# 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? – GiorgioP Dec 19 '18 at 6:31
• Yes it is. Sorry. I fixed it in the question. – lomby Dec 19 '18 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)? – lomby Dec 19 '18 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? – lomby Dec 20 '18 at 12:54