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.

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}
