Help in filling in the gaps from a contour integration in QFT [duplicate]

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Evaluating $\int_{-\infty}^\infty \frac{pe^{ipr}}{\sqrt{p^2 + m^2}} dp$ using complex contours (1 answer)

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In a QFT class, we were calculating the following integral $$\int_{-\infty}^{\infty}dk \frac{ke^{ikx}}{\sqrt{k^2+m^2}}$$

and we decided to do a contour calculation. We chose a branch cut at negative $k^2+m^2$, which introduces branch cuts at $\Re(k)=0$ ${\Im(k)\geq m}$ and $\Im(k)\leq m$, that is, vertical half-lines on the imaginary axis. The contour was kind of just drawn, but it contains the real axis, and then circles around the upper half-plane, while avoiding the uppermost branch cut. Then, standard contour arguments are employed, even if with little care, the parts away from the real axis and the branch cut go to zero, and only the parts near the branch contribute. The lecturer then wrote $$\int_{i\infty+\varepsilon}^{im+\varepsilon}(...)+\int_{im-\varepsilon}^{i\infty-\varepsilon}(...)$$ in the limit as the real positive $\varepsilon$ vanishes, as the next step of the calculation. I feel a little cautious about this.

First, it seems from the notation that these correspond to integration along straight lines, parallel to the branch cut, situated at a horizontal distance $\varepsilon$ from it, either at the right, where the integration runs from up to down, or at the left, where it runs from down to up, but always staying at the height of $im$. My problem is with how the lecturer did away with the connective part between these two lines. I can think of two possibilities.

First, it could be that it's just the line parallel to the real axis connecting $im+\varepsilon$ and $im-\varepsilon$. I don't think this should be the case because $im$ is a pole of the integrand, so the contour should never run over it, but maybe there's some workaround that I'm not seeing.

Second, the connective part is a lower semi-circle of radius $\varepsilon$ centered at $im$. I feel tempted to believe this is the case, but I have some doubts. Surely, the integrand is holomorphic in the punctured lower semi-disk $D_\varepsilon=\{z\in \mathbb{C}; |z-im|\leq \varepsilon, \Im(z)\leq im, z\neq im\}$, for every positive $\varepsilon$. Then I think that $|\int_C f|<l(C)\sup_C |f|\sim M_\varepsilon \varepsilon$ for some constant $M$ dependent on $\varepsilon$, and this would vanish in the limit. However, this function is unbounded near $im$ right? So I'm not sure that in the limiting process, done in this way, this part can properly vanish because of the pole.

Am I overthinking this, does this work? Or is there some clearer contour?

Sketch of the contour

• Lourenco Entrudo: It seems this question already has an answer here, it may be a duplicate, I'll delete after reviewing
• Steven Clark: I don't believe your integral converges in the usual sense. Doesn't the result contain a distributional term involving the Dirac delta function $\delta(x)$?
• Lourenco Entrudo: @StevenClark oh god not again. QFT is hell
• Lourenco Entrudo: @DanieleTampieri Seeing as the answer touches upon the finer details of the vanishing of one part of the contour, and adds a comment about the convergence of the integral, I think its best to keep this question open

Answer 1

Using a lower semicircular contour about $im$, you can check that $$\left| \int_C f(z) ,円\mathrm{d}z\right| \leq M \sqrt{\varepsilon}$$

for some constant $M$. Indeed, let $M' = \sup \left\{ \left| \frac{ke^{ikx}}{\sqrt{k+im}} \right| : |k-im|\leq m\right\}$ and note that $M'$ is finite by the continuity. So, if $k$ is such that $|k - im| = \varepsilon$ and if $\varepsilon$ is small enough so that $\varepsilon \leq m$, then

$$ \left| \frac{ke^{ikx}}{\sqrt{k^2+m^2}} \right| = \frac{1}{\sqrt{\varepsilon}} \left| \frac{ke^{ikx}}{\sqrt{k+im}} \right| \leq \frac{M'}{\sqrt{\varepsilon}}. $$

Hence,

$$ \left| \int_C \frac{ke^{ikx}}{\sqrt{k^2+m^2}} ,円 \mathrm{d}z \right| \leq \text{length}(C) \cdot \frac{M'}{\sqrt{\varepsilon}} = \pi M' \sqrt{\varepsilon}. $$

A similar trick shows that the contour integral of the form $\int_C \frac{f(z)}{(z-z_0)^p} ,円 \mathrm{d}z$, where $f$ is analytic near $z_0$, $p < 1$, and $C$ is a circular arc about $z_0$, vanishes as the radius of $C$ goes to 0ドル$.

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By the way, the integral

$$ \int_{-\infty}^{\infty} \frac{k e^{ikx}}{\sqrt{k^2 + m^2}} ,円 \mathrm{d}k $$

does not converge either in the ordinary sense or the principal value sense. For large $k$ the integrand behaves like $\operatorname{sgn}(k) e^{ikx}$, which oscillates periodically, and hence the same must be true for its antiderivative. Consequently, any attempt to computing this integral via complex-analytic technique must also fail.

Howerver, a meaningful value can be assigned to the integral via a suitable regularization. I strongly believe Cesàro summability would be enough, but Abel summability is easier to handle in this case:

$$ \begin{align*} \text{Regularized } \int_{-\infty}^{\infty} \frac{k e^{ikx}}{\sqrt{k^2 + m^2}} ,円 \mathrm{d}k &:= \lim_{\varepsilon \to 0^+} \int_{-\infty}^{\infty} \frac{k e^{ikx}}{\sqrt{k^2 + m^2}} e^{-\varepsilon|k|} ,円 \mathrm{d}k. \end{align*}$$

Now the trick discussed in OP's lecture works nicely, yielding

$$ = 2i \int_{m}^{\infty} \frac{k e^{-kx}}{\sqrt{k^2 - m^2}} ,円 \mathrm{d}k. $$

• Lourenco Entrudo: Thanks for clearing that up. About the divergence of the integral: where could the regularization have occurred in my lecture's approach? Because it isn't mentioned, it's claimed that contour integration can bring the integral into that form. Maybe I should ask a more dedicated question about that