Franklin Pezzuti Dyer

A Half-Exponential Integral

2018 May 19

Find some upper and lower bounds for the integral $\int_0^1 f(x)dx$ given that $f:\mathbb R\mapsto\mathbb R$ is continuous, increasing, and that $f(f(x))=e^x$.

This is an interesting problem that I came up with while investigating half-exponential functions. I last investigated this type of function in this post about fractional iterates of functions, and I posed this puzzle to the users of Math Stack Exchange. In this short post, I will present my original solution to the problem.

As I showed in the previous post, the function $f(x)$ must be bounded above by $e^x$ and below by $x$ for all $x\in\mathbb R$. This implies that $0\lt f(0)\lt 1$. For convenience, I will let $f_0=f(0)$. To solve the problem, I will first split up the integral in question into $\int_0^1 f(x)dx=\int_0^{f_0} f(x)dx+\int_{f_0}^1 f(x)dx$ Now, by making the substitution $x\to f(x)$ and using integration by parts in the second integral, we see that $\begin{align} I:=\int_{f_0}^1 f(x)dx &=\int_{0}^{f^{-1}(1)} f(f(x))f'(x)dx\\ &=\int_{0}^{f_0} e^xf'(x)dx\\ &=\big[e^xf(x)\big]_0^{f_0}-\int_{0}^{f_0} e^xf(x)dx\\ &=e^{f_0}-f_0-\int_{0}^{f_0} e^xf(x)dx\\ \end{align}$ and so our original integral is equal to $I=e^{f_0}-f_0-\int_0^{f_0} (e^x-1)f(x)dx\tag{1}$ Now consider the following integral: $\int_0^{f_0} (e^x-1)f(x)dx$ It is always positive, because of the previously mentioned bounds for $f$. Furthermore, since $f$ is increasing, for all $x\in [0,f_0]$, we have that $f(x)\in [f_0,1]$. Thus, we have that $f_0\int_0^{f_0} (e^x-1)dx\lt\int_0^{f_0} (e^x-1)f(x)dx\lt \int_0^{f_0} (e^x-1)dx$ ...or, after simplifying the integrals in the upper and lower bounds, $f_0e^{f_0}-f_0^2-f_0\lt\int_0^{f_0} (e^x-1)f(x)dx\lt e^{f_0}-f_0-1$ By combining this inequality with equation (1), we have that $1 \lt I\lt (1-f_0)e^{f_0}+f_0^2$ It can be shown using basic calculus that for all $x\in [0,1]$, the quantity $(1-x)e^x+x^2$ is always less than or equal to $\ln^2(2)-2\ln(2)+2\approx 1.0942$. Thus, since $f_0\in [0,1]$, we have that $1 \lt I\lt \ln^2(2)-2\ln(2)+2$ Not only are these bounds correct, but they are also the best possible bounds, in that there exist functions $f$ making the integral arbitrarily close to $1$ or $\ln^2(2)-2\ln(2)+2$. In order to make the integral arbitrarily close to $1$, one need only choose a function $f(x)$ such that $f(0)$ is arbitrarily close to $1$, and such that $f(x)$ is arbitrarily close to $1$ for $x\in [0,f(0)]$, so that the graph of $f$ for $0\le x\le 1$ looks very much like a horizontal line segment. Similarly, to make $I$ arbitrarily close to $\ln^2(2)-2\ln(2)+2$, one should choose $f$ such that $f(0)=\ln(2)$ and such that $f(x)$ is very close to $\ln(2)$ for $x\in [0,\ln(2)]$.

The solution to this puzzle concludes this (short) blog post! I shall leave a final puzzle for the reader which is slightly more difficult than the one solved in this post:

Prove that $\frac{23}{27}\lt \int_0^1 g(x)dx\lt\frac{55}{54}$ given that $g:\mathbb R\mapsto\mathbb R$ is continuous, increasing, and that $g(g(x))=x^2+1$.