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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.

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  • $\begingroup$ Is $d$ a real number? $\endgroup$ – GiorgioP Dec 19 '18 at 6:31
  • $\begingroup$ Yes it is. Sorry. I fixed it in the question. $\endgroup$ – lomby Dec 19 '18 at 10:31
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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}

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    $\begingroup$ 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)? $\endgroup$ – lomby Dec 19 '18 at 10:30
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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.

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  • $\begingroup$ 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? $\endgroup$ – lomby Dec 20 '18 at 12:54

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