# Regularized loop integral confusion when changing integration variable from $l_E\to l^2_E$

I'm currently reading lecture notes on Dimensional Regularization where Wick Rotations are being employed. I understand how to regularize and compute these loop integrals but I'm having trouble rewriting them so the integral can be ready for a table. Take the integral found in the lecture notes: $$I_d=\mu^{\epsilon}\int _0^1dx\int \frac{d^dl}{(2\pi)^d}\ \frac{1}{[l^2-a^2+i\epsilon]^2}$$ where $$l^{\mu}$$ is the shift variable $$l^{\mu}=k^{\mu}-xp^{\mu}$$ and $$a^2=m^2-x(1-x)p^2$$. Now wick rotation can be performed giving us $$I_d=i\mu^{\epsilon}\int _0^1dx\frac{2\pi^{d/2}}{\Gamma(d/2)}\int \frac{dl_E}{(2\pi)^d}\ \frac{l_E^{d-1}}{[l_E^2+a^2+i\epsilon]^2}.$$ All that is worked out in the notes and I understand it but the problem comes in when we prepare the now 1d integral for the use of a beta function. That is the integral $$\int dl_E\ \frac{l_E^{d-1}}{[l_E^2+a^2-i\epsilon]^2}.$$ Were supposed to use the fact that $$\int_0^{\infty}dt \frac{t^{m-1}}{(t^2+a^2)^n}=\frac{1}{(a^2)^{n-m}}\frac{\Gamma(m)\Gamma(n-m)}{\Gamma(n)}.$$ In order to use the formula above, the lecture notes showed $$\int_0^{\infty} dl_E\ \frac{l_E^{d-1}}{[l_E^2+a^2-i\epsilon]^2}=\int_0^{\infty} \frac{1}{2}dl^2_E\ \frac{(l^2_E)^{d/2-1}}{[l_E^2+a^2-i\epsilon]^2}$$ and then the formula can be employed. My confusion results from changing the integration variable from $$l_E \to l^2_E$$ and where that factor of $$\frac12$$ comes from. I figured it was just a simple substitution at first by letting $$l_E=l^2_E$$ but computing $$dl^2_E$$ left me confused. What am I missing here? I've tried doing the regular change of variables but I'm not understanding the factor of $$\frac12$$ and the $$dl^2$$ term. Other than that, I understand the rewriting of the integrand.

It is indeed just a change of variables, but working it in the $$l_E \to l_E^2$$ way is a bit messy. I'd rather do it backwards. Notice that $$\frac{d}{dl_E} l_E^2 = 2 l_E,$$ and hence $$dl_E^2 = 2 l_E dl_E.$$ Using this, \begin{align} \int_0^{\infty} \frac{1}{2}dl^2_E\ \frac{(l^2_E)^{d/2-1}}{[l_E^2+a^2-i\epsilon]^2} &= \int_0^{\infty} l_E dl_E\ \frac{(l^2_E)^{d/2-1}}{[l_E^2+a^2-i\epsilon]^2}, \\ &= \int_0^{\infty} l_E dl_E\ \frac{(l_E)^{d-2}}{[l_E^2+a^2-i\epsilon]^2}, \\ &= \int_0^{\infty} dl_E\ \frac{(l_E)^{d-1}}{[l_E^2+a^2-i\epsilon]^2}, \end{align} which is the desired result.
• So its exactly as I thought. I figured the factor of 1/2 came from the derivative. But why does $\frac12$ $dl^2_E=l_EdlE$ ? Solving for $dl_E$ I get $dl_E=\frac{dl^2_E}{2l_E}$ Where does that $l_E$ term in the denominator go? I assume we just plug in $l_E=1$ since thats the only solution to $l_E=l^2_E$? That and 0.
• @aygx I merged the $l_E$ coming from the change of variables with the $l_E$ in the integrand's numerator. I'll update the answer to add in an extra step Jun 26, 2023 at 1:58