quantum field theory - A certain regularization and renormalization scheme

In a certain lecture of Witten's about some QFT in $1+1$ dimensions, I came across these two statements of regularization and renormalization, which I could not prove,

(1) $\int ^\Lambda \frac{d^2 k}{(2\pi)^2}\frac{1}{k^2 + q_i ^2 \vert \sigma \vert ^2} = - \frac{1}{2\pi} ln \vert q _ i \vert - \frac{1}{2\pi}ln \frac{\vert \sigma\vert}{\mu}$

(..there was an overall $\sum _i q_i$ in the above but I don't think that is germane to the point..)

(2) $\int ^\Lambda \frac{d^2 k}{(2\pi)^2}\frac{1}{k^2 + \vert \sigma \vert ^2} = \frac{1}{2\pi} (ln \frac{\Lambda}{\mu} - ln \frac{\vert \sigma \vert }{\mu} )$

I tried doing dimensional regularization and Pauli-Villar's (motivated by seeing that $\mu$ which looks like an IR cut-off) but nothing helped me reproduce the above equations.

I would glad if someone can help prove these above two equations.

Answer

Let's just look at the integral

$$\int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2}.$$ The other integrals should follow from this one. Introduce the Pauli-Villars regulator, $$\begin{eqnarray*} \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} &\rightarrow& \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} - \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\Lambda^2} \\ &=& (\Lambda^2-\alpha^2)\int \frac{d^2k}{(2\pi)^2} \frac{1}{(k^2+\alpha^2)(k^2+\Lambda^2)} \\ &=& (\Lambda^2-\alpha^2)\int_0^1 dx\, \int\frac{d^2k}{(2\pi)^2} \frac{1}{(k^2 + \beta^2)^2} \\ &=& (\Lambda^2-\alpha^2)\int_0^1 dx\, \frac{1}{2} \frac{2\pi}{(2\pi)^2} \int_0^\infty dk^2\,\frac{1}{(k^2 + \beta^2)^2} \\ &=& (\Lambda^2-\alpha^2) \frac{1}{4\pi} \int_0^1 dx\, \frac{1}{\beta^2} \\ &=& (\Lambda^2-\alpha^2) \frac{1}{4\pi} \int_0^1 dx\, \frac{1}{\Lambda^2 - x(\Lambda^2-\alpha^2)} \\ &=& -\frac{1}{2\pi} \ln \frac{|\alpha|}{\Lambda} \end{eqnarray*}$$ Where we have combined denominators with the Feynman parameter $x$, with the intermediate variable $\beta^2 = \Lambda^2 - x(\Lambda^2-\alpha^2)$. Of course, this could also be approached with dimensional regularization with the same result.

Addendum: After regularization we must renormalize. Using the minimal subtraction prescription we find

$$\int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} \rightarrow -\frac{1}{2\pi} \ln \frac{|\alpha|}{\mu},$$ as required.