# An Integral involving solid angle in Peskin and Schroeder

I cannot figure out an integral in the textbook Quantum Field Theory by Peskin and Schroeder.

On P.201 the integral above Eq.(6.70), the relevant part in question reads

$$\int\frac{\mathrm d\Omega_k}{4\pi}\frac{1}{[\xi\hat{k}\cdot p'+(1-\xi)\hat{k}\cdot p]^2}=\frac{1}{[\xi p'+(1-\xi)p]^2}$$

I tried by writing $\mathrm d\Omega_k=\sin\theta_k ~\mathrm d\theta_k~\mathrm d\phi_k$, as well as assuming $p'$ in the $\hat{z}$ direction, and $p$ in the $x-z$ plane, the resulting integral becomes complicated and involves

$$\int \sin\theta~\mathrm d\theta~\mathrm d\phi \frac{1}{\xi\cos\theta p'+(1-\xi)[\cos\theta p_z+\sin\theta\cos\phi p_x]}$$

Then one has to integral with respect to $\phi$ and $\theta$, neither of them is straightforward. Since the resulting expression in the textbook looks rather simple, I am wondering whether there is a straightforward way to obtain the result in the textbook?

Any comment is really appreciated.

I advise you to factor out $\hat{k}$ and align your $z$ axis with the vector $\xi p' + (1-\xi)p$.