Asymptotic rate of decay of the integrals

Question

For each $n$ define the sequence $y_n= \int_{0}^\infty (1-e^{-\frac{x^2}{4}})^n e^{-x}dx$. By dominated convergence theorem, it is clear that $y_n$ converges to $0$. I am interested in finding the rate at which $y_n$ decays. Mainly I am interested to know if the decay is exponential in $n$ or not i.e $y_n$ is roughly $c^n$ for some n ?

I have encountered this in a physics problem. I am unsure how to study the asymptotic behaviour. Any help is appreciated.

Answer 1

EDIT. After reading @Zarrax's answer, I realized that my original, heuristic guess cannot hold true. Below is a new answer.

---

Performing integration by parts,

$$ y_n = \frac{n}{2} \int_{0}^{\infty} x e^{-x^2/4} (1 - e^{-x^2/4})^{n-1} e^{-x} \, \mathrm{d}x. $$

Then, applying the substitution $y = e^{-x^2/4}$ (i.e., $x = 2\sqrt{\log(1/y)}$) followed by $y = t/n$, we get

\begin{align*} y_n &= n \int_{0}^{1} (1-y)^{n-1} e^{-2\sqrt{\log(1/y)}} \, \mathrm{d}y\\ &= \int_{0}^{n} \left( 1 - \frac{t}{n} \right)^{n-1} e^{-2\sqrt{\log n - \log t}} \, \mathrm{d}t. \end{align*}

Using this, we prove:

$$ \bbox[color:blue; padding:8px; border:1px dotted navy;]{ y_n = e^{-2\sqrt{\log n}}(1 + o(1)) } $$

Lower bound. Assume that $n \geq 2$. Noting that $\sqrt{1+x} \leq 1 + \frac{x}{2}$ for all $x \geq -1$, we get

\begin{align*} y_n &\geq \int_{0}^{n} \left( 1 - \frac{t}{n} \right)^{n-1} e^{-2\sqrt{\log n} + \frac{\log t}{\sqrt{\log n}}} \, \mathrm{d}t \\ &= e^{-2\sqrt{\log n}} \biggl( \int_{0}^{n} t^{1/\sqrt{\log n}} \left( 1 - \frac{t}{n} \right)^{n-1} \, \mathrm{d}t \biggr) \end{align*}

To analyze the asymptotic behavior of this lower bound, we use the inequality $1 + x \leq e^{-x}$ to find that

$$ t^{1/\sqrt{\log n}} \left( 1 - \frac{t}{n} \right)_+^{n-1} \leq (1 + t) e^{-t/2} $$

for all $t \in [0, n]$ and $n \geq 2$. So by the dominated convergence theorem,

$$ \int_{0}^{n} t^{1/\sqrt{\log n}} \left( 1 - \frac{t}{n} \right)^{n-1} \, \mathrm{d}t \to \int_{0}^{\infty} e^{-t} \, \mathrm{d}t = 1. $$

This shows that

$$ \color{blue}{ y_n \geq e^{-2\sqrt{\log n}}(1 + o(1)). } $$

Upper bound. Noting that $\sqrt{1+x} \geq 1 + \frac{x}{2+x}$ for all $x \geq -1$, we get

\begin{align*} y_n &\leq \int_{0}^{n} \left( 1 - \frac{t}{n} \right)^{n-1} e^{-2\sqrt{\log n} ( 1 - \frac{\log t}{2\log n - \log t})} \, \mathrm{d}t \\ &= e^{-2\sqrt{\log n}} \biggl( \int_{0}^{n} t^{\frac{2\sqrt{\log n}}{2\log n - \log t}} \left( 1 - \frac{t}{n} \right)^{n-1} \, \mathrm{d}t \biggr) \end{align*}

Arguing similarly as before, we find that

$$ t^{\frac{2\sqrt{\log n}}{2\log n - \log t}} \left( 1 - \frac{t}{n} \right)^{n-1} \leq (1 + t)e^{-t/2} $$

for all $t \in [0, n]$ and $n \geq 2$, hence by the dominated convergence theorem

$$\int_{0}^{n} t^{\frac{2\sqrt{\log n}}{2\log n - \log t}} \left( 1 - \frac{t}{n} \right)^{n-1} \, \mathrm{d}t \to \int_{0}^{\infty} e^{-t} \, \mathrm{d}t = 1. $$

Therefore

$$ \color{blue}{ y_n \leq e^{-2\sqrt{\log n}}(1 + o(1)). } $$

Answer 2

$$y_n= \int_{0}^\infty \Big(1-e^{-\frac{x^2}{4}})\Big)^n e^{-x}dx$$ $$y_n=1+ \sum_{k=1}^n (-1)^k \,\binom{n}{k}\int_{0}^\infty e^{-\frac{k x^2}{4}-x}\,dx$$ $$y_n=1+\sqrt \pi \sum_{k=1}^n (-1)^k \,\binom{n}{k}\,\frac{e^{\frac{1}{k}}\, \text{erfc}\left(\frac{1}{\sqrt{k}}\right)}{\sqrt{k}}$$

I am stuck at this point for the asymptotics.

Answer 3

The $(1 - e^{-{x^2 \over 4}})^n$ factor starts becoming relevant when $e^{-{x^2 \over 4}}$ is about ${1 \over n}$, since then $(1 - e^{-{x^2 \over 4}})^n = (1 - {1 \over n})^n \sim {1 \over e}$. In fact, for large enough $n$, $(1 - e^{-{x^2 \over 4}})^n > {1 \over 2e}$ beyond that value of $x$. If you solve $e^{-{x^2 \over 4}} = {1 \over n}$ you obtain $x = 2\sqrt{\ln n}$, so for large $n$ your integral is at least

$${1 \over 2e} \int_{2 \sqrt{\ln n}}^{\infty} e^{-x}\,dx$$ $$= {1 \over 2e} e^{-2\sqrt{\ln{n}}}$$ This gives a lower bound for your integral, which shows it decays slower than any polynomial or exponential power.

If you want to bound it above, it's a little trickier. But you can churn out the calculation to show that the derivative of $(1 - e^{-{x^2 \over 4}})^ne^{-x}$ is positive on $[0,2\sqrt{\ln{n}}]$. Thus your integral is at most $2\sqrt{\ln{n}}$ times the value of $(1 - e^{-{x^2 \over 4}})^ne^{-x}$ at $x = 2\sqrt{\ln{n}}$, which is of order $\sqrt{\ln{n}}\,e^{-2\sqrt{\ln{n}}}$.

So the overall rate of decay of your integral is somewhere between the orders of $e^{-2\sqrt{\ln{n}}}$ and $\sqrt{\ln{n}}\,e^{-2\sqrt{\ln{n}}}$. My guess would be it is $O(e^{-2\sqrt{\ln{n}}})$.

If you want to get more refined asymptotics I'd suggest trying to see what Laplace's method gives you. Your domain of integration is unbounded but sometimes that method will still work.