# show that $\lim_{n \to \infty} n\int_0^1 e^{-tn}(\frac{\sinh t}{t})^n \, dt=1$

I tried to solve $$\lim_{n \to \infty} n\int_0^1\ldots \int_0^1\frac{1}{x_1+x_2+\ldots+x_n}\,dx_1\ldots dx_n$$

I started with

$$\frac{1}{x_1+x_2+\ldots+x_n}=\int_0^{\infty}e^{-t(x_1+x_2+\ldots+x_n)}\,dt$$

$$\Rightarrow I=\int_0^1\ldots\int_0^1\frac{1}{x_1+x_2+\ldots+x_n}dx_1\ldots dx_n=\int_0^{\infty}\int_0^1\ldots\int_0^1e^{-t(x_1+x_2+\ldots+x_n)}dx_1\ldots dx_n\,dt$$

$$\Rightarrow I=\int_0^{\infty}\int_0^1e^{-t(x_1)}dx_1\int_0^1e^{-t(x_2)} \, dx_2 \ldots\int_0^1e^{-t(x_n)}dx_n\,dt$$

$$\Rightarrow I=\int_0^\infty \left ( \frac{1-e^{-t}}{t}\right )^n\,dt$$ so our lim should be $$\Rightarrow \lim_{n \to \infty} n\int_0^{\infty} \left (\frac{1-e^{-t}} t \right )^n \, dt$$

and with some steps :

$$\Rightarrow \lim_{n \to \infty} 2n\int_0^\infty e^{-tn}\left (\frac{\sinh t} t \right )^n\,dt$$ using mathematica i got numerically that the result is $2$.

i could see that $$\lim_{n \to \infty} \int_1^\infty e^{-tn}\left (\frac{\sinh t}{t}\right )^n\,dt=0$$

so how to prove that

$$\lim_{n \to \infty} n\int_0 ^{1} e^{-tn}\left (\frac{\sinh t}{t}\right )^n\,dt=1$$

• $n e^{-tn}$ is an approximation of a Dirac delta distribution and $\frac{\sinh t}{t} = 1+O(t^2)$ in a neighbourhood of the origin. – Jack D'Aurizio Jun 23 '17 at 15:01

We can express the integral of interest as

$$n\int_0^1 e^{-nt}\left(\frac{\sinh(t)}{t}\right)^n\,dt=n\int_0^1 \left(\frac{1-e^{-2t}}{2t}\right)^n\,dt \tag1$$

Next, applying the estimates

$$1-t\le \frac{1-e^{-2t}}{2t}\le 1-t+\frac23 t^2\le e^{-t+\frac23t^2}$$

to the integral on the right-hand side of $(2)$ reveals

$$n\int_0^1 (1-t)^n\,dt\le n\int_0^1 e^{-nt}\left(\frac{\sinh(t)}{t}\right)^n\,dt\le n\int_0^1 e^{-nt}e^{2nt^2/3}\,dt \tag 2$$

The left-hand side of $(2)$ is easy to evaluate and we find that

$$\frac{n}{n+1}\le n\int_0^1 e^{-nt}\left(\frac{\sinh(t)}{t}\right)^n\,dt$$

For the right-hand side of $(2)$, we enforce the substitution $t\to t/n$ and find that

\begin{align} n\int_0^1 e^{-nt}e^{2nt^2/3}\,dt &=\int_0^n e^{-t} e^{2t^2/3n}\,dt\\\\ &=\int_0^\infty e^{-t}e^{2t^2/3n}\xi_{[0,n]}(t)\,dt \end{align}

Inasmuch as $e^{-t}e^{2t^2/3n}\xi_{[0,n]}(t)\le e^{-t/3}$ and $\int_0^\infty e^{-t/3}\,dt<\infty$, the Dominated Convergence Theorem guarantees that

\begin{align} \lim_{n\to \infty }n\int_0^1 e^{-nt}e^{2nt^2/3}\,dt&=\int_0^\infty \lim_{n\to \infty}\left(e^{-t}e^{2t^2/3n}\xi_{[0,n]}(t)\right)\,dt\\\\ &=\int_0^\infty e^{-t}\,dt\\\\ &=1 \end{align}

Putting everything together, we assert that

$$\lim_{n\to \infty}n\int_0^1 e^{-nt}\left(\frac{\sinh(t)}{t}\right)^n\,dt=1$$

• Hey Mark, this looks good! Might you like to explain to me why $e^{-t}e^{2 t^2/3n}\theta(n-t)\leq e^{-t/3}$ – tired Jun 24 '17 at 14:26
• @tired Thank you. It should be easy to show that $e^{\frac23 (t^2/n-t)}\le 1$ for $t\in [0,n]$. – Mark Viola Jun 24 '17 at 14:35
• outch, sure . I get a bit rusty those days ...:-) – tired Jun 24 '17 at 14:36
• I'm rusty too. I took me a while to figure out a rigorous way forward to this quite simple problem. – Mark Viola Jun 24 '17 at 14:37

Let us stick to the original limit

$$\lim_{n\to\infty} I_n \stackrel{?}{=} 2 \qquad\text{for}\qquad I_n := n \int_{0}^{\infty} \left( \frac{1-e^{-t}}{t}\right)^n \, dt.$$

Heuristics. Notice that $\log\left(\frac{1-e^{-t}}{t}\right) = -\frac{1}{2}t + \mathcal{O}(t^2)$ near $t = 0$. So it is not unreasonable to expect that

$$I_n \approx n \int_{0}^{\infty} e^{-\frac{n}{2}t} \, dt = 2.$$

Formal proof. Notice that the function $g : [0,\infty) \to \mathbb{R}$ defined by

$$g(t) = \frac{1-e^{-t}}{t} = \int_{0}^{1} e^{-ts} \, ds$$

is decreasing in $t$. Let $f : (0, 1] \to [0, \infty)$ be the inverse of $g$. Then with the substitution $t = f(u)$, we have

$$I_n = n \int_{0}^{1} u^n |f'(u)| \, du.$$

Now set $h(u) = u^2|f'(u)|$ for $u \in (0, 1]$. By the inverse function theorem, we find that

$$h(1) = -\frac{1}{g'(0)} = 2 \qquad\text{and}\qquad \lim_{u \to 0^+} h(u) = - \lim_{t \to +\infty} \frac{g(t)^2}{g'(t)} = 1.$$

So $h$ is continuous and bounded on $[0, 1]$ by setting $h(0) = 1$. Then the conclusion follows by applying the approximation-to-the-identity property (see Lemma below if you are new to this) yields

$$\lim_{n\to\infty} I_{n+1} = \lim_{n\to\infty} \frac{n+1}{n} \int_{0}^{1} nu^{n-1}h(u) \, du = h(1) = 2.$$

Lemma. Let $\varphi \in C([0, 1])$. Then $\displaystyle \lim_{n\to\infty} \int_{0}^{1} nu^{n-1} \varphi(u) \, du = \varphi(1)$.

Proof of Lemma. Using the identity $\int_{0}^{1} nu^{n-1} \, du = 1$, we may assume that $\varphi(1) = 0$. For each $\epsilon > 0$, pick $\delta > 0$ such that $|\varphi(x)| < \epsilon$ for $x \in [1-\delta, 1]$. Then

\begin{align*} \left| \int_{0}^{1} nu^{n-1} \varphi(u) \, du \right| &\leq \left( \sup_{x\in[0,1-\delta]} |\varphi(x)| \right) \int_{0}^{1-\delta} nu^{n-1} \, du + \epsilon \int_{1-\delta}^{1} nu^{n-1} \, du \\ &\leq (1-\delta)^n \sup_{x\in[0,1-\delta]} |\varphi(x)| + \epsilon \end{align*}

and hence

$$\limsup_{n\to\infty} \left| \int_{0}^{1} nu^{n-1} \varphi(u) \, du \right| \leq \epsilon.$$

Since this is true for all $\epsilon > 0$, letting $\epsilon \to 0^+$ proves the claim. ////

For $x\in[-1,1]$, \begin{align} \frac1{x^2}\left(\frac{\sinh(x)}{x}-1\right) &=\sum_{k=1}^\infty\frac{x^{2k-2}}{(2k+1)!}\\ &\le\sum_{k=1}^\infty\frac1{6^k}\\[3pt] &=\frac15\tag{1} \end{align} Therefore, on $[0,1]$, $1\le\frac{\sinh(x)}{x}\le1+\frac{x^2}5\le e^{x^2/5}$. \begin{align} &\lim_{n\to\infty}n\int_0^1e^{-tn}\left(\frac{\sinh(t)}{t}\right)^n\,\mathrm{d}t\\ &=\lim_{n\to\infty}\int_0^ne^{-t}\left(\frac{\sinh(t/n)}{t/n}\right)^n\,\mathrm{d}t\\ &=\color{#090}{\lim_{n\to\infty}\int_0^ke^{-t}\left[1,e^{\frac{t^2}{5n}}\right]\,\mathrm{d}t} +\color{#C00}{\lim_{n\to\infty}\int_k^ne^{-t}\left[1,e^{\frac{t}{5}}\right]\,\mathrm{d}t}\\[6pt] &=\color{#090}{1-e^{-k}}+\color{#C00}{O\!\left(e^{-4k/5}\right)}\\[12pt] &=1+O\!\left(e^{-4k/5}\right)\tag{2} \end{align} Since $(2)$ holds for all $k\ge0$, we have $$\lim_{n\to\infty}n\int_0^1e^{-tn}\left(\frac{\sinh(t)}{t}\right)^n\,\mathrm{d}t=1\tag{3}$$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\lim_{n \to \infty}\pars{n\int_{0}^{1}\cdots\int_{0}^{1} {\dd x_{1}\ldots\dd x_{n} \over x_{1} + \cdots + x_{n}}} = \lim_{n \to \infty}\bracks{n\int_{0}^{\infty} \exp\pars{n\ln\pars{1 - \expo{-t} \over t}}\,\dd t} \end{align}

$$\left\{\begin{array}{l} \ds{\mbox{Note that}\quad\left.{1 - \expo{-t} \over t}\right\vert_{\ t\ >\ 0} = {\expo{0} - \expo{-t} \over 0 - \pars{-t}} = \expo{\xi}\ \mbox{with} -t < \xi < 0 \implies \left.{1 - \expo{-t} \over t} \right\vert_{\ t\ >\ 0} < 1} \\[5mm] \ds{\mbox{Moreover,}\quad\totald{}{t}\pars{1 - \expo{-t} \over t} = {1 \over \expo{t} - 1} - {1 \over t} < 0\quad \mbox{because}\quad \pars{\expo{t} - 1}_{\ t\ >\ 0} > t} \end{array}\right.$$
With these considerations, the $\ds{\ln}$-prefactor $\ds{n}$ 'claims' by an application of the Laplace Method: \begin{align} &\lim_{n \to \infty}\pars{n\int_{0}^{1}\cdots\int_{0}^{1} {\dd x_{1}\ldots\dd x_{n} \over x_{1} + \cdots + x_{n}}} = \lim_{n \to \infty}\bracks{n\int_{0}^{\infty}\exp\pars{-\,{nt \over 2}} \pars{1 + {nt^{2} \over 24}}\,\dd t} \\[5mm] = &\ 2\lim_{n \to \infty}\int_{0}^{\infty}\expo{-t} \pars{1 + {t^{2} \over 6n}}\,\dd t = 2\lim_{n \to \infty}\pars{1 + {1 \over 3n}} = \bbx{\large 2} \end{align}