$\sigma_{i}$ to $\sigma_c$ is written as $$N_{\mathrm{H}}
=
\int^{\sigma_{i}}_{\sigma_{c}}d\sigma
\frac{V}{V'}
=
\int^{\sigma_{i}}_{\sigma_{c}}d\sigma
\frac{1}{\frac{8}{27}\frac{\left(\mu M\right)^2}{\sigma^5}+\frac{1}{2}\sigma^3}.$$ The amplitude of the curvature perturbation in comoving gauge $\cal R$ and the spectral index $n_s$ at the pivot scale are written as $${\cal R}
 \simeq
\frac{1}{2\pi\sqrt{3}}\frac{V^{3/2}(\sigma_{i})}{|V'(\sigma_{i})|}
=
\frac{\mu^2}{
    2\pi\sqrt{3}\left[
        \frac{8}{27}\frac{\left(\mu M\right)^2}{\sigma_i^5}
        +\frac{1}{2}\sigma_i^3
    \right]
}
=
\frac{\mu^2}{
\pi\sqrt{3}\left[
    \frac{\sigma_d^8}{\sigma_i^5}   +\sigma_i^3
\right]
}
,$$ and $$n_s-1
=
-\frac{80}{27}\frac{\left(\mu M\right)^2}{\sigma_{i}^6}
+
3\sigma_{i}^2
=
-\frac{1}{5}\frac{\sigma_d^8}{\sigma_i^6}
+3\sigma_i^2
.$$ In this model, when the value of the inflaton at the pivot scale $\sigma_{i}$ is larger than $\sigma_{d}$, the spectral index becomes blue.

New inflation
-------------

Second, we describe the new inflation model [@Izawa:1996dv]. This model has one superfield ${ \Phi}$. ${ \Phi}$ has $R$ charge $\frac{2}{5}$, and it is assumed that $ U(1)_{R}$ symmetry is dynamically broken down to a discrete $ Z_{8R}$ at a scale $ v$. The superpotential is given by $${ W_{N} }= { v}^{{}2}{ \Phi} - \frac{ g}{5}{\Phi}^{5},$$ where $g$ is a coupling constant. The $R$-invariant Kähler potential is given by $$K_{N} = |{ \Phi}|^{2} + \frac{C_{N}}{4}|{ \Phi}|^{4}.$$ Here, $ C_{N}$ is a constant of order 1.

The scalar potential yields a vacuum, $$\braket{\Phi} \simeq \left(\frac{ v^{2}}{ g}\right)^{{1/4}}.$$ At this vacuum, the potential has the negative vacuum 