Question on page 339 of Shiryaev's Probability Suppose that for each $n\geq1$ there is a given sequence of independent random variables
$$\xi_{n1},\xi_{n2},\ldots,\xi_{nn}$$
with $E\xi_{nk}=0$, $V\xi_{nk}=\sigma_{nk}^2$, $\sum_{k=1}^n\sigma_{nk}^2=1$. Let $S_n=\xi_{n1}+\ldots+\xi_{nn},$
$$F_{nk}(x)=P\{\xi_{nk}\leq x\},\ \Phi(x)=(2\pi)^{-1/2}\int_{-\infty}^x e^{-y^2/2}dy,\ \Phi_{nk}(x)=\Phi\left(\frac{x}{\sigma_{nk}}\right)$$
On page 339 there is an inequality:
$$\sum_{k=1}^n\left| t\int_{-\infty}^{\infty}(e^{itx}-1-itx)(F_{nk}(x)-\Phi_{nk}(x))dx\right|\leq\frac{|t|^3}{2}\varepsilon\sum_{k=1}^n\int_{|x|\leq\varepsilon}|x||F_{nk}(x)-\Phi_{nk}(x)|dx+2t^2\sum_{k=1}^n\int_{|x|>\varepsilon}|x||F_{nk}(x)-\Phi_{nk}(x)|dx$$
How can I get it?
Then the book says we can use
$$E|\xi|^n=\int_{-\infty}^{\infty}|x|^ndF(x)=n\int_0^\infty x^{n-1}[1-F(x)+F(-x)]dx$$
to prove
$$\int_{|x|\leq\varepsilon}|x||F_{nk}(x)-\Phi_{nk}(x)|dx\leq2\sigma_{nk}^2$$
But I can't figure why.
Thank you!
 A: We have that $|e^{itx}-1-itx|\leqslant \frac{|tx|^2}2$, which can be obtained by Taylor's formula, and this gives an upper bound for $\int_{-\varepsilon}^{\varepsilon}$. 
For the other part, note that 
$$|e^{itx}-1-itx|\leqslant|e^{itx}-1|+|tx|\leqslant\left|\int_0^{|tx|}ie^{is}ds\right|+|tx|\leqslant 2|tx|,$$
and we can control $\int_{\{|x|>\varepsilon\}}$.
A: For the second question, rewrite
$$
\mathbb E|\xi|^n=\int_{-\infty}^{\infty}|x|^n\mathrm dF(x)=\int_{0}^{\infty}x^n\mathrm dF(x)+
\int_{0}^{\infty}x^n\mathrm dG(x),
$$
with $G:x\mapsto-F(-x)$ and use an integration by parts formula to compute
$$
\int_0^{+\infty}u(x)\mathrm dv(x),\qquad u:x\mapsto x^n,\quad v=F+G-1.$$
To use this identity, note that for any random variables $\zeta_1$ and $\zeta_2$ with PDF $G_1$ and $G_2$,
$$
\int_{-\infty}^{+\infty}|x|\,|G_1(x)-G_2(x)|\mathrm dx\leqslant(*),
$$
with
$$
(*)=\int_{-\infty}^0(-x)\,(G_1(x)+G_2(x))\mathrm dx+\int_0^{+\infty}x\,(1-G_1(x)+1-G_2(x))\mathrm dx.
$$
One sees that
$(*)=A(G_1)+A(G_2)$, with
$$
A(G)=\int_0^{+\infty}x\,(1-G(x)+G(-x))\mathrm dx.
$$
Applying the first identity one gets $A(G)=\frac12\mathbb E(\zeta^2)$ for every random variable $\zeta$ with PDF $G$, hence $(*)=\frac12(\mathbb E(\zeta_1^2)+\mathbb E(\zeta_2^2))$. In particular, the inequality mentioned in the post holds, without the restriction to $|x|\leqslant\varepsilon$ (this was expected) and without the factor $2$ on the RHS (this was less expected).
