Franklin Pezzuti Dyer


A Half-Exponential Integral

2018 May 19

Find some upper and lower bounds for the integral 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 Now, by making the substitution $x\to f(x)$ and using integration by parts in the second integral, we see that and so our original integral is equal to Now consider the following integral: 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 ...or, after simplifying the integrals in the upper and lower bounds, By combining this inequality with equation (1), we have that 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 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 given that $g:\mathbb R\mapsto\mathbb R$ is continuous, increasing, and that $g(g(x))=x^2+1$.

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