Typical value of totient function Can anyone tell me what the expected value of Euler's totient function φ$(n)$ is (roughly) if you choose a random integer $n$ in the range $[N,N+M]$, where $M$ is large and $N$ is larger than $M$? (I think of $M$ as being $cN$ for some small constant $c$, which, if one wanted an answer accurate to $1+o(1)$, would in reality be a slowly decreasing function of $N$.) 
 A: Let me also mention the following:
You can adapt Schoenberg's result to prove that
1/M * {N <= n <= N + M : phi(n) / n <= t} --> F(t) 
uniformly in t, where F is a distribution function. The proof goes
by computing the moments sum((phi(n)/n)^k , N <= n <= N + M).
You can probably get a O(loglog N / log N) rate of convergence
(as was done by Levin ... if I recall correctly). 
A: I've just realized I was being a little bit slow. I had already found on the internet that $n^{-2}\sum\limits_{k=1}^n\phi(k)$ is roughly $3/\pi^2$ and stupidly didn't notice that I could "differentiate" this to get exactly what I want. That is, $\sum\limits_{k=1}^N \phi(k)$ is about $3N^2/\pi^2$, so the difference between the sum to $N+M$ and the sum to $N$ is around $6NM/\pi^2$, from which it follows that the average value near $N$ is around $6N/\pi^2$, which is entirely consistent with the well-known fact that the probability that two random integers are coprime is $6/\pi^2$.
I'm adding this paragraph after Greg's comment. To argue that the probability that two random integers are coprime, you observe that the probability that they do not have $p$ as a common factor is $(1-1/p^2)$. If you take the product of that over all $p$ then you've got the reciprocal of the Euler product formula for $\zeta(2)$, or $1^{-2}+2^{-2}+\ldots= \pi^2/6$. It's not that hard to turn these formal arguments into a rigorous proof, since everything converges nicely.
