Logarithm of an operator

Question

For a positive operator $A$ we can write $A^{\alpha}=\frac{2\sin(\pi \alpha)}{\pi}\int_0^{+\infty}w^{\alpha-1}(A+w)^{-1}Adw$ for all $\alpha\in[0,1]$.

Is there a similar formula for $\log(A)$ where $A\ge 2$?

Thanks in advance.