Problem with Infinite number of iterated integrals

Question

I have been working on a problem from a past high school Math Competition and I have been stumped. I am asked to compute $$\lim_{n\rightarrow\infty} \int_{0}^{1}\int_{0}^{1}\cdot\cdot\cdot\int_{0}^{1}\frac{x_{1}^{505}+x_{2}^{505}+...+x_{n}^{505}}{x_{1}^{2020}+x_{2}^{2020}+...+x_{n}^{2020}}\,dx_{1}\,dx_{2}\,...\,dx_{n}$$ At first I tried a lot of direct evaluations of the integral to find a simplification i.e. solving the general integrals $$\int_{0}^{1}\frac{x^{505}+a}{x^{2020}+b}\,dx$$ but this really only complicated the matter, so I began to look for more clever manipulations. I noticed for $x_i \in [0,1]$ that we would have the integrand $$ f(x_1,x_2,...,x_n) = \frac{\frac{1}{n}}{\frac{1}{n}}\cdot\frac{\sum_{i=1}^{n}x_{i}^{505}}{\sum_{i=1}^{n}x_{i}^{2020}}$$ which can be seen as a Riemann sum so at infinity, $$\lim_{n\rightarrow\infty}f(x_1,x_2,...,x_n)=\frac{\int_{0}^{1}x^{505}\,dx}{\int_{0}^{1}x^{2020}\,dx}=\frac{2021}{506}$$ My thought process was that the region in $\mathbb{R}^n$ being integrated over is the unit cube, so I could try to view the integral as the product of the volume of the region by some average value of the function in the region, which would maybe approach a constant value because of the Riemann sum? I was just unsure if any of this can be justified, especially when the number of integrals approaches infinity. Any help or advice is appreciated.

Answer 1

Here's a solution that's entirely analytical.

Let $I_n=\int_{0}^{1}\int_{0}^{1}\cdot\cdot\cdot\int_{0}^{1}\frac{x_{1}^{505}+x_{2}^{505}+...+x_{n}^{505}}{x_{1}^{2020}+x_{2}^{2020}+...+x_{n}^{2020}}\,dx_{1}\,dx_{2}\,...\,dx_{n}$

Then $$\begin{split} I_n &= \int_{0}^{1}\cdot\cdot\cdot\int_{0}^{1}\left(x_{1}^{505}+...+x_{n}^{505}\right)\int_0^{+\infty}e^{-(x_1^{2020}+...+x_n^{2020})t}\,dt\,dx_{1}\,...\,dx_{n}\\ &=\int_0^{+\infty}\int_{0}^{1}\cdot\cdot\cdot\int_{0}^{1}\left(x_{1}^{505}+...+x_{n}^{505}\right)e^{-(x_1^{2020}+...+x_n^{2020})t}\,dx_{1}\,...\,dx_{n}\,dt\\ &=\int_0^{+\infty}\int_{0}^{1}\cdot\cdot\cdot\int_{0}^{1}\left(x_{1}^{505}+...+x_{n}^{505}\right)e^{-tx_1^{2020}}...e^{-tx_n^{2020}}\,dx_{1}\,...\,dx_{n}\,dt\\ &= n\int_0^{+\infty}\int_0^1x_1^{505}e^{-x_1^{2020}t}dx_1\left( \int_0^1 e^{-ty^{2020}}dy\right)^{n-1}dt \,\,\text{(by symmetry)}\\ &= \int_0^{+\infty}\int_0^1 x^{505}e^{-\frac u n x^{2020}}dx\left( \int_0^1 e^{-\frac u n y^{2020}}dy\right)^{n-1} du\,\,\text{ (with }u=nt\text{)}\\ &= \int_0^{+\infty}\frac{\int_0^1 x^{505}e^{-\frac u n x^{2020}}dx}{\int_0^1 e^{-\frac u n y^{2020}}dy}\left( \int_0^1 e^{-\frac u n y^{2020}}dy\right)^n du \end{split}$$ Now, for a given $u>0$ $$\begin{split} \left( \int_0^1 e^{-\frac u n y^{2020}}dy\right)^n &= \exp\left( n\ln \int_0^1e^{-\frac u n y^{2020}}dy\right)\\ &= \exp\left( n\ln \int_0^1\left(1-\frac u n y^{2020}+\mathcal O(\frac 1 {n^2})\right)dy\right)\\ &=\exp\left( n\ln \left(1 - \frac u {2021n} + \mathcal O(\frac 1 {n^2}) \right)\right)\\ &=e^{-\frac u {2021}}+o(1) \end{split}$$ and the same integral is dominated by 1. Therefore by the Dominated Convergence Theorem: $$ \lim_{n\rightarrow+\infty}I_n = \int_0^{+\infty}\frac{\int_0^1 x^{505}dx}{\int_0^1 1dy}e^{-\frac u {2021}} du = \frac{2021}{506} $$

Answer 2

I was the author of this question! Glad to see it is still engaging minds ~1 year after the initial contest. :) Your answer is correct, but as you noted your reasoning is not really rigorous, although you are on the right track.

To give you a pointer, see if you can find a statistical interpretation of that integral instead of geometric.