Given positive real numbers $a_k, k=1,2,...,$ let $f:[0,\infty) \to \mathbb R$ where we take $f(x)$ to be equal to $a_k$ on the interval $(k-1, k]$ for $1\le k < \infty$. If $f(ny)$ is replaced by $y$ in $$\exp\left\{ \int_0^1 \log f(ny) dy\right\},$$ then $$e = \exp\left\{ -\int_0^1 \log y dy\right\}.$$
Why is this true?
That is because $\displaystyle\int_0^1 \log y\,\mathrm d y$ is a convergent improper integral, as $$\int \log y\,\mathrm d y=y\log y -y\quad\text{and}\quad y\log y -y\,\biggr|_0^1= (0\cdot\log 1-1)-0=-1,$$ so $\quad\displaystyle\exp\Bigl(\int_0^1 \log y\,\mathrm d y\Bigr)=\exp(-1)$.