Im reading this. On page 170, $X$ is a Banach Space , $BIP(X)$ is mean bounded imaginary power. The Mellin transform for one function $f(t)$ is: $$M(f)(\tau)= \int_{0}^{\infty}f(t)t^{\tau - 1}dt $$
and the Mellin's inversion formula: $$f(t)=\frac{1}{2\pi i}\int_{c-\infty}^{c+\infty}Mf(\tau)t^{-\tau}d\tau.$$
Then, using properties of Mellin transform and inverse Mellin transform I did:
$$M(t^{\epsilon}(\epsilon +t)^{-2\epsilon})= \epsilon^{\tau -\epsilon } \frac{\Gamma(\tau +\epsilon )\Gamma(\epsilon-\tau)}{\Gamma(2\epsilon)}$$
and using functional calculus (I don't know so much, is something like change one operator for one variable, I'm learning) so $$A^{\epsilon}(\epsilon +A)^{-2\epsilon}x= \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}A^{-\epsilon}\epsilon^{\tau -\epsilon }\frac{\Gamma(\tau +\epsilon )\Gamma(\epsilon-\tau)}{\Gamma(2\epsilon)}x d\tau.$$
So my idea is take limit under integral when $\epsilon\to 0$ but what do I do with the integral: $$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\Gamma(\tau)\Gamma(-\tau)}{\Gamma(0)}x dx?$$
I think $A^{0}=I$ and I'm not sure if the gamma function is continuous? Thanks