Berry-Esseen Smoothing Inequality from Feller Volume 2; Lemma 2 XVI.4 Fellow Feller fans,
I have a question concerning Fellers (An introduction to probability and its Applications, Volume 2 1971) treatment of the Berry-Esseen inequality obtained by smoothing which is Lemma 2 on page 538.
Specifically it is about going from (3.11) to (3.12) on page 538.
To make the question (more or less) self contained denote by $X\sim F_X$, $Y\sim F_Y$ and $Z \sim F_Z$ three absolutely continuous and independent random variables, where $Z$ has a density $f_Z(x) = \frac{1-\cos(Tx)}{\pi T x^2}$ for $T > 0$.
Characteristic functions are denoted by $C_X$, $C_Y$ and $C_Z$.
For example $C_X(\zeta) = \mathbb{E}[\exp(i \zeta X)]$.
With the inverse transformation for densities and $C_{X+Z}(\zeta)=C_X(\zeta)C_Z(\zeta)$ we then have
\begin{align}\tag{3.11}
f_{X+Z}(x) - f_{Y+Z}(x) = \frac{1}{2\pi} \int_{\mathbb{R}} e^{-i \zeta x} \left[C_X(\zeta) - C_Y(\zeta)\right] C_Z(\zeta) d\zeta.
\end{align}
This is (3.11) in Feller on page 538 (where $C_Z(\zeta) = \omega_T(\zeta)I_{[-T,T]}(\zeta)$ in Fellers notation).
The other relevant (in-) equalities are
\begin{align}\tag{3.12}
F_{X+Z}(y) - F_{Y+Z}(y) = \frac{1}{2\pi} \int_{\mathbb{R}} e^{-i \zeta y} \frac{C_X(\zeta) - C_Y(\zeta)}{-i\zeta} C_Z(\zeta) d\zeta.
\end{align}
and just for reference the final goal (in combination with Lemma 1, pager 537)
\begin{align}\tag{3.13}
\left\vert F_{X}(y) - F_{Y}(y) \right\vert \leq \frac{1}{\pi} \int_{-T}^T  \left\vert \frac{C_X(\zeta) - C_Y(\zeta)}{\zeta}\right\vert d\zeta + \frac{24 m}{\pi T},
\end{align}
where $m$ is constant depending only on $F_Y$.
Now to the question.
Starting from (3.11) we are interested in going to (3.12) and hence need to integrate with respect to $x$.
There are now two options:
1) What I would do: Integrate  both sides of (3.11) with respect to $x$. Then we have
\begin{align}
F_{X+Z}(y)-F_{X+Z}(a) - F_{Y+Z}(y)+F_{Y+Z}(a) &= \int_{a}^y f_{X+Z}(x) - f_{Y+Z}(x) dx\\
&= \frac{1}{2\pi} \int_{\mathbb{R}} \frac{ie^{-i \zeta x}}{\zeta} \Big|_{x=a}^{x=y} \left[C_X(\zeta) - C_Y(\zeta)\right] C_Z(\zeta) d\zeta.
\end{align}
On the right the factor $\frac{ie^{-i \zeta x}}{\zeta} \big|_{x=a}^{x=y}$ in the integral is related to $e^{-i \zeta x}$ when integrating. To deal with this factor (because there is no convergence for $a \to -\infty$) take absolute values on both sides, pull them into the integral on the right. Now we have $\left\vert \frac{ie^{-i \zeta x}}{\zeta} \big|_{x=a}^{x=y} \right\vert \leq \frac{2}{\vert \zeta \vert}$ for all $y$ and $a$. Now we can take the limit $a \to -\infty$ to get
\begin{align}\tag{3.12*}
\left\vert F_{X+Z}(y) - F_{Y+Z}(y) \right\vert \leq \frac{1}{2\pi} \int_{\mathbb{R}} 2 \left\vert \frac{C_X(\zeta) - C_Y(\zeta)}{\zeta}\right\vert \vert C_Z(\zeta) \vert d\zeta.
\end{align}
This method does not give (3.12) - the absolute values are already in play, but it is still good enough to get to something like (3.13). However, this yields an additional factor of $2$ that is not present in Fellers treatment.
I have finally
\begin{align}\tag{3.13*}
\left\vert F_{X}(y) - F_{Y}(y) \right\vert \leq \frac{2}{\pi} \int_{-T}^T  \left\vert \frac{C_X(\zeta) - C_Y(\zeta)}{\zeta}\right\vert d\zeta + \frac{24 m}{\pi T}.
\end{align}
2) What Feller does: Essentially guess the anti-derivative of (3.11) and argue that there are no integration constants. Then one gets (3.12) directly ("Integrating with respect to $x$ we obtain" (3.12) in Fellers words) and this saves the factor $2$ that we otherwise buy in. But I don't really understand the argument: "No integration constant appears because both sides tend to $0$ as $\vert x \vert \to \infty$, the left because $F(x) - G(x) \to 0$, the right by the Riemann-Lebesgue lemma 4 of XV, 4."
My question now is how Fellers argument can be formalized (or explained in more detailed).
I would like to get (3.12) instead of (3.12*) to save the factor $2$ in (3.13) versus (3.13*).

It is precisely the Riemann-Lebesgue Lemma which will reduce the constant.
The exponential factor may not converge alone, but the whole integral vanishes:
$$
\int \frac{\mathrm{e}^{-i \zeta a}}{-i\zeta}\cdot (C_X(\zeta)-C_Y(\zeta))C_Z(\zeta)\mbox{d}\zeta = \int {\mathrm{e}^{-i \zeta a}}\cdot \frac{C_X(\zeta)-C_Y(\zeta)}{-i\zeta}C_Z(\zeta)\mbox{d}\zeta =^{\text{R-L Lemma}} o(1) \quad \text{ as } a\to-\infty,
$$
for the lemma we need $R(\zeta)=\frac{C_X(\zeta)-C_Y(\zeta)}{\zeta}$ to be absolutely integrable, fortunately otherwise 3.13 would be trivial.
Then you are left with only one integral
$$
\int {\mathrm{e}^{-i \zeta y}}\cdot \frac{C_X(\zeta)-C_Y(\zeta)}{-i\zeta}C_Z(\zeta)\mbox{d}\zeta 
$$
and get the upper bound $\int |R(\zeta)| C_Z(\zeta)\mbox{d}\zeta $, with the factor of $1$ not $2$.
Note: For a more quantitative form, see The method of cumulants for the normal approximation
