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\begin{document}
\title{\bf{2 parte }}
\author{Pepe Gimenez}

\date{\today}
\maketitle

\noindent
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\section{An Estimation}
\section{A convergent series for $a_m$}
We now evaluate $Q_m$, under the condition

\begin{equation}
m+max\{m_1,m_2\}>0
\end{equation}
todavia no se muy bien porque, parece ser que sera necesario para
la convergencia de cierta integral

Now if we make the substitution

\begin{equation}
\omega=N^{-2}-i\varphi
\end{equation}

We have,
\begin{equation}\begin{split}
Q_m^t=\sum_{\tiny{\begin{array}{c}
  h,k \\
  0 \leq h < k \leq N\\
  (h,k)=1 \\
\end{array}}}\Omega_{h,k}e^{-2\pi  i h\frac{m}{k}}\frac{1}{i}\int_{N^{-2}-\theta_{h,k}^{''}}^{N^{-2}+\theta_{h,k}^{'}}\Psi_k(k\omega)
P\left(e^{\frac{2\pi}{k}\left(ih'-k^{-1}\omega^{-1}\right)}\right)^t
e^{2\pi  m \omega}d\omega
\end{split}\end{equation}
Now since

\begin{align}
P(x)=\left(%
\begin{array}{c}
  \sum_{\nu=1}^{-min(\beta_1,\beta_2)}a_{-\nu} (1)x^{-\nu} \\
  \sum_{\nu=1}^{-min(\beta_1,\beta_2)}a_{-\nu} (2)x^{-\nu} \\
\end{array}%
\right), &&\Psi_k(z)=z^{r} \left(%
\begin{array}{cc}
  e^{2\pi m_1\frac{z-1/z}{k}} & 0 \\
  0 & e^{2\pi m_2\frac{z-1/z}{k}} \\
\end{array}%
\right)
\end{align}

Then

\begin{equation}
\Psi_k(k\omega)=k^{r} \omega^{r} \left(%
\begin{array}{cc}
  e^{2\pi m_1\omega}e^{-\frac{2\pi m_1}{k^2\omega}} & 0 \\
  0 & e^{2\pi m_2\omega}e^{-\frac{2\pi m_2}{k^2\omega}} \\
\end{array}%
\right)
\end{equation}
then

\begin{eqnarray*}
\lefteqn{Q_m}\\
&&\begin{split}
=\sum_{\tiny{\begin{array}{c}
  h,k \\
  0 \leq h < k \leq N\\
  (h,k)=1 \\
\end{array}}}\Omega_{h,k}e^{-2\pi  i h\frac{m}{k}}\frac{1}{i}\int_{N^{-2}-\theta_{h,k}^{''}}^{N^{-2}+\theta_{h,k}^{'}}k^{r} \omega^{r} \left(%
\begin{array}{cc}
  e^{2\pi m_1\omega}e^{-\frac{2\pi m_1}{k^2\omega}} & 0 \\
  0 & e^{2\pi m_2\omega}e^{-\frac{2\pi m_2}{k^2\omega}} \\
\end{array}%
\right)\\
\left(%
\begin{array}{c}
  \sum_{\nu=1}^{-min(\beta_1,\beta_2)}a_{-\nu} (1)e^\frac{2\pi i h' \nu}{k}e^\frac{2\pi  \nu}{k^2\omega} \\
  \sum_{\nu=1}^{-min(\beta_1,\beta_2)}a_{-\nu} (2)e^\frac{2\pi i h' \nu}{k}e^\frac{2\pi  \nu}{k^2\omega} \\
\end{array}%
\right) e^{2\pi  m \omega}d\omega
\end{split}\\
&&\begin{split} =\sum_{\tiny{\begin{array}{c}
  h,k \\
  0 \leq h < k \leq N\\
  (h,k)=1 \\
\end{array}}}\Omega_{h,k}e^{-2\pi  i h\frac{m}{k}}k^{r}\\
\left(%
\begin{array}{c}
  \sum_{\nu=1}^{-min(\beta_1,\beta_2)}e^\frac{2\pi i h' \nu}{k}\frac{1}{i}\int_{N^{-2}-\theta_{h,k}^{''}}^{N^{-2}+\theta_{h,k}^{'}} \omega^{r}a_{-\nu} (1)e^\frac{2\pi  (\nu-m_1)}{k^2\omega}e^{2\pi  (m+m_1) \omega} d\omega \\
  \sum_{\nu=1}^{-min(\beta_1,\beta_2)}e^\frac{2\pi i h' \nu}{k}\frac{1}{i}\int_{N^{-2}-\theta_{h,k}^{''}}^{N^{-2}+\theta_{h,k}^{'}} \omega^{r}a_{-\nu} (2)e^\frac{2\pi  (\nu-m_2)}{k^2\omega}e^{2\pi  (m+m_2) \omega} d\omega\\
\end{array}%
\right)
\end{split}\\
&&=\sum_{\tiny{\begin{array}{c}
  h,k \\
  0 \leq h < k \leq N\\
  (h,k)=1 \\
\end{array}}}\Omega_{h,k}e^{-2\pi  i h\frac{m}{k}}k^{r}\sum_{\nu=1}^{-min(\beta_1,\beta_2)}e^\frac{2\pi i h'
\nu}{k}\left(%
\begin{array}{cc}
  a_{-\nu} (1) & 0 \\
  0 & a_{-\nu} (2) \\
\end{array}%
\right)I_k(m,\nu)
\end{eqnarray*}

where


\begin{eqnarray*}
\lefteqn{I_k(m,\nu)}\\
&&=\left(%
\begin{array}{c}
  \frac{1}{i}\int_{N^{-2}-i\theta_{h,k}^{''}}^{N^{-2}+i\theta_{h,k}^{'}} \omega^{r}e^\frac{2\pi  (\nu-m_1)}{k^2\omega}e^{2\pi  (m+m_1) \omega} d\omega \\
  \frac{1}{i}\int_{N^{-2}-i\theta_{h,k}^{''}}^{N^{-2}+i\theta_{h,k}^{'}} \omega^{r}e^\frac{2\pi  (\nu-m_2)}{k^2\omega}e^{2\pi  (m+m_2) \omega} d\omega \\
\end{array}%
\right)\\
&&=\left(%
\begin{array}{c}
  I_k^1(m,\nu) \\
  I_k^2(m,\nu) \\
\end{array}%
\right)
\end{eqnarray*}

Now we cut the complex plane from $0$ to $-\infty$ along the
negative real axis, and consider the path shown in the figure
below

\includegraphics[scale=0.3]{path.eps}

Then we can write

\begin{eqnarray*}
\lefteqn{I_k^j(m,\nu)}\\
&&=\frac{1}{i}\int_{-\infty}^{(0+)}-\frac{1}{i}\int_{-\infty}^{-\varepsilon}-\frac{1}{i}\int_{-\varepsilon}^{-\varepsilon-i\theta_{h,k}^{''}}-\frac{1}{i}\int_{-\varepsilon-i\theta_{h,k}^{''}}^{N^{-2}-i\theta_{h,k}^{''}}-\frac{1}{i}\int_{N^{-2}+i\theta_{h,k}^{'}}^{-\varepsilon+i\theta_{h,k}^{'}}-\frac{1}{i}\int_{-\varepsilon+i\theta_{h,k}^{'}}^{-\varepsilon}-\frac{1}{i}\int_{-\varepsilon}^{-\infty}\\
&&=L_k^j(m,\nu)-J_1^j-J_2^j-J_3^j-J_4^j-J_5^j-J_6^j
\end{eqnarray*}

Where the integrand in all the integrals is

\begin{equation}
 \omega^{r}e^\frac{2\pi  (\nu-m_j)}{k^2\omega}e^{2\pi  (m+m_j) \omega}
\end{equation}


We will also assume that $0<\varepsilon<N^{-2}$. Now in the
integral$J_2$ we have

\begin{equation}
\begin{array}{cc}
  \omega=-\varepsilon+i\upsilon, & 0\geq\upsilon\geq -\theta_{h,k}^{''},\\
  \Re(\omega)=-\varepsilon, & \Re\left( \frac{1}{\omega}\right)=\frac{-\varepsilon}{\varepsilon^2+\upsilon^2}<0, \\
  |\omega|=\left(\varepsilon^2+\upsilon^2\right)^{\frac{1}{2}}\leq \left(N{-4}+k^{-2}N{-2}\right)^{\frac{1}{2}}\leq2^{\frac{1}{2}}k^{-1}N^{1}  \\
\end{array}
\end{equation}

\begin{eqnarray*}
\lefteqn{}\\
&&
\end{eqnarray*}


\begin{equation}
\end{equation}

\end{document}