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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!

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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.

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  • $\begingroup$ This may help: math.stackexchange.com/questions/1565896/… $\endgroup$ – Dzoooks Jun 24 '18 at 3:17
  • $\begingroup$ Yes, I know $ \int_{(0,1)^n} \frac{n}{x_1+ \cdots +x_n} dx_1 \cdots dx_n = \frac{n(-1)^n}{(n-1)!} \sum_{i=0}{n}(-1)^i \binom{n}{i} i^{n-1} \log i $ But, I can't calcurate from it. $\endgroup$ – S. Green Jun 24 '18 at 3:28
  • $\begingroup$ Are you allowed to use any theorems regarding convergence of random variables? $\endgroup$ – JimmyK4542 Jun 24 '18 at 4:20
  • $\begingroup$ I allowed to use to theorems related to Lebesgue integral , such as Fubini's theorem and dominated convergence theorem. If theorems regarding convergence of random variables is necessary, I will search the proof. $\endgroup$ – S. Green Jun 24 '18 at 4:45
  • $\begingroup$ If you got this problem in a Real Analysis class, you probably should stick to theorems from the class. If you just saw this problem somewhere and are doing it for fun, then there might be other methods for it. $\endgroup$ – JimmyK4542 Jun 24 '18 at 5:00
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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 $\lim_{n\to\infty} n\mathcal{I}_n = 2$.

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  • $\begingroup$ Wonderful solution! Thanks. $\endgroup$ – S. Green Jun 26 '18 at 1:26
  • $\begingroup$ @S.Green I suspect this question is supposed to be solved by DCT but I've trouble in finding a suitable dominanting function. That's why I switch to heavier tool like Watson's lemma. I'm glad you didn't mind the overkill. $\endgroup$ – achille hui Jun 26 '18 at 1:43
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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: $$\lim_{n\to\infty} n \int^\infty_0 \left( \frac {1-e^{-t}}{t}\right)^n\,dt = \lim_{n\to\infty} n \frac{2}{n} = 2$$

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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*}

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