contour intrgration, what's the right answer?

Question

There exists an integral as follow: $$ \bar G(t)=\int_{-\infty}^{\infty}\frac{dE}{2\pi\hbar}e^{-iEt/\hbar}\frac{1}{E-\epsilon+i0^{+}} $$ My solution is: $$ {2\pi\hbar}\bar G(t)=-i\pi e^{-i\epsilon t/\hbar} + P.V.\int_{-\infty}^{\infty} dE\frac{e^{-iEt/\hbar}}{E-\epsilon}\\=-i\pi e^{-i\epsilon t/\hbar} + 2\pi i \frac{1}{2} Res\Big(\frac{e^{-iEt/\hbar}}{E-\epsilon},\epsilon\Big)=-i\pi e^{-i\epsilon t/\hbar} + i\pi e^{-i\epsilon t/\hbar}=0 $$ where I have used Principle Value formla(a.k.a. Sokhotski–Plemelj theorem). What's wrong? someone say the answer is $\frac{-i}{\hbar}\theta(t)e^{-i0^{+}t}\exp(-i\epsilon t/\hbar)$.