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An advanced Green function has support only before its source time. It is selected by a future boundary condition, rather than an initial causal response. For $L={\partial }_t^2+2\lambda {\partial }_t+{\omega }_0^2$ with $\lambda <0$ and ${\Omega }^2={\omega }_0^2-{\lambda }^2>0$, the real-axis inverse Fourier transform of $(-{\omega }^2-2i\lambda \omega +{\omega }_0^2{)}^{-1}$ is $-H(-t)e^{-\lambda t}\sin (\Omega t)/\Omega$. Its first derivative jumps by one at zero. A growing retarded solution exists with a different inverse contour; this illustrates causality versus stability of a Green function.

Let $\tau =t-t^{\prime }$ and $\Omega =\sqrt{K^2-b^2}>0$. Because $|e^{-i(u+iv)\tau }|=e^{v\tau }$, close below for $\tau >0$ and above for $\tau <0$. The lower closure is clockwise. The two residues sum to $i{e}^{b\tau }\sin (\Omega \tau )/\Omega$. Hence, for nonzero $b$,In particular, the printed sign with $e\gamma >0$ gives zero for $t>t^{\prime }$ and an advanced Green function for $t<t^{\prime }$. Reversing that sign puts the poles below and gives a retarded Green function. Both solutions have the unit derivative jump required by $({\partial }_{\tau }^2-2b{\partial }_{\tau }+K^2)G=\delta (\tau )$. If $b=0$, the bare integral has real-axis poles and requires an additional contour prescription; a unique retarded or advanced answer cannot be inferred without one.

For the inverse transform $G(\tau )=\int d\omega e^{-i\omega \tau }g(\omega )/(2\pi )$, positive $\tau$ requires closing a decaying contour below the real axis, clockwise; negative $\tau$ requires closing above, counterclockwise. Poles strictly below give a retarded Green function, while poles strictly above give an advanced Green function. Real-axis poles need a prescription. The sign convention in the Fourier exponential is essential.