Let $M = \{ G(x) = 0 \} \subseteq \mathbb{P}^4$ be a [quintic Calabi-Yau](https://en.wikipedia.org/wiki/Quintic_threefold) 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](https://en.wikipedia.org/wiki/Method_of_steepest_descent) or [stationary phase](https://en.wikipedia.org/wiki/Stationary_phase_approximation). In that case what are the Leftschetz thimble and Morse flow? Perhaps set $G$ bo the the Fermat quintic or something for an example.




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**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](https://en.wikipedia.org/wiki/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.




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




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**Clarifications**


The notation $(1 + \frac{5}{6} \mathbf{e}^2)^{1/2}$ is taken from an arXiv.hep-th paper from [2013](http://arxiv.org/abs/1308.2217) 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 $\int\_M \mathbb{e}^3 = 5$?



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
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.


