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Let $ R_\lambda (x,y) = \frac{1}{2\sqrt{-\lambda}} e^{-\sqrt{-\lambda}|x-y|}$ .

Now according to the book I am following if $\lambda > 0$ and we shift $\lambda \rightarrow \lambda + i\epsilon$, we get

$ R_\lambda (x,y) = \frac{1}{2\sqrt{-\lambda}} e^{-\sqrt{-\lambda}|x-y|} \rightarrow \frac{i}{2\sqrt{\lambda}}e^{i\sqrt{\lambda}|x-y| -\epsilon|x-y|/2\sqrt{\lambda}}$

To show this I had $\sqrt{-\lambda}=\sqrt{|\lambda|}e^{-i\pi/2} = -i \sqrt{|\lambda|}$, and then I expanded $\sqrt{-\lambda -i\epsilon} = -i\sqrt{|\lambda|} + \frac{i\epsilon}{-2i\sqrt{\lambda}}$. However, this gives a positive sign for the $\epsilon$ term of the exponential in the resolvent.

Is there anything I am missing here? Any help will be appreciated.

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