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 
$$\lim_{n \to \infty} n\int_0 ^{1} e^{-tn}\left (\frac{\sinh t}{t}\right )^n\,dt=1$$
 A: 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 
$$\lim_{n\to \infty}n\int_0^1 e^{-nt}\left(\frac{\sinh(t)}{t}\right)^n\,dt=1$$
A: 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. ////
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
A: 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
$$
\lim_{n\to\infty}n\int_0^1e^{-tn}\left(\frac{\sinh(t)}{t}\right)^n\,\mathrm{d}t=1\tag{3}
$$
