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\begin{document}

\begin{lemma}
Suppose $N\ge 0$, $h\ge 2$ is an integer, and
$\min (x,x+hy)>0$. Then 
\begin{multline}
\int^y_0\dots\int^y_0 C(x+y_1+\dots +y_h)\,dy_1\dots dy_{h} 
\\
=L(0,\sym^2f)y^h+\sum_{n\le N}c_n
   \int^y_0\ldots\int^y_0 I_{k,0}(n,x+y_1+\ldots +y_h)\,dy_1\ldots dy_{h} 
\\
+\sum^h_{l=0}(-1)^{h-\ell} \binom{h}{\ell} 
   \sum_{n>N}c_nI_{k,h}(n,x+\ell y). 
\end{multline}
\begin{multline}
\int^y_0\dots\int^y_0 C(x+y_1+\dots +y_h)\,dy_1\dots dy_{h} 
\\
=L(0,\sym^2f)y^h+\sum_{n\le N}c_n
   \int^y_0\ldots\int^y_0 I_{k,0}(n,x+y_1+\ldots +y_h)\,dy_1\ldots dy_{h} 
+\sum^h_{l=0}(-1)^{h-\ell} 
\end{multline}
%\setlength{\multlinegap}{0pt}
\begin{multline*}
\int^y_0\dots\int^y_0 C(x+y_1+\dots +y_h)\,dy_1\dots dy_{h} 
\\
=L(0,\sym^2f)y^h+\sum_{n\le N}c_n
   \int^y_0\ldots\int^y_0 I_{k,0}(n,x+y_1+\ldots +y_h)\,dy_1\ldots dy_{h} 
\\
+\sum^h_{l=0}(-1)^{h-\ell} \binom{h}{\ell} 
   \sum_{n>N}c_nI_{k,h}(n,x+\ell y). 
\end{multline*}
\begin{multline*}
\int^y_0\dots\int^y_0 C(x+y_1+\dots +y_h)\,dy_1\dots dy_{h} 
\\
\begin{aligned}
=L(0,\sym^2f)y^h&+\sum_{n\le N}c_n
   \int^y_0\ldots\int^y_0 I_{k,0}(n,x+y_1+\ldots +y_h)\,dy_1\ldots dy_{h} 
\\
&+\sum^h_{l=0}(-1)^{h-\ell} \binom{h}{\ell} 
   \sum_{n>N}c_nI_{k,h}(n,x+\ell y). 
\end{aligned}
\end{multline*}
\begin{align*}
\int^y_0\dots\int^y_0 C(x&+y_1+\dots +y_h)\,dy_1\dots dy_{h} 
\\
&\begin{aligned}
=L(0,\sym^2f)y^h&+\sum_{n\le N}c_n
   \int^y_0\ldots\int^y_0 I_{k,0}(n,x+y_1+\ldots +y_h)\,dy_1\ldots dy_{h} 
\\
&+\sum^h_{l=0}(-1)^{h-\ell} \binom{h}{\ell} 
   \sum_{n>N}c_nI_{k,h}(n,x+\ell y). 
\end{aligned}
\end{align*}
\end{lemma}

\begin{proof} 
We denote by $C$, $C'$, $C''$ the oriented polygonal
paths with the vertices
\begin{align}
C&:&2-i\infty&,  &   \, 2-i&,  &  \, -h-\frac 12-i&, &\, -h-\frac 12+i&, &\, 
2+i&, &\,    2+i\infty &;
\\
C'&:&-\frac{1}{12}-i\infty &, & \,-\frac{1}{12}-i&, & \, -h-\frac 12-i&, &\,
   -h-\frac 12+i&, &\, -\frac{1}{12}+i&, &\, -\frac{1}{12}+i\infty &;
\end{align}
Using the well-known formula 
$$ \frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}
   \frac{u^{-s}\Gamma(s)}{\Gamma(s+q+1)}\,ds=\begin{cases}
   \frac{(1-u)^q}{\Gamma(q+1)}&\text{if}\quad 0<u\le 1,\\
   0 & \text{if}\quad u>1,   \end{cases} $$
where $q>0$ and $c>0$ (see \cite{15}), we can write 
\begin{gather*}
\int^x_0C(u)du=\int^x_0\bigg(\sum_{n\le u}c_n\bigg)\, du= 
   \sum_{n\le x}c_n(x-n),\\
\sum^{\infty}_{n=1}\frac{c_n}{2\pi i}\int^{2+i\infty}_{2-i\infty}
   \frac{x^{s+1}}{n^s}\frac{1}{s(s+1)}\, ds=
   \frac{1}{2\pi i}\int^{2+i\infty}_{2-i\infty}
   L(s,\sym^2f)\frac{x^{s+1}}{s(s+1)}\,ds;
\end{gather*}
\begin{align*} 
\int^y_0C(x+y_1)dy_1&=\int^{x+y}_{0}C(u)du-\int^x_0C(u)\,du \\
&= \frac{1}{2\pi i} \int^{2+i\infty}_{2-i\infty}
   L(s,\sym^2f) \frac{(x+y)^{s+1}-x^{s+1}}{s(s+1)} \,ds \\[1pt]
&= L(0,\sym^2f)y+\frac{1}{2\pi i} \int_C L(s,\sym^2f)
   \frac{(x+y)^{s+1}-x^{s+1}}{s(s+1)} \,ds. 
\end{align*} 
Similarly,
\begin{align}
\int^y_0\ldots\int^y_0 &C(x+y_1+\dots +y_h)dy_1\ldots 
   dy_h-L(0,\sym^2f)y^h 
\\
&=\frac{1}{2\pi i} \int_C L(s,\sym^2f)\bigg\{ \int^y_0\dots \int^y_0
   \big[(x+y+y_2+\dots +y_h)^{s+1} 
\notag
\\
&\qquad\qquad 
- (x+y_2+\dots +y_h)^{s+1}\big]\, dy_2\dots dy_h   
   \bigg\} \frac{1}{s(s+1)}  \,ds 
\\
& =\sum^h_{\ell=0} (-1)^{h-\ell}\binom{h}{\ell}
   \frac{1}{2\pi i} \int_{C} L(s,\sym^2f)(x+\ell y)^{s+h}
   \frac{\Gamma (s)}{\Gamma(s+h+1)}\,ds.
\notag
%   \eqno{(2.3)} 
\end{align}
\begin{equation}
\begin{aligned}
\int^y_0\ldots\int^y_0 &C(x+y_1+\dots +y_h)dy_1\ldots 
   dy_h-L(0,\sym^2f)y^h 
\\
&=\frac{1}{2\pi i} \int_C L(s,\sym^2f)\bigg\{ \int^y_0\dots \int^y_0
   \big[(x+y+y_2+\dots +y_h)^{s+1} 
\\
&\qquad\qquad 
- (x+y_2+\dots +y_h)^{s+1}\big]\, dy_2\dots dy_h   
   \bigg\} \frac{1}{s(s+1)}  \,ds 
\\
& =\sum^h_{\ell=0} (-1)^{h-\ell}\binom{h}{\ell}
   \frac{1}{2\pi i} \int_{C} L(s,\sym^2f)(x+\ell y)^{s+h}
   \frac{\Gamma (s)}{\Gamma(s+h+1)}\,ds.
%   \eqno{(2.3)} 
\end{aligned}
\end{equation}
To evaluate  the integral in (2.3), we use the functional equation (2.2)
and the estimate
$$ \int^{2+it}_{-\frac{1}{12}+it}L(s,\sym^2f)u^{s+h}
   \frac{\Gamma (s)}{\Gamma(s+h+1)}\,ds =o(1). $$
We have
$$ \frac{1}{2\pi i}\int_{C}L(s,\sym^2f)u^{s+h} 
   \frac{\Gamma (s)}{\Gamma (s+h+1)} \,ds 
   =\frac{1}{2\pi i}\int_{C'}L(s,\sym^2f)u^{s+h}
   \frac{\Gamma (s)}{\Gamma (s+h+1)} \,ds $$
$$ =\frac{1}{2\pi i}\int_{C''}L(1-s,\sym^2f)u^{h+1-s}
   \frac{\Gamma (1-s)}{\Gamma (h+2-s)} \,ds $$
$$ =\int_{C''}L(s,\sym^2f)u^{h+1-s}f(k,s)
   \frac{\Gamma (1-s)}{\Gamma (h+2-s)} \,ds $$
$$ = \sum^{\infty}_{n=1}c_n  \int_{C''}\frac{u^{h+1-s}}{n^s}f(k,s)
   \frac{\Gamma (1-s)}{\Gamma (h+2-s)} \,ds $$
$$ = \sum^{\infty}_{n=1}c_n  \int_{C_h}\frac{u^{h+1-s}}{n^s}f(k,s)
   \frac{\Gamma (1-s)}{\Gamma (h+2-s)} \,ds 
   = \sum^{\infty}_{n=1}c_nI_{k,h} (n,u),
   \eqno{(2.4)} $$
where we have also used the estimate
$$ \int^{\frac{13}{12}+it}_{it}
   \frac{u^{h+1-s}}{n^s}f(k,s) \frac{\Gamma(1-s)}{\Gamma(h+2-s)}\,ds=o(1). $$
For $j=0,1,2,\ldots$, if $\min (u,u+y)>0$, then
$$ \int^{y}_{0}I_{k,j}(n,u+y_1)dy_1 
   =\int^y_0\left[ \int_{C_j}\frac{(u+y_1)^{j+1-s}}{n^s}f(k,s)
   \frac{\Gamma(1-s)}{\Gamma(j+2-s)}\,ds\right]dy_1 $$
$$ =\int^y_0\left[ \int_{C_{j+1}}\frac{(u+y_1)^{j+1-s}}{n^s}f(k,s)
   \frac{\Gamma(1-s)}{\Gamma(j+2-s)}\,ds\right]dy_1 $$
$$ =\int_{C_{j+1}}\frac{(u+y)^{j+2-s}-u^{j+2-s}}{n^s}f(k,s)
   \frac{\Gamma(1-s)}{\Gamma(j+3-s)}\,ds 
   =I_{k,j+1}(n,u+y)-I_{k,j+1}(n,u).
   \eqno{(2.5)} $$
With the help of (2.5), we prove the relation
$$ \int^y_0\ldots\int^y_{0} I_{k,0}(n,x+y_1+\dots+y_h)\, dy_1\ldots dy_h $$
$$ = \sum^h_{\ell=0}(-1)^{h-\ell} \binom{h}{\ell} I_{k,h} (n,x+\ell y).
   \eqno{(2.6)} $$
The lemma follows from (2.3), (2.4), and (2.6). 
\end{proof}

\begin{lemma} 
We have 
$$ I_{k,j}(n,x)= b_j \frac{x^{\frac 23j+\frac 13}}{n^{\frac 13j+\frac 23}}
   \cos\Big(6\pi(nx)^{\frac 13}-\frac{\pi}{2}j\Big)
   + O\Big(\frac{x^{\frac 23j}}{n^{\frac 13j+1}}\Big) \quad (n=1,2,3,\ldots),
$$  
where 
$$ b_j=\frac{1}{2^j\pi^{j+1}3^{\frac 12}}. $$
\end{lemma}

If we take the functional equation for  $L(s,\sym^2f)$ in the form
(2.2), then the proof of Lemma 2 is completely similar to that of
Theorem  2 in \cite[I]{18}.

We also note that, in a different form, Lemma 2 was formulated  in the
author's paper \cite{6} as a particular case of a general result by 
Hafner; see Theorem~A, Theorem B,  and Lemma 2.1 in Hafner's paper
\cite{21}.

Now we pass to the proof of (2.1). Suppose $x\le X\le 2x$,
$1\le u\le x$, and $h\ge 2$ is an integer. By Lemmas ~1 and
2, relation (2.5), and the inequality $L(0, \sym^2f)>0$, we have
\begin{align*}
\int^u_0&\left\{ \int^y_0\dots\int^y_0 C (X+y_1+\dots+y_h)dy_1\dots dy_h
   \right\}\, dy 
\\
&=\int^u_0\bigg\{L(0,\sym^2f)y^h+\sum^h_{l=0}(-1)^{h-l}
   \binom{h}{\ell}\sum^{\infty}_{n=1} c_nI_{k,h} (n,X+\ell y)\bigg\}\, dy 
\\
&=\frac{1}{h+1}L(0,\sym^2f)u^{h+1}
  + u\bigg\{(-1)^hI_{k,h} (1,X)+(-1)^h\sum^{\infty}_{n=2} c_nI_{k,h}(n,X)\bigg\}
\\
&\qquad \qquad\qquad
+ \sum^{h}_{\ell =1}\frac{(-1)^{h-\ell}}{\ell}\binom{h}{\ell}
   \sum^{\infty}_{n=1} c_n\big\{ I_{k,h+1}(n,X+\ell u)-I_{k,h+1}(n,X)\big\}
\\
&\ge u\bigg\{(-1)^hb_hX^{\frac 13+\frac 23 h}\cos \Big(6\pi X^{\frac 13}-
  \frac{\pi}{2}h\Big)-b_h X^{\frac 13 +\frac 23 h}\sum^{\infty}_{n=2}
  \frac{|c_n|}{n^{\frac 23+\frac 13 h}} + O(x^{\frac 23 h})\bigg\}
\\
&\qquad\qquad\qquad
- \bigg\{\sum^{h}_{\ell=1}\binom{h}{\ell}
   \sum^{\infty}_{n=1} \frac{|c_n|}{n^{\frac 23+\frac 13(h+1)}}2b_{h+1}
   (X+\ell u)^{\frac 13+\frac 23(h+1)} + O(x^{\frac23(h+1)})\bigg\}
\\
&\ge X^{\frac13+\frac23(h+1)}
   \bigg\{ \frac{u}{X^{\frac23}} b_h \Big[1+(-1)^h\cos\Big(6\pi X^{\frac 13}
   -\frac{\pi}{2}h\Big) - \zeta^3\Big(\frac23+\frac13h\Big)\Big] 
\\
&\qquad\qquad\qquad
-b_{h+1}2^{h+1}(h+1)^{h+1}\zeta^3\Big(1+\frac 13h\Big)
   + O (x^{-\frac13-\frac23}u) + O(x^{-\frac 13})\bigg\}. 
\end{align*}
For natural $m$, let $X_m$ denote the root of the equation
$6X^{1/3}-h/2=m$; we have 
$$ 3\big(X^{\frac13}_{m+2}-X^{\frac13}_{m}\big)=1,\quad X_{m+2}-X_m\sim
  X^{\frac23}_{m}. $$
For $x\gg 1$ there always exists $m$ such that
$$ x+1\le X_m< X_{m+1} \le x+2x^{\frac23}. $$
Out of two numbers $X_m$ and $X_{m+1}$, belonging to the interval
$[x+1,x+2x^{2/3}]$, we choose the number (to be denoted by $X_q$)
for which  
$$ (-1)^h\cos \Big(6\pi X^{\frac13}_q-\frac{\pi}{2}h\Big)=1. $$
We let $h\ge 2$ satisfy the condition
$$ \zeta^3\Big(\frac23+\frac13 h\Big)< 2. $$
Set $u=c'x^{2/3}$, where $c'$ is a positive constant such that 
$$ c'b_h\Big\{2-\zeta^3\Big(\frac23+\frac13 h\Big)\Big\}
   > b_{h+1}2^{h+1}(h+1)^{h+1}\zeta^3\Big(1+\frac13 h\Big). $$
We have proved the inequality 
$$ \int^{c'x^{2/3}}_{0}\bigg\{\int^y_0 \ldots \int^y_0
   C(X_q+y_1+\ldots y_h)dy_1\ldots dy_h\bigg\}\, dy \gg
   x^{\frac13+\frac23(h+1)}. 
   \eqno{(2.7)} $$
Set $\widetilde c=hc'+2$. Using (2.7), we obtain
$$ x^{\frac13+\frac23(h+1)}\ll \int^{\widetilde cx^{2/3}}_{0} |C(x+y_1)|\, dy_1
   \int^{c'x^{2/3}}_{0} \bigg\{ \int^y_0\, dy_2\ldots
   \int^y_0dy_h\bigg\}\, dy $$
$$ \ll x^{\frac23 h}\int^{x+\widetilde c x^{2/3}}_{x} |C(y)|\, dy, $$
which implies the inequality 
$$ x\ll \int^{x+\widetilde c x^{2/3}}_{x} |C(y)|dy. $$
Next, we have 
$$ \int^x_0|C(y)|\, dy \ge
   \sum^{[\frac12 \widetilde c^{-1}x^{1/3}]}_{j=0} \int^{\frac
   x2+(j+1)\widetilde{c}x^{2/3}}_{\frac x2+j\widetilde{c}x^{2/3}} |C(y)|\, dy 
   \gg x^{\frac 43}. $$
We have prove that 
$$ \int^x_0|C(y)|dy \ge K_3x^{\frac 43} $$
if $x>x_1$ and $x_1>0$ is sufficiently large. Since $c_1=1$,
for $2\le x\le x_1$ we have 
$$ \int^x_0|C(y)|\, dy \ge \int^2_0|C(y)|\, dy =1. $$
It follows that 
$$ \int^x_0|C(y)| dy\ge K_4 x^{\frac 43} 
   \eqno{(2.8)} $$
for $x\ge 2$. Using (2.8) and the Cauchy inequality, we arrive at
(2.1).

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\end{document}
