Use of Schwinger Feynman parameters in a complex integral

So what went wronggg?

• For a factor $1/(\omega^2-v^2k^2)$ on the real $\omega$ axis, you cannot use a Laplace $\int_0^\infty d\alpha,e^{-\alpha(\omega^2-v^2k^2)}$ because $\omega^2-v^2k^2$ is not positive definite. You need the causal (Feynman) representation $$ \frac{1}{\omega^2-v^2k^2+i0}=i\int_0^\infty d\alpha e^{i\alpha(\omega^2-v^2k^2+i0)}. $$ It fixes convergence and the contour.

• For the double pole you should use $$ \frac{1}{(\omega-qk+i0)^2}=-\int_0^\infty d\beta \beta e^{i\beta(\omega-qk+i0)}. $$ Your laplace for $\omega^{-2}$ on $[0,\infty)$ is not the same object and it’s incompatible with integrating $\omega$ over $\mathbb R$.

• With the correct Schwinger forms, the exponent has $+i\alpha \omega^2$ and $-i\alpha v^2 k^2$. That’s a quadratic in (k). $ -k(,v\alpha+q\beta - ix - v\gamma,) $ treats the $k$ dependence as linear and then $\int dk e^{ik(\cdots)}\sim 2\pi\delta(\cdots)$. That would only be ok if the exponent had no $k^2$ piece.

• $\delta(-iv\alpha-iq\beta-x+iv\gamma)$ is not a thing in standard distribution theory on $\mathbb R$.

So buddy, you start from $$ \mathcal I = -\int \frac{dk}{2\pi}\int \frac{d\omega}{2\pi} \frac{\omega^2}{(\omega^2-v^2k^2+i0) (\omega-qk+i0)^2} \Big[1-\cos(k\Delta x-\omega\Delta t)\Big], $$ with $\Delta x=x-x'$ and $\Delta t=t-t'$. I will split the cosine as $\tfrac12(e^{i\Phi}+e^{-i\Phi})$, and then we'll treat the $I_1$ (the “1”), $I_2\propto e^{i\Phi}$, $I_3\propto e^{-i\Phi}$ separately and $I_3$ is the $(\Delta t,\Delta x)\mapsto-(\Delta t,\Delta x)$ mirror of $I_2$. Using the causal Schwinger representations $$ \frac{1}{\omega^2-v^2k^2+i0} = i\int_0^\infty d\alpha e^{,i\alpha(\omega^2-v^2k^2+i0)} \qquad \frac{1}{(\omega-qk+i0)^2} = -\int_0^\infty d\beta \beta e^{,i\beta(\omega-qk+i0)}. $$

Also convert the numerator by differentiation with respect to $\Delta t$: $$ \omega^2 e^{-i\omega\Delta t} = -\partial_{\Delta t}^2 e^{-i\omega\Delta t}, $$ which is ig the cleanest way to take $\omega^2$ through the Gaussians. For $I_2$ (the $+i\Phi$ phase), the integrand’s exponential is $$ \exp(i\alpha\omega^2 - i\alpha v^2 k^2 + i\beta\omega - i\beta qk + i k\Delta x - i\omega\Delta t). $$

Now the inner integrals are gaussian!

• $\omega$ integral: complete the square in $\omega$: $$ \int_{-\infty}^{\infty}\frac{d\omega}{2\pi} e^{i\alpha\omega^2 + i(\beta-\Delta t)\omega} = \frac{1}{2\sqrt{\pi}}\frac{e^{-i\frac{(\beta-\Delta t)^2}{4\alpha}}}{\sqrt{-i\alpha}}. $$ Then apply $-\partial_{\Delta t}^2$ afterward to account for $\omega^2$.

• $k$ integral: likewise, $$ \int_{-\infty}^{\infty}\frac{dk}{2\pi} e^{-i\alpha v^2 k^2 + i(\Delta x-\beta q)k} = \frac{1}{2\sqrt{\pi}}\frac{e^{i\frac{(\Delta x-\beta q)^2}{4\alpha v^2}}}{\sqrt{i\alpha v^2}}. $$

Multiply them, then act with $-\partial_{\Delta t}^2$ $$ I_2 = i\int_0^\infty d\alpha\int_0^\infty d\beta;\big(-\partial_{\Delta t}^2\big)\left[\frac{1}{4\pi}\frac{1}{\alpha v} e^{-i\frac{(\beta-\Delta t)^2}{4\alpha}} e^{,i\frac{(\Delta x-\beta q)^2}{4\alpha v^2}}\right]. $$

Do the $\alpha$ integral with the standard kernel $$ \int_0^\infty \frac{d\alpha}{\alpha} \exp (-i\frac{A}{\alpha}) = -i\pi,\text{sgn}(\Im A) - \ln A + \text{const}, $$ after completing the quadratic form in $\beta$. The phase in the exponent combines to $$ -\frac{i}{4\alpha}\left[(\beta-\Delta t)^2 - \frac{(\Delta x-\beta q)^2}{v^2}\right] = -\frac{i}{4\alpha}\left[(1-\frac{q^2}{v^2})\beta^2 -2\beta\Big(\Delta t+\frac{q \Delta x}{v^2}\Big) + \Delta t^2 - \frac{\Delta x^2}{v^2}\right]. $$

Provided $q^2\ne v^2$, we complete the square in $\beta$, integrate $\alpha$, then integrate $\beta$ using $\int_0^\infty \beta e^{-i s \beta^2} d\beta$ type formulae or by differentiating w.r.t. $q$. The outcome is a sum of logs whose branch points sit at the characteristic lines $$ \Delta t=\pm \frac{\Delta x}{v},\qquad \Delta t = q,\Delta x, $$ exactly the three kinematic loci you expect from the poles $\omega=\pm v k$ and $\omega=qk$ (with the double pole generating the stronger singularity along $\Delta t=q,\Delta x)$. Doing the mirror piece $I_3$ flips $(\Delta t,\Delta x)\to-(\Delta t,\Delta x)$. The $I_1$ (the “1” in $1-\cos$) gives the usual UV divergent constant you regulate with a cutoff or $\overline{\text{MS}}$ a d it sets the additive constant in front of the logs.

Why your result only “saw one pole”?

I believe this is because you effectively replaced the gaussian $k$ integral by a delta constraint that keeps a linear combination of $\alpha,\beta,\gamma$ to a complex number. That projection collapses the $\omega^2-v^2k^2$ to a single kinematic condition, so you only harvest logarithms associated with one branch cut. The real calculation keeps the full quadratic form, and the $\alpha$ integration generates the log kernel that knows about all three characteristic directions.