Evaluate $\int _0^{\infty}d\lambda \left(\lambda ^2 + 2\lambda b + c\right)^{-\frac{\epsilon}{2}}$ with $b<0,\epsilon>0$ and $\epsilon$ is very small $\epsilon\to 0$.
I see this in the book Heavy Quark Physics(the book, it is on Page.79). The original integration is $$-\left(\frac{4}{3}\right)g^2\mu^{\epsilon}\int \frac{d^nq}{(2\pi)^n}\frac{1}{(q^2-m^2)v\cdot (q+p)}$$ Where $q$ and $p$ are in Minkowski space that $q=(q_0,q_1,q_2,q_3),q^2=q_0^2-q_1^2-q_2^2-q_3^2$. After combine the denominator, Wick rotation and then use dimensional regularization, $$n=4-\epsilon, \int\frac{d^n q}{(2\pi)^n}\to q^{3-\epsilon}\frac{2\pi^{\frac{4-\epsilon}{2}}}{\Gamma (\frac{4-\epsilon}{2})}\int d q$$ and some other calculations, finally, 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 $$\int _0^{\infty}d\lambda \left(\lambda ^2 + 2\lambda b + c\right)^{-\frac{\epsilon}{2}}=\frac{1}{1-\epsilon}\left(\color{blue}{c^{-\frac{\epsilon}{2}}b}-\epsilon (c-b^2)\int _0^{\infty}d\lambda \left(\lambda ^2 + 2\lambda b + c\right)^{-1-\frac{\epsilon}{2}}\right)$$
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!
It is answered here: https://physics.stackexchange.com/questions/239690/detail-of-a-trick-evaluating-hqet-self-energy/239970#239970 For those who are interested I also post the answer here. It turns out that I have a misunderstanding about the dimensional regulation(DR). DR does not mean to set the dimension $n$ to a number close to the real dimension $4$. Which means it is not necessary to require $\epsilon \to 0$ in $n=4-\epsilon$ when integral. It is only required at last. So one can use $\epsilon > 1$ to integral and vanish the $\lim _{\lambda\to\infty}\lambda ^z$ term, and then use $\epsilon \to 0$ in the final result.