Problem with loop Integral (HQET)

I have come across the Integral:

$$\int_0^{\infty}dx [x^2-ixa+c]^{n-\frac{d}{2}}e^{-bx},$$

where $$n = 1,2 ; a,b,c,d \in \mathbb{R}; b,d > 0$$. This integral should contain some divergences for $$d \rightarrow 4$$ (for $$c=0$$). I guess one must be able to write it as some combination of Gamma functions.

• Is $d$ a real number? Commented Dec 19, 2018 at 6:31
• Yes it is. Sorry. I fixed it in the question. Commented Dec 19, 2018 at 10:31

\begin{align} I&=\int_0^\infty (x-iax+c)^{n-\frac{d}{2}} e^{-bx}\,dx \\ &=\int_0^\infty \left[\left(x-\frac{ia}{2}\right)^2+\frac{a^2+4c}{4}\right]^{n-\frac{d}{2}} e^{-bx}\,dx \\ &=\int_0^\infty \sum_{k=0}^{n-\frac{d}{2}} {n-\frac{d}{2}\choose k} \left(x-\frac{ia}{2}\right)^{2k} \left(\frac{a^2+4c}{4} \right)^{n-\frac{d}{2}-k} e^{-bx}\,dx \\ &=\int_0^\infty \sum_{k=0}^{n-\frac{d}{2}}\sum_{m=0}^{2k} {n-\frac{d}{2}\choose k} {2k \choose m} \left(\frac{a^2+4c}{4} \right)^{n-\frac{d}{2}-k} \left(-\frac{ia}{2} \right)^{2k-m} x^m e^{-bx}\,dx \\ &= \sum_{k=0}^{n-\frac{d}{2}}\sum_{m=0}^{2k} {n-\frac{d}{2}\choose k} {2k \choose m} \left(\frac{a^2+4c}{4} \right)^{n-\frac{d}{2}-k} \left(-\frac{ia}{2} \right)^{2k-m} \int_0^\infty x^m e^{-bx}\,dx \\ &= \sum_{k=0}^{n-\frac{d}{2}}\sum_{m=0}^{2k} {n-\frac{d}{2}\choose k} {2k \choose m} \left(\frac{a^2+4c}{4} \right)^{n-\frac{d}{2}-k} \left(-\frac{ia}{2} \right)^{2k-m}\frac{m!}{b^{m+1}} \\ \end{align} After using Wolfram Alpha to simplify the inner sum, we obtain \begin{align} I&= e^{-iab/2}\sum_{k=0}^{n-\frac{d}{2}} \frac{\Gamma\left(2k+1, -\frac{iab}{2}\right)}{b^{2k+1}} {n-\frac{d}{2}\choose k} \left(\frac{a^2+4c}{4} \right)^{n-\frac{d}{2}-k} \\ \end{align}
• Thanks for you answer! I have a question though, because I forgot to mention explicitly that $d$ is in general not an integer. So in this formula the sum would be not defined, right? However this derivation should also work if you use the binomial series (see wiki)? Commented Dec 19, 2018 at 10:30
I don't see why there should be any divergences at $$d=4$$ since (I assume that $$b>0$$) the expression $$x^2−iax+c$$ is never zero on the real positive $$x$$ axis --- so there is no danger of dividing by zero anywhere in the integral. If we factor $$x^2-iax+c= (x-\alpha)(x-\beta)$$ then the integral will only misbehave if either of the the branch points at $$x=\alpha$$ or $$x=\beta$$ hit the endpoint at $$x=0$$, or if they become equal while at the same time pinching the contour.
• Ah yes. I did enter the constant $c$ for generality. Actually I am looking for the Integral with $c=0$. Then it should be divergent at $d=4$. right? Commented Dec 20, 2018 at 12:54