text: # Sophomore's Eternal Dream $\int_0^1\cdots\int_0^1 (t_1t_2\cdots t_n)^{t_1t_2\cdots t_n} dt_1 dt_2\cdots dt_n$
  
  Given
  
  $$I_n=\int_0^1\int_0^1\cdot\cdot\cdot\int_0^1 (t_1t_2\cdot\cdot\cdot t_n)^{t_1t_2\cdot\cdot\cdot t_n} dt_1 dt_2\cdot\cdot\cdot dt_n$$
  
  Prove: $$I_1=I_2
  
  For $$n=1$$, it is Sophomore's dream: https://en.wikipedia.org/wiki/Sophomore%27s_dream
  
  $$I_1=\int_0^1 t_1^{t_1} dt_1=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}$$
  
  For $$n=2$$, $$I_2=\int_0^1\int_0^1 (t_1t_2)^{t_1t_2}dt_1 dt_2$$
  
  let $$x=t_1, y=t_1 t_2$$, where the integral region is mapping to the triangle region $$0, and Jacobian $$J=\frac{1}{x}$$
  
  $$I_2=\int_0^1\int_0^x y^y \frac{1}{x}dy dx=\int_0^1\int_y^1 y^y \frac{1}{x} dx dy=-\int_0^1 y^y\ln(y) dy$$
  
  Series expansion: $$y^y=e^{y\ln(y)}=\sum_{n=0}^\infty \frac{1}{n!}y^n \ln^n(y)$$
  
  $$I_2=-\sum_{n=0}^\infty \frac{1}{n!}\int_0^1 y^n \ln^{n+1}(y) dy$$
  
  $$\int_0^1 y^n \ln^{n+1}(y) dy=\frac{(-1)^{n+1}n!}{(n+1)^{n+1}}$$
  
  After simplify the result we got:
  
  $$I_2=I_1$$
  
  Next, for $$I_3$$, let $$x=t_1, y=t_1t_2, z=t_1t_2t_3$$, Jacobian is $$J=\frac{1}{z}$$
  
  $$I_3=\int_0^1\int_0^x\int_0^y z^z\frac{1}{z} dzdydx$$
  
  Question.1: Is there an analytic result for $$I_3$$?
  
  Question.2: How to generalize the calculation to $$I_n$$, as well as the limit of $$I_n$$?
  
  • How about "Sophomore's Eternal Dream" Cause there it's like an infinite integral? Jul 2, 2022 at 19:15
  • Repeated integration by parts gives $\int_0^1 \ln(x)^n x^x dx/n!$ for $n+1$ variables. Jul 2, 2022 at 21:15
  • how to separate them? @eyeballfrog Jul 2, 2022 at 21:40
  
  Let $$(U_n)_{n \ge 1}$$ be a sequence of independent uniform random variables. By removing null sets, we may assume that for every $$n \ge 1$$, $$U_n>0$$ everywhere For every $$n \ge 1$$, set $$X_n = -\ln(U_n)$$, $$R_n = U_1 \cdots U_n$$ and $$S_n = -\ln R_n = X_1 + \cdots +X_n$$. Then $$I_n = E[(U_1 \cdots U_n)^{U_1 \cdots U_n}] = E[R_n^{R_n}] = E[\exp(R_n \ln(R_n))] = E[\exp(-S_ne^{-S_n})].$$ The non-negative sequence $$(R_n)_{n \ge 1}$$ is non-decreasing, so it converges necessarily to $$0$$ (since Borel Cantelli lemma prevents $$(U_n)_{n \ge 1}$$ from converging to $$1$$). Hence $$R_n^{R_n} \to 1$$ as $$n \to +\infty$$. Since $$0 \le R_n^{R_n} \le 1$$. Lebesgue dominated convergence applies so $$I_n \to 1$$ as $$n \to +\infty$$.
  By construction, the random variables $$(X_n)_{n \ge 1}$$ are independent random variables with distribution Exponential$$(1)$$, so the distribution of $$S_n$$ is Gamma$$(n,1)$$. Hence $$I_n = \int_0^\infty \exp(-se^{-s}) \frac{s^{n-1}e^{-s}}{(n-1)!}ds.$$ For all $$s \ge 0$$, $$\exp(-se^{-s}) = \sum_{k=0}^\infty \frac{(-se^{-s})^k}{k!}.$$ This series is normally convergent with regard to $$s \ge 0$$. Since we integrate with regard to a probability measure, we derive $$\begin{eqnarray*} I_n &=& \sum_{k=0}^\infty \int_0^\infty \frac{(-se^{-s})^k}{k!} \frac{s^{n-1}e^{-s}}{(n-1)!}ds \\ &=& \sum_{k=0}^\infty \frac{(-1)^k}{k!(n-1)!} \int_0^\infty s^{k+n-1}e^{-(k+1)s}ds \\ &=& \sum_{k=0}^\infty \frac{(-1)^k}{k!(n-1)!} \frac{(k+n-1)!}{(k+1)^{k+n}}.\end{eqnarray*}$$ $$\begin{eqnarray*} I_{n+1}-I_n &=& \sum_{k=0}^\infty \frac{(-1)^k}{k!} \frac{(k+n-1)!}{n!(k+1)^{k+n+1}} (k+n - n(k+1)) \\ &=& \frac{n-1}{n!}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{(k-1)!} \frac{(k+n-1)!}{(k+1)^{k+n+1}} \end{eqnarray*}$$ The last sum of alternating series is strictly positive since for all $$k \ge 1$$, $$\begin{eqnarray*} \Big(\frac{1}{k!} \frac{(k+n)!}{(k+2)^{k+n+2}} \Big) \Big/ \Big( \frac{1}{(k-1)!} \frac{(k+n-1)!}{(k+1)^{k+n+1}} \Big) &=& \frac{k+n}{k(k+2)} \Big/\Big(1+\frac{1}{k+1}\Big)^{k+n+1} \\ &\le& \frac{k+n}{k(k+2)} \Big/\Big(1+\frac{k+n+1}{k+1}\Big) \\ &=& \frac{k+n}{k(k+2)} \times\frac{k+1}{2k+n+1}\\ &<& 1. \end{eqnarray*}$$ We are done.
  • Awesome! I have checked all the calculations. But I have one question here: $I_n=\sum_{k=0}^\infty \frac{(-1)^k}{k!(n-1)!} \frac{(k+n-1)!}{(k+1)^{k+n}}$ How to show from here that $\lim_{n\rightarrow\infty} I_n=1$ ? Only the first term (when $k=0$) gives 1 and all other terms vanish after taking the limit term by term. But can I exchange the limit and series? and what is the theorem to permit this, thank you! Jul 2, 2022 at 22:43
  • You can try to apply Lebesgue dominated convergence theorem (view the sum as an integral with regard to the counting measure on $\mathbb{N}]$). Jul 3, 2022 at 13:10
  We can get the result from the fact that for any function $$f$$, we have $$\int_0^{x}\frac{1}{y}\int_0^{y}\ln\left(\frac{z}{y}\right)^nf(z)dz dy =\int_0^x\int_0^1\ln(u)^nf(y u)du dy = \int_0^1\frac{\ln(u)^n}{u}\int_0^{xu}f(v)dv du \\ = \left.\left[\frac{\ln(u)^{n+1}}{n+1}\int_0^{x u}f(v)dv\right]\right|^1_0 - \int_0^1\frac{\ln(u)^{n+1}}{n+1}f(x u)du = -\frac{1}{n+1}\int_0^x\ln\left(\frac{y}{x}\right)^{n+1}f(y)dy.$$ $$I_{n+1}$$ can be written as $$I_{n+1} = \int_0^1\frac{1}{x_n}\int_0^{x_n}\frac{1}{x_{n-1}}\int_0^{x_{n-1}}...\int_0^{x_2}\frac{1}{x_1}\int_0^{x_1}x_0^{x_0}dx_0dx_1...dx_{n-1}dx_n.$$ Applying the above identity $$n$$ times then gives $$I_{n+1} = \frac{(-1)^{n}}{n!}\int_0^1\ln(x)^n x^xdx.$$ Changing variables to $$x = \exp(-u)$$ gives an alternate way to write it that makes it more clear why $$I_{n+1}\rightarrow 1$$: $$I_{n+1} = \frac{1}{n!}\int_0^\infty x^ne^{-x}e^{-xe^{-x}}dx.$$ $$e^{-xe^{-x}}$$ approaches $$1$$ as $$x\rightarrow\infty$$, and $$x^n e^{-x}$$ has its largest contribution to the integral near $$x = n$$. So as $$n$$ gets larger, this integral effectively becomes the gamma integral, equal to $$n!$$.
  I suppose there is one more thing to this. If we want a fancy sum, we have $$I_{n+1} = \frac{1}{n!}\int_0^\infty x^ne^{-x}e^{-xe^{-x}}dx = \sum_{k=1}^\infty \int_0^\infty \frac{(-1)^{k-1} x^{n+k-1} e^{-k x}}{n!(k-1)!}dx \\= -\sum_{k=1}^\infty\frac{(-1)^k(n+k-1)!}{k^{n+k}n! (k-1)!} = -(-1)^n\sum_{k=1}^\infty \binom{n+k-1}{n}(-k)^{-(n+k)}$$
  • Thank you, this trick is fantastic! but how to show the limit of $I_n$ is equal to 1? To do this I need to switch the limit and series, but what is the theorem to guarantee this? Jul 2, 2022 at 23:45
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Sophomore's Eternal Dream $\int_0^1\cdots\int_0^1 (t_1t_2\cdots t_n)^{t_1t_2\cdots t_n} dt_1 dt_2\cdots dt_n$
  
  Given
  
  $$I_n=\int_0^1\int_0^1\cdot\cdot\cdot\int_0^1 (t_1t_2\cdot\cdot\cdot t_n)^{t_1t_2\cdot\cdot\cdot t_n} dt_1 dt_2\cdot\cdot\cdot dt_n$$
  
  Prove: $$I_1=I_2
  
  For $$n=1$$, it is Sophomore's dream: https://en.wikipedia.org/wiki/Sophomore%27s_dream
  
  $$I_1=\int_0^1 t_1^{t_1} dt_1=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^n}$$
  
  For $$n=2$$, $$I_2=\int_0^1\int_0^1 (t_1t_2)^{t_1t_2}dt_1 dt_2$$
  
  let $$x=t_1, y=t_1 t_2$$, where the integral region is mapping to the triangle region $$0, and Jacobian $$J=\frac{1}{x}$$
  
  $$I_2=\int_0^1\int_0^x y^y \frac{1}{x}dy dx=\int_0^1\int_y^1 y^y \frac{1}{x} dx dy=-\int_0^1 y^y\ln(y) dy$$
  
  Series expansion: $$y^y=e^{y\ln(y)}=\sum_{n=0}^\infty \frac{1}{n!}y^n \ln^n(y)$$
  
  $$I_2=-\sum_{n=0}^\infty \frac{1}{n!}\int_0^1 y^n \ln^{n+1}(y) dy$$
  
  $$\int_0^1 y^n \ln^{n+1}(y) dy=\frac{(-1)^{n+1}n!}{(n+1)^{n+1}}$$
  
  After simplify the result we got:
  
  $$I_2=I_1$$
  
  Next, for $$I_3$$, let $$x=t_1, y=t_1t_2, z=t_1t_2t_3$$, Jacobian is $$J=\frac{1}{z}$$
  
  $$I_3=\int_0^1\int_0^x\int_0^y z^z\frac{1}{z} dzdydx$$
  
  Question.1: Is there an analytic result for $$I_3$$?
  
  Question.2: How to generalize the calculation to $$I_n$$, as well as the limit of $$I_n$$?
  
  • How about "Sophomore's Eternal Dream" Cause there it's like an infinite integral? Jul 2, 2022 at 19:15
  • Repeated integration by parts gives $\int_0^1 \ln(x)^n x^x dx/n!$ for $n+1$ variables. Jul 2, 2022 at 21:15
  • how to separate them? @eyeballfrog Jul 2, 2022 at 21:40
  
  Let $$(U_n)_{n \ge 1}$$ be a sequence of independent uniform random variables. By removing null sets, we may assume that for every $$n \ge 1$$, $$U_n>0$$ everywhere For every $$n \ge 1$$, set $$X_n = -\ln(U_n)$$, $$R_n = U_1 \cdots U_n$$ and $$S_n = -\ln R_n = X_1 + \cdots +X_n$$. Then $$I_n = E[(U_1 \cdots U_n)^{U_1 \cdots U_n}] = E[R_n^{R_n}] = E[\exp(R_n \ln(R_n))] = E[\exp(-S_ne^{-S_n})].$$ The non-negative sequence $$(R_n)_{n \ge 1}$$ is non-decreasing, so it converges necessarily to $$0$$ (since Borel Cantelli lemma prevents $$(U_n)_{n \ge 1}$$ from converging to $$1$$). Hence $$R_n^{R_n} \to 1$$ as $$n \to +\infty$$. Since $$0 \le R_n^{R_n} \le 1$$. Lebesgue dominated convergence applies so $$I_n \to 1$$ as $$n \to +\infty$$.
  By construction, the random variables $$(X_n)_{n \ge 1}$$ are independent random variables with distribution Exponential$$(1)$$, so the distribution of $$S_n$$ is Gamma$$(n,1)$$. Hence $$I_n = \int_0^\infty \exp(-se^{-s}) \frac{s^{n-1}e^{-s}}{(n-1)!}ds.$$ For all $$s \ge 0$$, $$\exp(-se^{-s}) = \sum_{k=0}^\infty \frac{(-se^{-s})^k}{k!}.$$ This series is normally convergent with regard to $$s \ge 0$$. Since we integrate with regard to a probability measure, we derive $$\begin{eqnarray*} I_n &=& \sum_{k=0}^\infty \int_0^\infty \frac{(-se^{-s})^k}{k!} \frac{s^{n-1}e^{-s}}{(n-1)!}ds \\ &=& \sum_{k=0}^\infty \frac{(-1)^k}{k!(n-1)!} \int_0^\infty s^{k+n-1}e^{-(k+1)s}ds \\ &=& \sum_{k=0}^\infty \frac{(-1)^k}{k!(n-1)!} \frac{(k+n-1)!}{(k+1)^{k+n}}.\end{eqnarray*}$$ $$\begin{eqnarray*} I_{n+1}-I_n &=& \sum_{k=0}^\infty \frac{(-1)^k}{k!} \frac{(k+n-1)!}{n!(k+1)^{k+n+1}} (k+n - n(k+1)) \\ &=& \frac{n-1}{n!}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{(k-1)!} \frac{(k+n-1)!}{(k+1)^{k+n+1}} \end{eqnarray*}$$ The last sum of alternating series is strictly positive since for all $$k \ge 1$$, $$\begin{eqnarray*} \Big(\frac{1}{k!} \frac{(k+n)!}{(k+2)^{k+n+2}} \Big) \Big/ \Big( \frac{1}{(k-1)!} \frac{(k+n-1)!}{(k+1)^{k+n+1}} \Big) &=& \frac{k+n}{k(k+2)} \Big/\Big(1+\frac{1}{k+1}\Big)^{k+n+1} \\ &\le& \frac{k+n}{k(k+2)} \Big/\Big(1+\frac{k+n+1}{k+1}\Big) \\ &=& \frac{k+n}{k(k+2)} \times\frac{k+1}{2k+n+1}\\ &<& 1. \end{eqnarray*}$$ We are done.
  • Awesome! I have checked all the calculations. But I have one question here: $I_n=\sum_{k=0}^\infty \frac{(-1)^k}{k!(n-1)!} \frac{(k+n-1)!}{(k+1)^{k+n}}$ How to show from here that $\lim_{n\rightarrow\infty} I_n=1$ ? Only the first term (when $k=0$) gives 1 and all other terms vanish after taking the limit term by term. But can I exchange the limit and series? and what is the theorem to permit this, thank you! Jul 2, 2022 at 22:43
  • You can try to apply Lebesgue dominated convergence theorem (view the sum as an integral with regard to the counting measure on $\mathbb{N}]$). Jul 3, 2022 at 13:10
  We can get the result from the fact that for any function $$f$$, we have $$\int_0^{x}\frac{1}{y}\int_0^{y}\ln\left(\frac{z}{y}\right)^nf(z)dz dy =\int_0^x\int_0^1\ln(u)^nf(y u)du dy = \int_0^1\frac{\ln(u)^n}{u}\int_0^{xu}f(v)dv du \\ = \left.\left[\frac{\ln(u)^{n+1}}{n+1}\int_0^{x u}f(v)dv\right]\right|^1_0 - \int_0^1\frac{\ln(u)^{n+1}}{n+1}f(x u)du = -\frac{1}{n+1}\int_0^x\ln\left(\frac{y}{x}\right)^{n+1}f(y)dy.$$ $$I_{n+1}$$ can be written as $$I_{n+1} = \int_0^1\frac{1}{x_n}\int_0^{x_n}\frac{1}{x_{n-1}}\int_0^{x_{n-1}}...\int_0^{x_2}\frac{1}{x_1}\int_0^{x_1}x_0^{x_0}dx_0dx_1...dx_{n-1}dx_n.$$ Applying the above identity $$n$$ times then gives $$I_{n+1} = \frac{(-1)^{n}}{n!}\int_0^1\ln(x)^n x^xdx.$$ Changing variables to $$x = \exp(-u)$$ gives an alternate way to write it that makes it more clear why $$I_{n+1}\rightarrow 1$$: $$I_{n+1} = \frac{1}{n!}\int_0^\infty x^ne^{-x}e^{-xe^{-x}}dx.$$ $$e^{-xe^{-x}}$$ approaches $$1$$ as $$x\rightarrow\infty$$, and $$x^n e^{-x}$$ has its largest contribution to the integral near $$x = n$$. So as $$n$$ gets larger, this integral effectively becomes the gamma integral, equal to $$n!$$.
  I suppose there is one more thing to this. If we want a fancy sum, we have $$I_{n+1} = \frac{1}{n!}\int_0^\infty x^ne^{-x}e^{-xe^{-x}}dx = \sum_{k=1}^\infty \int_0^\infty \frac{(-1)^{k-1} x^{n+k-1} e^{-k x}}{n!(k-1)!}dx \\= -\sum_{k=1}^\infty\frac{(-1)^k(n+k-1)!}{k^{n+k}n! (k-1)!} = -(-1)^n\sum_{k=1}^\infty \binom{n+k-1}{n}(-k)^{-(n+k)}$$
  • Thank you, this trick is fantastic! but how to show the limit of $I_n$ is equal to 1? To do this I need to switch the limit and series, but what is the theorem to guarantee this? Jul 2, 2022 at 23:45
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract contains a detailed mathematical discussion and multiple questions. It starts with a given integral expression and proceeds to prove a specific equality for the cases \( n = 1 \) and \( n = 2 \). It then poses two questions regarding the analytic result for \( I_3 \) and the generalization to \( I_n \) and the limit of \( I_n \).
  
  ### Summary:
  - **Document Description**: The document discusses a specific integral expression \( I_n \) and provides a detailed proof for the equality \( I_1 = I_2 \). It then poses two questions about the analytic result for \( I_3 \) and the generalization to \( I_n \) and the limit of \( I_n \).
  - **Questions**:
    1. Is there an analytic result for \( I_3 \)?
    2. How to generalize the calculation to \( I_n \), as well as the limit of \( I_n \)?
  
  ### Label:
  Label: multiple-question
  
  ### Additional Note:
  The provided answers in the comments and the subsequent detailed solution by another user address the questions and provide a rigorous proof for the limit of \( I_n \) as \( n \to \infty \). However, the original