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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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up vote 3 down vote accepted

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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A new version of the post surreptitiously introduces a parameter $n$ in the integral. So now, I have no clue about what is $a$. – Did Jun 13 '11 at 23:15
@Didier, $n$ is just a constant, $a$ is the lower bound on the integral that is an arbitrarily small positive value so that the support doesn't include 0. – sigma.z.1980 Jun 13 '11 at 23:19
Right, thanks for the explanation--although I fail to see why you do not simply replace $\gamma_1\gamma_2/n^2$ by $c$ or something and I still do not understand the rôle of $a$. But what about the content of my answer? Is this the asymptotics you want to consider ($\mu\to+\infty$, everything else fixed)? – Did Jun 13 '11 at 23:39
What's $f(\frac{\alpha}{\mu})$? Is it $\Big(1-\frac{(\frac{\alpha}{\mu}(2 \mu-\frac{\alpha}{\mu}))^2 \gamma_1 \gamma_2}{2n^2 \mu^4}\Big)^{\frac{\lambda}{2}}$ – sigma.z.1980 Jun 14 '11 at 0:36
If you except the fact that when one replaces $x$ by $\alpha/\mu$ in $2-x$ one obtains $2-(\alpha/\mu)$ instead of $2\mu-(\alpha/\mu)$ and the fact that since there is no power of $\mu$ in the denominator of $f(x)$ one can wonder at the appearance of an extra $\mu^4$ in the denominator of the formula in your comment, yes. Plus, of course, the replacement of $2^2$ by $2n^2$ in the denominator which should be carried through in the definition of $f$. – Did Jun 14 '11 at 5:17

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