Peskin: 

> by taking $T$ to $0$ in **a** slightly imaginary direction: $T → \infty(1 - i\epsilon)$

where $\epsilon >0$. 

We can take $T → \infty(1 + i\epsilon)$, $\epsilon >0$ and this is still a slightly imaginary direction. And thus you can continue down the desired path. 

**Further explanation:**

> What all this means is that if we use $(1−i\epsilon)H$ instead of $H$, we can be cavalier about the boundary conditions on the endpoints of the path. Any reasonable boundary conditions will result in the ground state as both the initial and final state. Thus we have
[![enter image description here][1]][1]

Srednicki does this to $H$ rather that to $T$ but its equivalent, and for the same reason. 

Down the line, we learn that taking $ H \rightarrow H(1-i\epsilon)$ is the same as taking $ m^2 \rightarrow m^2-i\epsilon$ which results in this Feynman propagator
$$\Delta (x-x') = \int \frac{d^4k}{(2\pi)^4}\frac{e^{i(k-k')}}{k^2+m^2-i\epsilon}$$

and the poles look like this: 

[![enter image description here][3]][3]

However, if we take $ H \rightarrow H(1+i\epsilon)$ and so $ m^2 \rightarrow m^2+i\epsilon$ the poles will shift to 

[![enter image description here][4]][4]

but we have to do the Wick rotation the opposite of show in the picture (i.e clockwise) but this doesn't change the calculation.  

  [1]: https://i.sstatic.net/R3iZG.png

  [3]: https://i.sstatic.net/wPx9l.png
  [4]: https://i.sstatic.net/NjyOQ.png