On deriving the standard form of the Klein-Gordon propagator

[Nauj Onerom]

I'm trying to make sense of the derivation of the Klein-Gordon propagator in Peskin and Schroeder using contour integration. It seems the main step in the argument is that $e^{-i p^0(x^0-y^0)}$ tends to zero (in the $r\rightarrow\infty$ limit) along a semicircular contour below (resp. above) the real line for $x^0 > y^0$ (resp. $x^0< y^0$). It is this that I'm having trouble making sense of. If we consider $$ \int_{C_{below}} \frac{dp^0}{2\pi}\frac{e^{ip(x--y)}}{i(p^2-m^2)} = \int_{0}^\pi \frac{d\theta}{2\pi}(-ire^{i\theta})\frac{e^{-i[re^{i\theta}(x^0-y^0) - \vec{p}\cdot(\vec{x}-\vec{y})]}}{i((re^{i\theta})^2 - \vec{p}^2 - m^2)},$$ the integrand seems to tend to zero as $r\rightarrow\infty$ regardless of whether $x^0 > y^0$ or $y^0 > x^0$ since the exponential is bounded. So I guess my question is what stops us from choosing this same contour for the $y^0 > x^0$ case, which would give a nonzero result?

 

[frenchfryburrito]

To use contour integration you need to pick a contour where the extra half-semicircle will go to 0 for large imaginary p0 (due to the exponential). The upper half plane corresponds to +ip0, while the lower half plane corresponds to -ip0. For x0>y0, -ip0 will converge to 0 along the semicircle at infinity (this is the lower half plane), while ip0 blows up along the semicircle at infinity (this is the upper half plane. The argument works exactly in reverse for x0<y0.