TITLE: Geometric mean of reals between 0 and 1
QUESTION [32 upvotes]: What is the geometric mean of all reals between $0$ and $1$?
I was thinking over this, but could not come up with anything useful. Please help me out.

REPLY [11 votes]: This is a more general result. First consider the Generalised mean:
$$\operatorname{GM}(n, m)=\sqrt[n]{\frac1m\sum_{k=1}^mx^n_k}$$
We can extend this definition to a function rather than a sequence. Let $I=[m_0,m_1]$ be a bound such that it is the domain $f(x)$. Now we can write the Generalised mean of the values of $f(x)$ as $$\operatorname{GM}(n, f)=\sqrt[n]{\frac1{m_1-m_0}\int_{m_0}^{m_1}f^n(x)\ \mathrm dx}$$
A property of the discrete Generalised mean was that $n=1$ was the arithmetic mean, $n=2$ was the quadratic mean, $n=0$ was the geometric mean, $n=-1$ was the harmonic mean, as $n\to \infty$ it approached the maximum of the values and as $n\to-\infty$ it approched the minimum of the values. 
If we take $f(x)=x$, $[m_0, m_1]=[0, 1]$ and limit $n\to 0$ (just like the geometric mean for the discrete case) we get $$\lim_{n\to 0}\sqrt[n]{\int_0^1 x^n \mathrm dx}=\lim_{n\to0}\frac1{\sqrt[n]{n+1}}=\lim_{n\to\infty}\left(1+\frac1n\right)^{-n}=e^{-1}$$ as expected. This approach also allows you to find the other means. If you limit $n\to \infty$ or $\to -\infty$ you will get the upper and lower bounds of the domain.