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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}
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.
