# Three integrals in Peskin's Textbook

Peskin's QFT textbook

1.page 14

$$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}$$

when $x^2\gg t^2$, how do I apply the method of stationary phase to get the book's answer.

2.page 27

$$\int_{-\infty}^{\infty} \mathrm{d}p\frac{p\ e^{ipr}}{\sqrt{p^2+m^2}}$$

where $r>0$

3.page 27

$$\int_{m}^{\infty}\mathrm{d}E \sqrt{E^2-m^2}e^{-iEt}$$

where $m>0$

I'm crazy about these integrals, but the textbook doesn't give the progress.

• Hey, what about Mr. Schroeder? Commented Mar 25, 2014 at 13:29

1. Since $x\gg p$, we see that $\sin(px)$ is highly oscillatory. In fact, the integral becomes

$$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}\sim \int_{-\infty} ^\infty \mathrm{d}p\ p\ e^{ipx-it\sqrt{p^2 +m^2}}$$

modulo some factor of $\pm2/i$. Observe now this integral resembles $\int f(p)\exp(g(p))\,\mathrm{d}p$. We find the point $\tilde{p}$ such that

$$g'(\tilde{p})=0.$$

Then just replace $g(p)$ with $g(\tilde{p})+\frac{1}{2}g''(\tilde{p})(p-\tilde{p})^{2}$ and carry out the integral as a moment of a Gaussian. For more on this approximation, see e.g. the relevant chapter of Hunter and Nachtergaele's book (freely and legally available).

2. This is just a Fourier transform of $p\,(p^{2}+m^{2})^{-1/2}$.

3. I assume you are referring to Eq (2.51) on page 27. We write the integral as

$$I(t) = \int^{\infty}_{m}\sqrt{E^{2}-m^{2}}e^{-iEt}\,\mathrm{d}E.$$

Peskin and Schroeder consider this integral as $t\to\infty$. If we consider a change of variables to

$$E^{2}-m^{2}=\mu^{2}\quad\Longrightarrow\quad \mathrm{d}E = \frac{\mu}{\sqrt{m^{2}+\mu^{2}}}\mathrm{d}\mu$$

We have

\begin{align} I(t) &= \int^{\infty}_{0} \frac{\mu^{2}}{\sqrt{m^{2}+\mu^{2}}}e^{-it\sqrt{m^{2}+\mu^{2}}}\mathrm{d}\mu \\ &=\frac{1}{2}\int^{\infty}_{-\infty} \frac{\mu^{2}}{\sqrt{m^{2}+\mu^{2}}}e^{-it\sqrt{m^{2}+\mu^{2}}}\mathrm{d}\mu \end{align}

Let

$$f(\mu) = \frac{\mu^{2}}{\sqrt{m^{2}+\mu^{2}}},\quad\mbox{and}\quad \phi(\mu) = \sqrt{m^{2}+\mu^{2}}$$

so

$$I(t)=\frac{1}{2}\int^{\infty}_{-\infty}f(\mu)e^{-it\phi(\mu)}\mathrm{d}\mu.$$

Observe

$$f(\mu)=\mu\phi'(\mu).$$

As $t\to\infty$, the integral becomes highly oscillatory.

There are two ways to approach the problem from here. The first, unforgivably handwavy but faster: take the stationary phase approximation, and pretend that $f(\mu_{\text{crit}})$ is some arbitrary constant.

The critical points for $\phi$ are $\mu_{0}=0$ and $\mu_{\pm}=\pm im$. We only care about the real $\mu$, so we Taylor expand about $\mu_0$ to second order:

\begin{align} \phi(\mu)&=\phi(0)+\frac{1}{2!}\phi''(0)\mu^{2}\\ &=m + \frac{1}{2m}\mu^{2} \end{align}

We now approximate the integral as

$$I(t) \sim \int^{\infty}_{-\infty}f(c)e^{-itm}e^{-it\mu^{2}/2m}\mathrm{d}\mu \approx f(c)e^{-itm}\sqrt{\frac{4\pi m}{t}}.\tag{1}$$

The other approximation doesn't fix $f$. Observe $f(\mu)\sim|\mu|$, so we have

$$I(t) \sim e^{-itm} \int^{\infty}_{0}\mu e^{-it\mu^{2}/2m}\mathrm{d}\mu.$$

We have (using Fresnel integrals)

$$\int^{\infty}_{0}\mu e^{-it\mu^{2}/2m}\mathrm{d}\mu\sim \frac{im}{t}.$$

Hence

$$I(t)\sim\frac{im}{t}e^{-imt}.\tag{2}$$

• On the last integral, for the method of stationary phase, if you evaluate $f(c)$ you get 0 which makes the approximation 0, is there something I'm missing on why that doesn't happen Commented Jul 6, 2022 at 2:04
• @JoshuaPasa Like I said, "unforgivably handwavy but faster". Remember, the point is to show $I(t)\sim\exp(-imt)$. Besides, if you feel as nervous about it as I do (and I do!), then you can use the Fresnel integral approach. It's a good idea to verify "dodgy methods" by several means, especially in QFT. Commented Jul 6, 2022 at 5:22
• Thanks, I'm just going through Peskin and Schroeder's book myself, and I couldn't get the method of stationary phase to work it out. Commented Jul 6, 2022 at 13:43
• @JoshuaPasa As an aside, if you are self-studying QFT, I would encourage you to start with Brian Hatfield's "Quantum Field Theory of Point Particles And Strings". Contrary to its title, its overwhelmingly about vanilla QFT with a chapter on strings. It shows all calculations, and discusses functional methods not discussed elsewhere (e.g., the functional Schrodinger equation). It's a good prelude to Peskin & Schroeder, in my opinion. Commented Jul 6, 2022 at 16:01
• I've heard that Peskin and Schroeder is the book everyone uses as it contains a lot of detail. Does Hartfield's book skim over a lot of detail or is it good enough to get a solid understanding of QFT? Commented Jul 6, 2022 at 16:28

Just a note on the third integral. $$\frac{1}{4\pi^2}\int_m^\infty dE\sqrt{E^2-m^2}e^{-iEt}.$$ If you don't want to explicitly do the calculation, as in Alex's answer, there is a plausibility argument. In the limit where $$t$$ is very large, the exponential oscillates very rapidly. The oscillations will cancel each other except in the region where $$\sqrt{E^2-m^2}$$ has very large slope. In fact, at $$E=m$$, the slope of this function is infinite. Therefore, we can guess that the integral might be proportional to $$e^{-imt}$$.

This is just complementary to Alex's answer.

1. For the second integral the book provides an analysis in order to push the contour up to wrap around the upper branch cut. After some manipulation, it gives the following integral $$\frac{1}{4\pi^2 r}\int_m^\infty d\rho \frac{\rho e^{-\rho r}}{\sqrt{\rho^2-m^2}}$$ At the limit $r\rightarrow \infty$, the effect of the exponential suppression due to the factor $e^{-\rho r}$ overwhelms that of the singularity of $\frac{1}{\sqrt{\rho^2-m^2}}$ at $m$. As a result, one may roughly treat $\frac{\rho}{\sqrt{\rho^2-m^2}}$ as a constant, and this leads to (2.52).