Bohr–Mollerup theorem for $x\gt 1$

Question

The Bohr–Mollerup theorem states that $f(x)=\int_{0}^\infty t^{x-1}e^{-t}\, dt$ is the only function on $x\gt 0$ such that the following conditions are satisfied simultaneously:

1. $f(1)=1,$
2. $\forall x\gt 0:\, f(x+1)=xf(x),$
3. $\forall x\gt 0: f(x)$ is logarithmically convex.

But is $f(x)=\int_0^\infty t^{x-1}e^{-t}\, dt$ the only function on $x\ge 1$ such that

1. $f(1)=1,$
2. $\forall x\gt 1:\, f(x+1)=xf(x),$
3. $\forall x\gt 1: f(x)$ is logarithmically convex?

If not, what would be a counterexample?