has $\mathcal{L}^1$-measure zero. Applying the [dominated convergence theorem](/theorems/4) to the $w_0$ part in the $t$ variable and the displayed $L^1$ estimate to the $r_\delta$ part proves the claim.
[/proof]
Applying the claim gives
\begin{align*}
\lim_{\delta\downarrow 0}
\frac{1}{2\pi i}\int_{-R}^{R}
F(\delta+iy)e^{T(\delta+iy)}
\frac{1+(\delta+iy)^2/R^2}{\delta+iy}
\,i\,d\mathcal{L}^1(y)
=
\int_0^\infty A(t)K_R(T-t)\,d\mathcal{L}^1(t).
\end{align*}
Combining this identity with the contour formula yields
\begin{align*}
F(0)=
\int_0^\infty A(t)K_R(T-t)\,d\mathcal{L}^1(t)
+
E_R(T).
\end{align*}
[/step]
[step:Estimate Newman's kernel against the sharp cutoff]
[claim:Newman's kernel approximates the Heaviside cutoff with integrable error]
There exists an absolute constant $C_0>0$ such that, for every $R>0$ and every $u \neq 0$,
\begin{align*}
\left|K_R(u)-\mathbb{1}_{(0,\infty)}(u)\right|
\leq
C_0\min\left\{1,\frac{1}{R^2u^2}\right\}.
\end{align*}
[/claim]
[proof]
The identity
\begin{align*}
\int_0^\infty \frac{\sin v}{v}\,d\mathcal{L}^1(v)=\frac{\pi}{2}
\end{align*}
gives, for $u>0$ and $x:=Ru$,
\begin{align*}
1-K_R(u)
&=
\frac{1}{\pi}
\left[
\int_R^\infty \frac{\sin(yu)}{y}\,d\mathcal{L}^1(y)
+
\frac{1}{R^2}\int_0^R y\sin(yu)\,d\mathcal{L}^1(y)
\right] \\
&=
\frac{1}{\pi}
\left[
\int_x^\infty \frac{\sin v}{v}\,d\mathcal{L}^1(v)
+
\frac{1}{x^2}\int_0^x v\sin v\,d\mathcal{L}^1(v)
\right].
\end{align*}
Since
\begin{align*}
\int_0^x v\sin v\,d\mathcal{L}^1(v)=\sin x-x\cos x,
\end{align*}
and two integrations by parts give
\begin{align*}
\int_x^\infty \frac{\sin v}{v}\,d\mathcal{L}^1(v)
=
\frac{\cos x}{x}+\frac{\sin x}{x^2}
-
2\int_x^\infty \frac{\sin v}{v^3}\,d\mathcal{L}^1(v),
\end{align*}
we obtain the estimate
\begin{align*}
\left|1-K_R(u)\right|
\leq
\frac{1}{\pi}
\left(
\frac{2}{x^2}
+
2\int_x^\infty \frac{1}{v^3}\,d\mathcal{L}^1(v)
\right).
\end{align*}
Therefore, for $x>0$,
\begin{align*}
|1-K_R(u)|
\leq
\frac{1}{\pi}
\left(
\frac{2}{x^2}
+
\frac{1}{x^2}
\right)
=
\frac{3}{\pi R^2u^2}.
\end{align*}
The bound by an absolute constant follows from the displayed $R^{-2}u^{-2}$ estimate when $|Ru|\geq 1$. When $0<|Ru|<1$, the definition of $K_R$ gives
\begin{align*}
\left|K_R(u)-\mathbb{1}_{(0,\infty)}(u)\right|
&\leq
1+\frac{1}{\pi}\int_0^R
\frac{|\sin(yu)|}{y}\left(1-\frac{y^2}{R^2}\right)d\mathcal{L}^1(y)\\
&\leq
1+\frac{|u|}{\pi}\int_0^R
\left(1-\frac{y^2}{R^2}\right)d\mathcal{L}^1(y)\\
&\leq
1+\frac{2}{3\pi}.
\end{align*}
For $u<0$, the identity $K_R(-u)=1-K_R(u)$ reduces the estimate to the case $u>0$. These estimates prove the claim with any constant $C_0$ satisfying
\begin{align*}
C_0\geq4.
\end{align*}
[/proof]
Using the claim and the bound $|A(t)|\leq M$, we estimate for every $T>0$:
\begin{align*}
&\left|
\int_0^\infty A(t)K_R(T-t)\,d\mathcal{L}^1(t)
-
\int_0^{\,T} A(t)\,d\mathcal{L}^1(t)
\right| \\
&\qquad \leq
M\int_0^\infty
\left|K_R(T-t)-\mathbb{1}_{(0,\infty)}(T-t)\right|
\,d\mathcal{L}^1(t) \\
&\qquad \leq
M C_0\int_{-\infty}^{\infty}
\min\left\{1,\frac{1}{R^2u^2}\right\}
\,d\mathcal{L}^1(u).
\end{align*}
The last integral is explicit:
\begin{align*}
\int_{-\infty}^{\infty}
\min\left\{1,\frac{1}{R^2u^2}\right\}
\,d\mathcal{L}^1(u)
=
\int_{|u|\leq 1/R}1\,d\mathcal{L}^1(u)
+
\int_{|u|>1/R}\frac{1}{R^2u^2}\,d\mathcal{L}^1(u)
=
\frac{4}{R}.
\end{align*}
Thus
\begin{align*}
\left|
\int_0^\infty A(t)K_R(T-t)\,d\mathcal{L}^1(t)
-
\int_0^{\,T} A(t)\,d\mathcal{L}^1(t)
\right|
\leq
\frac{4C_0M}{R}.
\end{align*}
[/step]
[step:Let the truncation length tend to infinity and then remove the kernel scale]
From the contour identity and the kernel estimate,
\begin{align*}
\left|
\int_0^{\,T} A(t)\,d\mathcal{L}^1(t)-F(0)
\right|
\leq
\frac{4C_0M}{R}+|E_R(T)|.
\end{align*}
For fixed $R>0$, the previous contour step gives $E_R(T)\to 0$ as $T\to\infty$. Hence
\begin{align*}
\limsup_{T\to\infty}
\left|
\int_0^{\,T} A(t)\,d\mathcal{L}^1(t)-F(0)
\right|
\leq
\frac{4C_0M}{R}.
\end{align*}
Since $R>0$ was arbitrary, letting $R\to\infty$ gives
\begin{align*}
\lim_{T\to\infty}
\int_0^{\,T} A(t)\,d\mathcal{L}^1(t)
=
F(0).
\end{align*}
Therefore the improper integral $\int_0^\infty A(t)\,d\mathcal{L}^1(t)$ converges, and its value is the holomorphic continuation value $F(0)$.
[/step]
