Question

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 $N$.)

Answer 1

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).

Answer 2

I've just realized I was being a little bit slow. I had already found on the internet that $n^{-2}\sum_{k=1}^nφ(k)$ is roughly $3/π^2$ and stupidly didn't notice that I could "differentiate" this to get exactly what I want. That is, $\sum_1^N φ(k)$ is about $3N^2/π^2$, so the difference between the sum to $N+M$ and the sum to $N$ is around $6NM/π^2$, from which it follows that the average value near $N$ is around $6N/π^2$, which is entirely consistent with the well-known fact that the probability that two random integers are coprime is $6/π^2$.

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 ζ(2), or 1^{-2}+2^{-2}+... = π^2/6. It's not that hard to turn these formal arguments into a rigorous proof, since everything converges nicely.