Limit: $ \lim_{n \to \infty} \int_{(0,1)^n} \frac{n}{x_1 + \cdots + x_n} dx_1 \cdots dx_n $

Question

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.

Answer 1

The limit is $2$.

This might be an overkill but it allow you to compute the asymptotic expansion instead of just the limit.

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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 $\lim_{n\to\infty} n\mathcal{I}_n = 2$.

Answer 2

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: $$\lim_{n\to\infty} n \int^\infty_0 \left( \frac {1-e^{-t}}{t}\right)^n\,dt = \lim_{n\to\infty} n \frac{2}{n} = 2$$

Answer 3

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*}

Answer 4

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).