Limit of integrals (18)

Integration, Limit & Continuity, One-Variable Calculus

The following problem looks weird at first but it’s really not. The problem basically comes from my (successful) attempt to prove the following nice-looking limit in multi-variable calculus

\displaystyle \lim_{n\to\infty} \int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1 + \cdots + x_n}{x_1^2 + \cdots + x_n^2} \ dx_1 \ \cdots \ dx_n=\frac{3}{2}.

I have explained that in the remark below the problem.

Problem. Show that \displaystyle \lim_{n\to\infty} n\int_0^{\infty}\frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^xe^{-t^2} dt \right)^n dx=\frac{3}{2}.

Solution. First notice that, since \displaystyle \int_0^{\infty}e^{-t^2}dt=\frac{\sqrt{\pi}}{2} (see this post!), we have

\displaystyle \begin{aligned} 0 < n\int_1^{\infty}\frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^x e^{-t^2} dt  \right)^n dx < n \int_1^{\infty} \frac{1}{x^{n+1}}\left(\int_0^{\infty}e^{-t^2} dt \right)^n dt=\left(\frac{\sqrt{\pi}}{2}\right)^n.\end{aligned}

Hence, since \displaystyle \frac{\sqrt{\pi}}{2} < 1, we get

\displaystyle \lim_{n\to\infty} n\int_1^{\infty}\frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^x e^{-t^2} dt \right)^n dx=0,

by the squeeze theorem, and thus

\displaystyle \lim_{n\to\infty}n\int_0^{\infty}\frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^xe^{-t^2} dt\right)^n dx = \lim_{n\to\infty}n \int_0^1 \frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^x e^{-t^2} dt \right)^n dx. \ \ \ \ \ \ \ \ \ \ \ (1)

Let

\displaystyle I_n:=\int_0^1 \frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^xe^{-t^2} dt\right)^n dx.

Using the Maclaurin series of e^{-x^2}, it’s clear that

\displaystyle x-\frac{x^3}{3} \le \int_0^x e^{-t^2} \ dt \le x-\frac{x^3}{3}+\frac{x^5}{10}

and so

\displaystyle n \int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}\right)^n dx \le nI_n \le n\int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}+\frac{x^4}{10}\right)^n dx. \ \ \ \ \ \ \ \ (2)

We now find the limits of the sequences on the LHS and the RHS of (2). First, the LHS one. The substitution \displaystyle 1-\frac{x^2}{3}=s gives

\displaystyle n \int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}\right)^n dx=\frac{n}{2}\int_{2/3}^1 \frac{1-e^{-3(1-s)}}{1-s}s^n ds

and hence

\displaystyle \lim_{n\to\infty}n \int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}\right)^n dx=\frac{1}{2}\lim_{s\to1}\frac{1-e^{-3(1-s)}}{1-s}=\frac{3}{2} \ \ \ \ \ \ \ \ \ \ \ (3)

because, as we showed here, \displaystyle \lim_{n\to\infty} n\int_c^1 f(x) \ x^n dx =f(1) for any c \in [0,1) and any continuous function f: [0,1] \to \mathbb{R}.

Now we find the limit of the sequence on the RHS of (2). The idea is the same. We first make the substitution x^2=s to get

\displaystyle n\int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}+\frac{x^4}{10}\right)^n dx=\frac{n}{2}\int_0^1 \frac{1-e^{-s}}{s}\left(1-\frac{s}{3}+\frac{s^2}{10}\right)^nds. \ \ \ \ \ \ \ \ \ (4)

The function \displaystyle y=1-\frac{s}{3}+\frac{s^2}{10}, \ \ s \in [0,1], is decreasing and so it has an inverse, i.e. we can find s in terms of y but to avoid the unnecessary mess, let’s just write s=g(y). Then the substitution \displaystyle y=1-\frac{s}{3}+\frac{s^2}{10} and (4) give

\displaystyle \begin{aligned} n\int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}+\frac{x^4}{10}\right)^n dx=-\frac{n}{2}\int_{23/30}^1 \frac{1-e^{-g(y)}}{g(y)}g'(y)y^n dy \\ =-\frac{n}{2}\int_{23/30}^1 \frac{1-e^{-g(y)}}{g(y)}\left(\frac{g(y)}{5}-\frac{1}{3}\right)^{-1}y^ndy\end{aligned}

and so, again using the fact that \displaystyle \lim_{n\to\infty} n\int_c^1 f(x) \ x^n dx =f(1) for any c \in [0,1) and any continuous function f: [0,1] \to \mathbb{R}, we have

\displaystyle \begin{aligned} \lim_{n\to\infty} n\int_0^1 \frac{1-e^{-x^2}}{x}\left(1-\frac{x^2}{3}+\frac{x^4}{10}\right)^n dx=-\frac{1}{2}\lim_{y\to1}\frac{1-e^{-g(y)}}{g(y)}\left(\frac{g(y)}{5}-\frac{1}{3}\right)^{-1} \\ =-\frac{1}{2}\lim_{z\to0} \frac{1-e^{-z}}{z}\left(\frac{z}{5}-\frac{1}{3}\right)^{-1}=\frac{3}{2}. \end{aligned} \ \ \ \ \ \ \ \ \ (5)

So (2), (3), (5) and the squeeze theorem together give \displaystyle \lim_{n\to\infty} nI_n=\frac{3}{2} and the result follows from (1). \ \Box

Remark (for those readers who are familiar with multi-variable calculus). It follows from the above problem that

\displaystyle \lim_{n\to\infty} \int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1 + \cdots + x_n}{x_1^2 + \cdots + x_n^2} \ dx_1 \ \cdots \ dx_n=\frac{3}{2}.

That’s because we can write \displaystyle \frac{1}{x_1^2 + \cdots + x_n^2}=\int_0^{\infty} e^{-(x_1^2+ \cdots + x_n^2)x} dx and thus

\displaystyle \begin{aligned} \int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1 + \cdots + x_n}{x_1^2 + \cdots + x_n^2} \ dx_1 \ \cdots \ dx_n= n\int_{0}^{1} \cdots \int_{0}^{1} \frac{x_1}{x_1^2 + \cdots + x_n^2} \ dx_1 \cdots \ dx_n \\ =n\int_{0}^{\infty} \int_0^1 \cdots \int_{0}^{1} x_1e^{-(x_1^2+ \cdots + x_n^2)x} dx_1 \cdots \ dx_n \ dx=n \int_0^{\infty}\left(\int_0^1x_1e^{-x_1^2x}dx_1\right)\left(\int_0^1e^{-t^2x}dt\right)^{n-1}dx \\ =n\int_0^{\infty}\frac{1-e^{-x}}{2x}\left(\int_0^1e^{-t^2x}dt\right)^{n-1}dx=n\int_0^{\infty}\frac{1-e^{-x}}{2x}\left(\int_0^{\sqrt{x}} \frac{e^{-t^2}}{\sqrt{x}} \ dx\right)^{n-1}dx \\ =n\int_0^{\infty}\frac{1-e^{-x^2}}{x^n}\left(\int_0^x e^{-t^2} dt \right)^{n-1} dx \end{aligned}

and the result follows because

\displaystyle \begin{aligned} \lim_{n\to\infty} n\int_0^{\infty}\frac{1-e^{-x^2}}{x^n}\left(\int_0^x e^{-t^2} dt \right)^{n-1} dx=\lim_{n\to\infty} n\int_0^{\infty}\frac{1-e^{-x^2}}{x^{n+1}}\left(\int_0^xe^{-t^2} dt \right)^n dx=\frac{3}{2},\end{aligned}

by the above problem.

Published by Y. Sharifi

Published

Leave a comment