<s>[INST] 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.
 [/INST] I'll first start by completing the square and using the Binomial theorem twice, before simplifying the resulting expression. Let's go.
\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}
</s>