Riemann integral substitution with $e^x$ For a Riemann-integrable function $f$: $[1, e]\longrightarrow\mathbb{R}$ we have
$$\int_1^e f(x)dx=\int_0^1 f(e^x)e^x dx$$
and this easily follows from more general results of Lebesgue integration theory. It's also not hard to see that $(f\circ\exp)\exp$ is Riemann-integrable since it's a.e. continuous whenever $f$ is.
Can we give an elementary proof of this equality, i.e. one that doesn't go beyond the toolbox of elementary analysis and standard Riemann integration theory?
I guess the main obstacle is that $f$ is not necessarily the derivative of an $F$, but there just might be some clever argument relying heavily on the niceness of the $\exp$-function.

The basics for a Riemann integrable function $f$ on $[a,b]$: Suppose $g:[c,d]\to [a,b]$ is strictly increasing and $C^1,$ with $g(c)= a, g(d) = b.$ Then
$$\int_a^bf = \lim \sum_{k=1}^{n}f(g(t_k))(g(t_k)-g(t_{k-1})).$$
By the MVT, the last sum is
$$\sum_{k=1}^{n}f(g(t_k))g'(s_k)(t_k-t_{k-1}).$$
If we had
$$\sum_{k=1}^{n}f(g(t_k))g'(t_k)(t_k-t_{k-1})$$
we'd be done. But $g'$ is uniformly continuous, so it's okay to put that in. Those are the main ideas; it's not too hard to fill in the gaps.
