A bounded infinite cycle as a product of bounded involutions Let $\sigma$ be a permutation of $\mathbf Q.$ We call $\sigma$ bounded (the term might be somewhat misleading, but however it is used in a couple of papers) if there is a real number $M$ such that
$$
\mathrm{dist}(\sigma x,x) =|\sigma x-x| \le M
$$
for all $x \in \mathbf Q.$ 
Now can a bounded infinite cycle
$$
\pi=(\ldots,-2,-1,0,1,2,\ldots)
$$
be written as a product of bounded involutions (permutations of order two)? Notice that "nice" bounded products of, say two, disjoint infinite cycles can be written as products of bounded involutions, e.g. the permutation
$$
(\ldots,-4,-2,0,2,4,\ldots) (\ldots,-3,-1,1,3,\ldots).
$$
EDIT: to clarify a bit chandok's argument below: suppose $\pi$ is a product of two involutions $\sigma$ and $\tau$: $\pi=\sigma \tau.$ Then
$$
\sigma \tau \sigma(n+1)=\sigma(n+1)+1.
$$
for all integers $n.$ On the other hand, $n+1=\pi(n)$ and $\tau \sigma=\pi^{-1},$ whence
$$
\sigma (\tau \sigma) \pi(n) = \sigma \pi^{-1} \pi(n)=\sigma(n)=\sigma(n+1)+1.
$$
Thus $\sigma(n+1)=\sigma(n)-1$ and $\sigma(n)=a-n$ for all naturals $n.$ Thus $\sigma$ is unbounded. 
The argument seems to be not easily adaptable for products of more than two involutions. 
 A: Suppose you have involutions $\sigma, \tau$ such that $\sigma \tau (n) = n+1$ forall $n \in \mathbb{Z}$, and $\sigma \tau (n) = n$ otherwise
Then forall $n \in \mathbb{Z}$, we have $\sigma \tau (n) = n+1$, so $\tau (n) = \sigma\sigma\tau(n) = \sigma(n+1)$,
and $\sigma\tau\sigma(n+1) = \sigma\tau\tau(n) = \sigma(n)$.
Moreover, $\sigma \tau (n-1) = n \neq n+1 = \sigma \tau (n)$, so $\sigma(n) = \tau(n-1) \neq \tau(n) = \sigma(n+1)$, so $\sigma(n+1)$ is not fixed by $\sigma \tau$.
Therefore $\sigma(n+1)$ is an integer, and $\sigma(n+1)+1 = \sigma(n)$, and likewise for $\tau$. So there exists integers $a$ and $b$ such that $\sigma(n) = a-n$ and $\tau(n) = b-n$.
And these involutions are not bounded.
Edit : So it can't be written as a product of only two involutions (I missed the part not being restricted to products of more than 2 involutions, I will think about it)
As you noticed, using a product of 2 bounded involutions you can form the product of two infinite cycles going opposite directions.
For $(a,b) \in \mathbb{R}$, call $\tau_{(a,b)} = \prod_{n \in \mathbb{Z}} ((n+a) (n+b))$.
$\tau_{(a,b)}$ is a bounded involution, and $\sigma_{(a,b)} = \tau_{(a+1,b)} \tau_{(a,b)}$ is a permutation that adds $1$ to numbers of the form $n+a$ and substracts $1$ to numbers of the form $n+b$.
Now consider the (infinite) product $\sigma = \sigma_{(0,1/2)} \sigma_{(1/2,1/3)} \sigma_{(1/3,1/4)} \ldots$.
$\sigma$ is just the permutation you want.
To write it as a finite product of bounded involutions, first rewrite it as a product of two permutations so that the things inside a product all have disjoint support so the involutions later can commute between themselves :
$\sigma = (\sigma_{(0,1/2)} \sigma_{(1/3,1/4)} \ldots) (\sigma_{(1/2,1/3)} \sigma_{(1/4,1/5)} \ldots)$
Then expand each $\sigma_{(a,b)}$ into its product of two involutions
$\sigma = ((\tau_{(1,1/2)}\tau_{(0,1/2)})(\tau_{(4/3,1/4)}\tau_{(1/3,1/4)})\ldots) ((\tau_{(3/2,1/3)}\tau_{(1/2,1/3)})(\tau_{(5/4,1/5)}\tau_{(1/4,1/5)}) \ldots)$
Commute them to get a product of 4 bounded involutions (with infinite support):
$\sigma = (\tau_{(1,1/2)}\tau_{(4/3,1/4)}\ldots)(\tau_{(0,1/2)}\tau_{(1/3,1/4)}\ldots) (\tau_{(3/2,1/3)}\tau_{(5/4,1/5)}\ldots)(\tau_{(1/2,1/3)}\tau_{(1/4,1/5)}\ldots)$
