Three integrals in Peskin's Textbook

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

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

This is just complementary to Alex's answer.

2. 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).

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 $e^{-imt}$.