Asymptotics of a sequences of integrals

Question

I would like to know the asymptotics of the following sequences of integrals: $$ I_n = \int _0 ^{+ \infty} e^{-t} \left ( \dfrac{t}{1 + t} \right )^n \ dt $$

I have tried using Laplace method ou saddle node method, but I have been unable to conclude anything.

I have tried to find out the behaviour using a software computation. Here is what I found: \begin{equation} \begin{array}{|c|c|c|} \hline \\ n & I_n & - \ln(I_n) \\ \hline 1 = 2^0 & 1.9269472 \cdot 10^{-1} & 1.646648079928304 \\ \hline 2 = 2^1 & 8.7215768 \cdot 10^{-2} & 2.4393701372220464 \\ \hline 4 = 2^2 & 2.6524946 \cdot 10^{-2} & 3.629669596451481 \\ \hline 8 = 2^3 & 4.7442047 \cdot 10^{-3} & 5.35083146190712 \\ \hline 16 = 2^4 & 4.1306898 \cdot 10^{-4} & 7.7918959541372 \\ \hline 32 = 2^5 & 1.3310510 \cdot 10^{-5} & 11.226956600782769 \\ \hline 64 = 2^6 & 1.0697730 \cdot 10^{-7} & 16.05064913913718 \\ \hline 128 = 2^7 & 1.2206772 \cdot 10^{-10} & 22.826445101120278 \\ \hline 256 = 2^8 & 8.8802107 \cdot 10^{-15} & 32.35495110837777 \\ \hline 512 = 2^9 & 1.3241607 \cdot 10^{-20} & 45.77092301635973 \\ \hline 1024 = 2^{10} & 8.1182635 \cdot 10^{-29} & 64.6808514171555 \\ \hline 2048 = 2^{11} & 2.1076879 \cdot 10^{-40} & 91.3578121482943 \\ \hline 4096 = 2^{12} & 9.3011756 \cdot 10^{-57} & 129.01720949348262 \\ \hline 8192 = 2^{13} & 7.3886987 \cdot 10^{-80} & 182.2068558012616 \\ \hline 16384 = 2^{14} & 1.7003331 \cdot 10^{-112} & 257.358706225986 \\ \hline 32768 = 2^{15} & 1.2703122 \cdot 10^{-158} & 363.5691819464147 \\ \hline 65536 = 2^{16} & 7.9749999 \cdot 10^{-224} & 513.7027491850932 \\ \hline 131072 = 2^{17} & 5.2817088 \cdot 10^{-316} & 725.9526396990342 \\ \hline 262144 = 2^{18} & 2.4716651 \cdot 10^{-446} & 1026.0480594180656 \\ \hline 524288 = 2^{19} & 1.2878115 \cdot 10^{-630} & 1450.3756643021711 \\ \hline \end{array} \end{equation} From this tabular, it seems that $\ln I_n \sim - 2 \sqrt{n}$.

Is there any method to study such sequences of integrals?

Answer 1

The integral $$I_n=\int_0^\infty e^{f(n,t)}\,dt,\;\;f(n,t)=n\ln t-n\ln(1+t)-t,$$ has a saddle point at $t^*$ where $\partial f(n,t)/\partial t=0$, $$t^\ast=-\tfrac{1}{2}+\tfrac{1}{2}\sqrt{1+4n}.$$ For $n\rightarrow\infty$ the integral tends to $$I_n\rightarrow e^{f(n,t^\ast)} = e^{-2\sqrt{n}+{\cal O}(1)},$$ so $\ln I_n\rightarrow -2\sqrt{n}$, as found numerically.

Answer 2

Let us assume the following conjecture:

Let $h: \mathbb{R}^+ \longrightarrow \mathbb{R}$ be a continuous function. Let also $g_n: \mathbb{R}^+ \longrightarrow \mathbb{R}$ be a $\mathcal{C}^2$ function over $\mathbb{R}^+$, depending of an positive integer $n$, with a unique maximum $\mathcal{U}_n$ such that $g_n''(\mathcal{U}_n) < 0$. Therefore, we have the following estimation: $$ \int _0 ^{+ \infty} h(t) e^{- g_n(t)} \ dt \underset{n \longrightarrow + \infty}{\sim} h(\mathcal{U}_n) \cdot \sqrt { \dfrac{2 \pi}{- g_n''(\mathcal{U}_n)} } \cdot e^{g_n(\mathcal{U}_n)} \ . $$

This is a little adaptation to the classical Laplace method (see [Wikipedia page on Laplace method, section "Other formulation"][1]

In our case, as pointed out by Carlo Beenakker, we have $h(t) = 1$, $g_n(t) = t - n \ln t + n \ln(1 + t)$ and $\mathcal{U}_n = \dfrac{\sqrt{4n + 1} - 1}{2}$. So, we finaly find that $$ I_n \underset{n \longrightarrow + \infty}{\sim} \sqrt{e \pi} n^{\frac{1}{4}} e^{-2 \sqrt{n}} \ . $$ [1]: https://en.wikipedia.org/wiki/Laplace%27s_method