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