I have already asked this question in math SE with no luck. I'm reading the HQET book, and I see this in the book Heavy Quark Physics(the book, it is on Page.79).
The original integration is
$$-\left(\frac{4}{3}\right)g^2\mu^{\epsilon}\int \frac{d^nq}{(2\pi)^n}\frac{1}{(q^2-m^2)v\cdot (q+p)}$$
After some calculations, it becomes
$$-\left(\frac{8}{3}\right)g^2 \mu^{\epsilon}\int _0^{\infty}d\lambda \frac{i}{(4\pi)^{2-\frac{\epsilon}{2}}}\Gamma (\frac{\epsilon}{2})\left(\lambda ^2 - 2\lambda v\cdot p + m^2\right)^{-\frac{\epsilon}{2}}$$
In the book, it says, define
$$I(a,b,c)\equiv \int _0^{\infty} d\lambda \left(\lambda ^2 +2 b\lambda +c\right)^a=\frac{1}{1+2a}\left(\left(\lambda ^2 +2b\lambda +c\right)^a\left(\lambda +b\right)\left.\right|_{0}^{\infty}+2a(c-b^2)I(a-1,b,c)\right)$$
And
$$\int _0^{\infty}d\lambda \left(\lambda ^2 + 2\lambda b + c\right)^{-\frac{\epsilon}{2}}=\frac{1}{1-\epsilon}\left(\color{blue}{c^{-\frac{\epsilon}{2}}b}-\epsilon (c-b^2)\int _0^{\infty}d\lambda \left(\lambda ^2 + 2\lambda b + c\right)^{-1}\right)$$
I'm stucked by the first term, it says in dimensional regulation, as long as $z$ depends on $\epsilon$ in a way that allow one to analytically continue $z$ to negative values
$$\lim _{\substack{\lambda \to \infty}}\lambda ^z = 0$$ I really don't understand what does this mean.
In fact, I know some other method to evaluate the original integration, I just want to understand and learn the trick to evaluate $\int _0^{\infty}d\lambda \left(\lambda ^2 + 2b\lambda + c\right)^{-\frac{\epsilon}{2}}$ here.
I need help. Thanks for reading!
Substitute $a=-\frac{\epsilon}{2}$ into your first expression for $I(a,b,c)$. The integral equals to $$\frac{1}{1-\epsilon}(\lim_{\lambda\to \infty}\frac{\lambda +b}{((\lambda+b)^2+c-b^2)^{\epsilon/2 }}-c^{-\frac{\epsilon}{2}}b-\epsilon(c-b^2)\int_0^{\infty}d\lambda \frac{1}{(\lambda^2+2b\lambda+c)^{1+\epsilon/2}})$$
Dimensional regularization means that we first compute the integral with some $\epsilon$ such that the integral converges. Then we take the limit $\epsilon\to 0$, with the poles properly subtracted by counter terms, to get a finite result. From your integral or from the above expression it is clear that when $\epsilon>1$ the integral converges. Therefore set $\epsilon>1$ first so that the term involving $\lambda\to\infty$ will vanish, and then let $\epsilon\to 0$. Based on what you wrote, I believe there is a sign error on the term $c^{-\frac{\epsilon}{2}}b$ in your expression.