content_image/303945.jpg content_image/303946.jpg content_image/303947.jpg content_image/303944.jpg content_image/303942.jpg content_image/303941.jpg content_image/303940.jpg
where 

<img src='content_image/303945.jpg'>

<img src='content_image/303946.jpg'>

<img src='content_image/303947.jpg'>

In terms of the dimensionless time T = t t$_{pl}$ this reads as: 

<img src='content_image/303944.jpg'>

where z$_{1}$ = α$_{1}$β$_{2}$ , z$_{2}$ = α$_{1}$γ$_{2}$ + β$_{1}$β$_{2}$ , z$_{3}$ = α$_{1}$δ$_{2}$ + β$_{1}$γ$_{2}$ + γ$_{1}$β$_{2}$ and z$_{4}$ = α$_{1}$ Σ$_{2}$ + β$_{1}$δ$_{2}$ + γ$_{1}$γ$_{2}$ + δ$_{1}$β$_{2}$ . Since at the end of inflation: 

z$_{1}$T 2 EI ≃ 1 . 32 × 10 − $^{12}$, z$_{2}$ ≃ − 1 . 57 × 10 − $^{36}$, z$_{3}$T − 2 EI ≃ 1 . 19 × 10 − $^{24}$, z$_{4}$T − 4 EI ≃ − 1 . 42 × 10 − $^{48}$, we can ignore z$_{2}$ , z$_{3}$T − $^{2}$, z$_{4}$T − 4 and the constant √ 3Λ$_{0}$ t$_{pl}$ . As a result the IMSE in this case is given by: 

where 

<img src='content_image/303942.jpg'>

<img src='content_image/303941.jpg'>

It has to be noted that the term 5 T comes from a d dt ln( a 2 √ 3 Λ ˙ Λ $_{2Λ}$) and has the most contribution in this model. In contrast, − a$_{0}$z$_{1}$T is a consequence of the running gravitational coupling. 

In a very similar way like the previous subsection the solution to (55) can be obtained as: 

<img src='content_image/303940.jpg'>

where K ′′ is a constant. It is notable that the parameter α in Hankel function H (1) α ( Q$_{q}$ T ) is 

