Asymptotic $\int_M \mathrm{exp}[\mathbf{e}\left(n -\frac{t}{2\pi i}\right)] \left( 1 + \frac{5}{6} \mathbf{e}^2 \right)^{1/2} $ on quintic Calabi-Yau Let $M = \{ G(x) = 0 \} \subseteq \mathbb{P}^4$ be a quintic Calabi-Yau and $\mathbf{e} \in H^2(M, \mathbb{Z})$ such that $\int_M \mathbf{e}^3 = 5$.  Then as $t \gg 1$:
$$
\int_M 
e^{n \mathbf{e}} 
e^{-\frac{t\mathbf{e}}{2\pi i}} 
\left( 1 + \frac{5}{6} \mathbf{e}^2 \right)^{1/2} = 
- \frac{5}{12}
\left( \frac{t}{2\pi i} - n\right)
\left( 2\left( \frac{t}{2\pi i} - n\right)^2 - 5 \right) \tag{$\ast$}$$
How does a high-frequency integral like this get computed?  This could be steepest descent or stationary phase.  In that case what are the Leftschetz thimble and Morse flow?  Perhaps set $G$ bo the the Fermat quintic or something for an example.

Question Because there are so many new elements for me here: the surface in $\mathbb{P}^4$, digesting what it means $\int_M \mathbf{e}^3 =5$.  In a deep way, this is hardly more than an Laplace transform. I would like help seeing how steepest descent / stationary phase is implemented and computed here.
The original paper talks about the pullback of the line bundle $\mathcal{O}_{\mathbb{P}^4}(n)$ to $\mathcal{O}_M(n)$ and details of $\hat{A}$-genus that I am relegating to a separate question.  
Withholding the physics context, the left side looks like particular case of a more general result, but this is already an involved computation.  I am struggling to see even this much and connections to more classical mathematics.

Tongue-in-cheek remark if the moments on the left side could be computed, then we get some analogue of Stirling formula $\log n! \approx n \log n$.

Clarifications
The notation $(1 + \frac{5}{6} \mathbf{e}^2)^{1/2}$ is taken from an arXiv.hep-th paper from 2013 which in turn cites others.  One possible interpretation is to use the Taylor series:
$$(1 + \frac{5}{6} \mathbf{e}^2)^{1/2}
= 1 + \binom{\tfrac{1}{2}}{1} \left(\tfrac{5}{6}\right)^1 \mathbf{e}^2
+  \binom{\tfrac{1}{2}}{1} \left(\tfrac{5}{6}\right)^2 \mathbf{e}^4
+  \binom{\tfrac{1}{2}}{1} \left(\tfrac{5}{6}\right)^3 \mathbf{e}^6
+ \dots $$
The remaining terms are $0$ since $M$ is 6-dimensional.  At least that's the gist of one of the comments.


*

*Another comment suggests the integral $\ast$ is exact: an equality for all values of $t,n$.

*Also $\mathbb{e}$ is just a 2-form like $x \, dx \vee dy$ restricted to the manifold $M$.  How are we assured these exists one with $\int_M \mathbb{e}^3 = 5$?  
 A: OK, let's see if I can put my money where my commenting mouth is. Let me say at the outset that I have no idea where such an integral comes from, but I claim that it doesn't matter to answer the question.
First of all, $\mathbf e$ is nothing mysterious: it is the 2-form dual to a hyperplane section of $M$. The equation $\int_M \mathbf e^3=5$ is just a fancy way of writing that the intersection of 3 hypeplanes in $\mathbf P^4$ (i.e. a line) intersects $M$ in 5 points.
Next, the coefficient $\left(n -\frac{t}{2\pi i}\right)$ never gets modified in any way, so let's just write it as $A$ for simplicity.
Finally, everything in the integrand is supposed to be an element of $H^*(M)$, so we should expand things as power series in $\mathbf e$. Since $\mathbf e \in H^2(M)$ we already have $\mathbf e^4=0$ (contrary to what was written in the edit). So we get
$$ (1+\frac{5}{6}\mathbf e^2)^{\frac12} = 1 + \frac{5}{12} \mathbf e^2 \\
 \operatorname{exp} (A \mathbf e) = 1 + A \mathbf e + \frac{A^2}{2} \mathbf e^2 + \frac{A^3}{6} \mathbf e^3$$.
Now multiply these together: the result is 
$$ \left( \frac{A^3}{6} + \frac{5A}{12} \right) \mathbf e^3 + \text{lower order terms}. $$ 
Applying $\int_M$ kills the lower order terms and turns the $\mathbf e^3$ into a 5, so we end up with 
$$\frac{5A^3}{6}+\frac{25A}{12}$$
The sign of the last term is opposite to what you wrote above, but agrees with the formula on p.37 of the paper you linked. 
By the way, I think you misinterpreted what the paper is saying about taking $t \rightarrow + \infty$. In the paper, they are saying that as $t \rightarrow +\infty$, some other integral expression (higher up the page) reduces to the left-hand side of your equation. They are then saying that the left-hand side equals the right-hand side, but that equality is valid for all $t$, as we just saw.
