# 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 $N$.)

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IIRC Hardy and Wright says a lot about average values of arithmetic functions such as as phi. (there's a chapter or two on it). – Kevin Buzzard Nov 28 '09 at 11:59

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.
It isn't difficult to argue $6N/\pi^2$ directly either. You should accept your own answer! You get a badge for that. – Greg Kuperberg Nov 28 '09 at 16:15