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I'm trying to approximate the sum $\sum_{\alpha=1}^{\mu} \Big(1-\frac{(\alpha(2 \mu-\alpha))^2 \gamma_1 \gamma_2}{2n^2 \mu^4}\Big)^{\frac{\lambda}{2}}$ with an integral $\int_{a}^{\infty}\exp\left(\lambda\Big(\frac{(-4 \alpha^2 \mu^2 +4 \alpha^3 \mu - \alpha^4) \gamma_1 \gamma_2}{4 n^2 \mu^4}\Big)\right)d \alpha$ with $a>0$ using Laplace method ($\alpha$ may be seen as a size of subset of species in a population size $\mu$, therefore the lower bound on summation is 1). Unfortunately, the assumptions for it are violated ($\alpha_0=\mu$ is a minimum point, and $\alpha_0= 2 \mu$ makes no sense; it also yields $f(\alpha_0)=0$).

I'd be grateful if anyone could help with this approximation.

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No Laplace method here (and I have no clue about what your $n$ and $a$ are) but basic Riemann sums. Define a function $f$ on $[0,1]$ by $ f(x)=\left(1-\frac14\gamma_1\gamma_2x^2(2-x)^2\right)^{\lambda/2}. $ You are looking at $ \sum_{\alpha=1}^{\mu}f\left(\frac{\alpha}{\mu}\right)=\mu I+o(\mu), $ when $\mu\to+\infty$, with $I$ positive and given by $ I=\int_0^1f(x)\mathrm{d}x. $

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    yes I$c$your point. Sorry the original post was written in$a$bit of$a$hurry. Though at the end of the day I used the Euler asymptotic approximation as given in 'Concrete Mathematics2011-06-14