Limit: $ \lim_{n \to \infty} \int_{(0,1)^n} \frac{n}{x_1 + \cdots + x_n} dx_1 \cdots dx_n $ I want to calcurate
$$ \lim_{n \to \infty} \int_{(0,1)^n} \frac{n}{x_1 + \cdots + x_n} \, dx_1 \cdots dx_n $$
I met this in studying Lebesgue integral. But, I don't know how to do at all. I would really appreciate if you could help me!
[Add]
Thanks to everybody who gave me comments, I can understand the following,
\begin{align*}
\lim_{n \to \infty} \int_{(0,1)^n} \frac{n}{x_1 + \cdots + x_n} dx_1 \cdots dx_n
&=\lim_{n \to \infty} n\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt 
\end{align*}
and
\begin{align*}
 \int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt 
  &=\frac{n}{(n-1)!}\sum_{i=0}^{n-1}{ n-1 \choose i} (-1)^{n-1-i} (i+1)^{n-2}\log(i+1)
\end{align*}
and
\begin{align*}
\int_0^\infty \bigg(\frac{1-e^{-t}}{t}\bigg)^n\,dt
&=\int_0^\infty \frac{z^{n-1}}{(n-1)!}\, \mathrm{Beta}(z,n+1)\,dz\\
&=n\,\int_0^\infty \frac{z^{n-1}}{z(z+1)\cdots(z+n)}\,dz
\end{align*}
But,I can't calcurate these integral and the limit. Please let me know if you find out.
 A: The limit is $2$. 
This might be an overkill but it allow you to compute the asymptotic expansion instead of just the limit.

Let $d^n x$ be a short hand for $dx_1\cdots dx_n$
Notice for $x_1,\ldots,x_n > 0$, we have
$\displaystyle\;\frac{1}{\sum_{k=1}^n x_k} = \int_0^\infty e^{-t\sum_{k=1}^n x_k} dt$
Using this, we can rewrite the integral at hand as
$$\begin{align} \mathcal{I}_n \stackrel{def}{=} \int_{(0,1)^n}\frac{d^n x}{\sum_{k=1}^n x_k}
= & \int_{(0,1)^{n}}\int_0^\infty e^{-t\sum_{k=1}^n x_k} dt  d^n x
=  \int_0^\infty \int_{(0,1)^{n}} e^{-t\sum_{k=1}^n x_k}d^n x dt\\
= & \int_0^\infty \left(\int_0^1 e^{-tx}\right)^n dt
=  \int_0^\infty \left(\frac{1-e^{-t}}{t}\right)^n dt
\end{align}
$$
Since all the integrands in above integrals (in given grouping) is non-negative,
we can use Tonelli's theorem to justify above manipulation.
Consider the function $\frac{1-e^{-t}}{t}$.
Over $(0,\infty)$, it is smooth and decreasing from $1$ at $t = 0^{+}$ to $0$ as $t \to \infty$. 
Over $\mathbb{C}$, it is entire and its zeros closest to origin are located at $\pm 2\pi i$. 
Let $s = -\log \frac{1-e^{-t}}{t} \iff \frac{1-e^{-t}}{t} = e^{-s}$.
As a function of $t$, $s(t)$ is smooth on $(0,\infty)$, increasing from $0$ at $t = 0^{+}$ to $\infty$ as $t \to \infty$.
Its power series expansion at $t = 0$ has radius of convergence $2\pi$.
$$s(t) = \frac{t}{2}-\frac{{t}^{2}}{24}+\frac{{t}^{4}}{2880}-\frac{{t}^{6}}{181440}+\frac{{t}^{8}}{9676800} + \cdots$$
This implies as a function of $s$, $t(s)$ has a power series expansion with finite convergence radius at $s = 0$. Changing variable to $s$, we obtain 
$$n\mathcal{I_n} = n\int_0^\infty e^{-ns} \frac{dt}{ds} ds \tag{*1}$$
Notice for large $s$, $t \sim e^s \implies \frac{dt}{ds} \sim e^s$. 
Together with above properties of $s$, we can apply 
Waton's lemma to extract the
asymptotic behavior of $\mathcal{I}_n$ as $n \to \infty$.
We help of a CAS, we can use Lagrange inversion theorem to obtain following expansion of $\frac{dt}{ds}$:
$$\frac{dt}{ds} = 2+\frac{2s}{3}+\frac{s^2}{3}+\frac{19s^3}{135}+\frac{17s^4}{324} + \cdots$$
Substitute this back into $(*1)$, we can read off the asymptotic expansion
of $n\mathcal{I}_n$ as
$$n\mathcal{I}_n \asymp 
2+ \frac{2}{3n} + \frac{2}{3n^2} + \frac{38}{45n^3} + \frac{34}{27n^4} + \cdots
$$
As a corollary, the desired limit is $\lim_{n\to\infty} n\mathcal{I}_n = 2$.
A: This answer will be based om Fubini's theorem and DCT.
Put
\begin{equation*}
I_n = \int_{V_n}\dfrac{x_1^p+x_2^p+\dots +x_n^p}{x_1^q+x_2^q+\dots +x_n^q}\, dx_1dx_2\dots dx_n
\end{equation*}
where $ V_n=(0,1)^n. $ We will try to prove that
\begin{equation*}
\lim_{n\to \infty}I_n = \dfrac{q+1}{p+1} \tag{1}
\end{equation*}
if $ p > -1$ and $ q \ge 1$.
With $p=0$ and $q=1$ this will answer OP's question.
Before the proof we need some preparations.
Put
\begin{equation*}
c_q = \int_{0}^{\infty}e^{-x^q}\, dx = \dfrac{\Gamma\left(\dfrac{1}{q}\right)}{q}
\end{equation*}
and
\begin{equation*}
\mathrm{eq}=\dfrac{1}{c_q}\int_{x}^{\infty}
e^{-y^q}\, dy.
\end{equation*}
In order to later find a majorant  we will use an inequality by $@$mickep (email communication in the case $q=1$).
\begin{equation*}
c_q\dfrac{1-\mathrm{eq}(t)}{t}\le \left(1+\dfrac{t^q}{2}\right)^{-\dfrac{1}{q}}, \quad t>0.\tag{2}
\end{equation*}
Proof of (2). Put
\begin{equation*}
f(t) = \dfrac{t}{\left(1+\dfrac{t^q}{2}\right)^{\dfrac{1}{q}}} - c_q(1-\mathrm{eq}(t)), \quad t \ge 0.
\end{equation*}
Then $f(0)=0$. We intend to prove that $ f'(t) \ge 0 $ if $ t \ge 0. $ We observe that 
\begin{gather*}
f'(t) = \dfrac{1}{\left(1+\dfrac{t^q}{2}\right)^{\dfrac{1}{q}+1}} -e^{-t^q} \ge 0
\\
\Longleftrightarrow\\
e^{t^q} \ge \left(1+\dfrac{t^q}{2}\right)^{\dfrac{1}{q}+1} \\
\Longleftrightarrow\\
t^{q} \ge \left(\dfrac{1}{q}+1\right)\ln\left(1+\dfrac{t^q}{2}\right)\\
\Longleftrightarrow\\
\dfrac{2q}{q+1}\dfrac{t^q}{2} \ge \ln\left(1+\dfrac{t^q}{2}\right)
\end{gather*}
Since $ x \ge \ln(1+x)$ and  $\dfrac{2q}{q+1} \ge 1  $ this is
 true and we have proved (2). 
Proof of (1).
 \begin{gather*}
I_n=[\mbox{ symmetry}] = n\int_{0}^{1}x_1^p\left(\int_{0}^{\infty}e^{-t(x_1^q+x_2^q+\dots +x_n^q)}\, dt\right)\, dx_1dx_2\dots dx_n =
\\[2ex] n\int_{0}^{1}\int_{0}^{\infty}x_1^pe^{-tx_1^q}\left(\int_{0}^{1}e^{-ty^q}\, dy\right)^{n-1}\, dx_1dt = \left[z=t^{1/q}y\right] =\\[2ex]n\int_{0}^{1}\int_{0}^{\infty}x_1^pe^{-tx_1^q}\left(\int_{0}^{t^{1/q}}\dfrac{e^{-z^q}}{t^{1/q}}\, dz\right)^{n-1}\, dx_1dt =\\[2ex]
n\int_{0}^{1}\int_{0}^{\infty}x_1^pe^{-tx_1^q}\left(\dfrac{1}{t^{1/q}}\left[-c_q\mathrm{eq}(z)\right]_{0}^{t^{1/q}}\right)^{n-1}\, dx_1dt = \left[t=\frac{s}{n-1}\right]=\\[2ex] \dfrac{n}{n-1 }\int_{0}^{1}\int_{0}^{\infty}x_1^pe^{-sx_1^q/(n-1)}\left(\dfrac{1}{s_{qn}}c_q(1-\mathrm{eq}(s_{qn}))\right)^{n-1}\, dx_1ds\tag{3}
\end{gather*}
where  $ s_{qn}= \left(\dfrac{s}{n-1}\right)^{1/q} $.
Now we will use Mickep's inequality (2) to find a majorant.
\begin{gather*}
e^{-sx_1^p/(n-1)}\left(\dfrac{1}{s_{qn}}c_q(1-\mathrm{eq}(s_{qn}))\right)^{n-1} \le 1 \cdot \left(\left(1+\dfrac{s}{2(n-1)}\right)^{-1/q}\right)^{n-1} =\\[2ex]
\left(\left(1+\dfrac{s}{2(n-1)}\right)^{n-1}\right)^{-1/q} \le \left(\left(1+\dfrac{s}{2(N-1)}\right)^{N-1}\right)^{-1/q} , \quad n \ge N
\end{gather*}
In the last inequality we have used that 
\begin{equation*}
\left(1+\dfrac{s}{2(n-1)}\right)^{n-1}
\end{equation*}
is increasing towards $ e^{s/2} $. If we choose  $ N $ such that $ \dfrac{N-1}{q}>1 $ the majorant 
\begin{equation*}
\left(\left(1+\dfrac{s}{2(N-1)}\right)^{N-1}\right)^{-1/q}
\end{equation*}
will belong to $L_1.$
Finally we will study the pointwise limit.
Put
\begin{equation*}
g(x) = c_q(1-\mathrm{eq}(x)).
\end{equation*}
Then
\begin{equation*}
g'(x) = e^{-x^{q}} = 1- x^q + x^{2q}\cdot B_{1}(x^q)
\end{equation*}
where $B_1$ is q bounded function in the neighbourhood of origin. We get that
\begin{equation*}
g(x) = x-\dfrac{x^{q+1}}{q+1}+x^{2q+1}B_2 \tag{4}
\end{equation*}
where $B_{2}$ is bounded for small $x^{q}$. However, from (4) we get
\begin{gather*}
\left(\dfrac{1}{s_{qn}}c_q(1-\mathrm{eq}(s_{qn}))\right)^{n-1}=\left(1-\dfrac{s}{(q+1)(n-1)}+\dfrac{s^2}{(n-1)^{2}}B_3\right)^{n-1} \to e^{-s/(q+1)},\quad n \to \infty
\end{gather*}
since $B_{3}$ is bounded.
Now we return to (3).
\begin{equation*}
\lim_{n\to \infty}I_n = \int_{0}^{1}\int_{0}^{\infty}x_1^pe^{-s/(q+1)}\, dx_1ds = \dfrac{q+1}{p+1}.
\end{equation*}
A: You say that we might as well find the following limit:
\begin{align}
\lim_{n\to\infty} n \int^\infty_0 \left( \frac {1-e^{-t}}{t}\right)^n\,dt
\end{align}
Set $u=e^{-t}$ to get:
\begin{align}
\int^\infty_0 \left( \frac {1-e^{-t}}{t}\right)^n\,dt = \int^1_0 \frac{1}{u} \left(\frac{u-1}{\log(u)}\right)^n\,du = \int^1_0 \frac{1}{u}\exp\left[n \log \left(\frac{u-1}{\log(u)}\right)\right]\,du
\end{align}
Now we can apply Laplace's Method to find the asymptotics of that as $n\to\infty$ and get:
$$\int^1_0 \frac{1}{u}\exp\left[n \log \left(\frac{u-1}{\log(u)}\right)\right]\,du \sim \frac{2}{n}$$
Hence:
$$\lim_{n\to\infty} n \int^\infty_0 \left( \frac {1-e^{-t}}{t}\right)^n\,dt = \lim_{n\to\infty} n \frac{2}{n} = 2$$
A: A somewhat more elementary solution. $f(t)=(1-e^{-t})/t$ is decreasing for $t>0$ since $$f'(t)=\frac{(1+t)e^{-t}-1}{t^2}<0\impliedby e^t>1+t.$$ Thus it has an inverse: $x=f(t)\iff t=g(x)$. At $x\to 0^+$ (i.e. $t\to+\infty$) we have $$\color{LightGray}{xg(x)=1-e^{-g(x)}\implies}g(x)\in\mathcal{O}(x^{-1})\text{ and }g^{(n)}(x)\in\mathcal{O}(x^{-n-1});$$ as $x\to 1^-$ (i.e. $t\to 0^+$) we have $g'(x)\to-2$ because of $f'(t)\to-1/2$.
So, doing the substitution $t=g(x)$ and integration by parts, for $n>1$ we obtain $$(n+1)\int_0^\infty\left(\frac{1-e^{-t}}{t}\right)^n dt=-(n+1)\int_0^1 x^n g'(x)\,dx=2+\int_0^1 x^{n+1}g''(x)\,dx.$$ The last term tends to $0$ as $n\to\infty$ (by DCT, with $x^3\big|g''(x)\big|$ as the dominating function).
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\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}}}}
\\[5mm] = & \
\lim_{n \to \infty}\
\braces{n\int_{0}^{1}\cdots\int_{0}^{1}
\bracks{\int_{0}^{\infty}
\expo{-\pars{x_{1}\ +\ \cdots\ +\ x_{n}}t}\,\,\,\,\,
\dd t}\dd x_{1}\ldots\dd x_{n}}
\\[5mm] = & \
\lim_{n \to \infty}\
\bracks{n\int_{0}^{\infty}\pars{\int_{0}^{1}
\expo{-tx}\,\,\dd x}^{n}\,\dd t}
\\[5mm] = & \
\lim_{n \to \infty}\
\bracks{n\int_{0}^{\infty}
\pars{1 - \expo{t} \over t}^{n}\,\dd t}
\\[5mm] = & \
\lim_{n \to \infty}\
\bracks{n\int_{0}^{\infty}
\exp\pars{n\ln\pars{1 - \expo{-t} \over t}}\,\dd t}
\\[5mm] = & \
\lim_{n \to \infty}\
\bracks{n\int_{0}^{\infty}\expo{-nt/2}\,\,\dd t} =
\bbx{\color{$44f}{2}} \quad Laplace's\ Method
\end{align}
