Franklin’s Blog

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 \(g(g(x))=x^2+1\).