Closed form for an integral involving the incomplete Gamma function? Let $\alpha>1$, $K>1$ and $n \in \mathbb{N}^+$. What is a closed form solution to a tough integral? $$I(\alpha,K,n)=-\int_{-\infty }^{\infty } \frac{2 i e^{-i K u} (K u-i) \Bigl[\alpha \, (-i u)^{\alpha } \,\Gamma (-\alpha ,-i u)\Bigr]^n}{u^2} \, du,$$
where $\Gamma(.,.)$ is the incomplete Gamma function:
$\Gamma (a,z)=\int _z^{\infty }d t\, t^{a-1} e^{-t}$.
I tried all manner of substitutions and various combinations of integration by parts.
 A: We may use the following integral representation for the incomplete gamma function (DLMF ref. see identity 8.6.4):

$$\Gamma{\left(\alpha,z\right)}=\frac{z^{\alpha}e^{-z}}{\Gamma{\left(1-\alpha\right)}}\int_{0}^{\infty}\frac{t^{-\alpha}e^{-t}}{z+t}\,\mathrm{d}t;~~~\small{\left|\arg{\left(z\right)}\right|<\pi,~\Re{\left(\alpha\right)}<1}.$$


Given $n\in\mathbb{N}^{+}\land a,b,\epsilon\in\mathbb{R}\land1<a\land1\le b\le n\land0<\epsilon$, define $I_{n}{\left(a,b;\epsilon\right)}$ via the integral
$$I_{n}{\left(a,b;\epsilon\right)}=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{2ie^{-ibx}\left(bx-i\right)\left[a\left(-ix\right)^{a}\,\Gamma{\left(-a,-ix\right)}\right]^{n}}{x^{2}+\epsilon^{2}}.$$
Then,
$$\begin{align}
I_{n}{\left(a,b;\epsilon\right)}
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{2ie^{-ibx}\left(bx-i\right)\left[a\left(-ix\right)^{a}\,\Gamma{\left(-a,-ix\right)}\right]^{n}}{x^{2}+\epsilon^{2}}\\
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{2ie^{-ibx}\left(bx-i\right)}{x^{2}+\epsilon^{2}}\left[\frac{e^{ix}}{\Gamma{\left(a\right)}}\int_{0}^{\infty}\frac{t^{a}e^{-t}}{t-ix}\,\mathrm{d}t\right]^{n}\\
&=\frac{1}{\left[\Gamma{\left(a\right)}\right]^{n}}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{2ie^{-ibx}\left(bx-i\right)}{x^{2}+\epsilon^{2}}e^{inx}\left[\int_{0}^{\infty}\frac{t^{a}e^{-t}}{t-ix}\,\mathrm{d}t\right]^{n}\\
&=\frac{2i}{\left[\Gamma{\left(a\right)}\right]^{n}}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(bx-i\right)e^{i\left(n-b\right)x}}{x^{2}+\epsilon^{2}}\left[\int_{0}^{\infty}\frac{t^{a}e^{-t}}{t-ix}\,\mathrm{d}t\right]^{n}\\
&=\frac{2i}{\left[\Gamma{\left(a\right)}\right]^{n}}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(bx-i\right)e^{i\left(n-b\right)x}}{x^{2}+\epsilon^{2}}\prod_{k=1}^{n}\int_{0}^{\infty}\mathrm{d}t_{k}\,\frac{t_{k}^{a}e^{-t_{k}}}{t_{k}-ix}\\
&=\small{\frac{2i}{\left[\Gamma{\left(a\right)}\right]^{n}}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(bx-i\right)e^{i\left(n-b\right)x}}{x^{2}+\epsilon^{2}}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,e^{-\sum_{k=1}^{n}t_{k}}\prod_{k=1}^{n}\left(\frac{t_{k}^{a}}{t_{k}-ix}\right)}\\
&=\small{\frac{2i}{\left[\Gamma{\left(a\right)}\right]^{n}}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\left(\prod_{k=1}^{n}t_{k}^{a}\right)e^{-\sum_{k=1}^{n}t_{k}}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(bx-i\right)e^{i\left(n-b\right)x}}{\left(x^{2}+\epsilon^{2}\right)\prod_{k=1}^{n}\left(t_{k}-ix\right)}}\\
&=\small{\frac{2i}{\left[\Gamma{\left(a\right)}\right]^{n}}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\frac{\left(\prod_{k=1}^{n}t_{k}^{a}\right)}{\exp{\left(\sum_{k=1}^{n}t_{k}\right)}}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(-bx-i\right)e^{-i\left(n-b\right)x}}{\left(x^{2}+\epsilon^{2}\right)\prod_{k=1}^{n}\left(t_{k}+ix\right)}}\\
&=:\frac{2i}{\left[\Gamma{\left(a\right)}\right]^{n}}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\frac{\left(\prod_{k=1}^{n}t_{k}^{a}\right)}{\exp{\left(\sum_{k=1}^{n}t_{k}\right)}}\,f_{n}{\left(b;\epsilon\right)},\\
\end{align}$$
where for $n\in\mathbb{N}^{+}\land b,\epsilon\in\mathbb{R}\land1\le b\le n\land0<\epsilon$ we've defined the auxiliary function denoting the innermost integration,
$$f_{n}{\left(b;\epsilon\right)}:=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(-bx-i\right)e^{-i\left(n-b\right)x}}{\left(x^{2}+\epsilon^{2}\right)\prod_{k=1}^{n}\left(t_{k}+ix\right)}.$$
For the integration over $x$, I will appeal to a well-suited proposition from Gradshteyn's Tables. Now, I usually try to make my posts as self-contained as possible by proving any non-trivial lemmas I plan on using, rather than taking the lazy way out and simply citing the result from an outside source. But I can't resist doing so, partly out of laziness, but mostly just to show off how Gradshteyn can on occasion be eerily clairvoyant.

Gradshteyn 3.386: Given the conditions
  $$-1<\Re{\left(\nu_{0}\right)}\land0<\Re{\left(\beta_{k}\right)}\land\sum_{k=0}^{n}\Re{\left(\nu_{k}\right)}<1\land0<p,$$
  we have the following two results:
  $$\int_{-\infty}^{\infty}\frac{\left(ix\right)^{\nu_{0}}e^{-ipx}\prod_{k=1}^{n}\left(\beta_{k}+ix\right)^{\nu_{k}}}{\beta_{0}-ix}\mathrm{d}x=2\pi e^{-\beta_{0}p}\beta_{0}^{\nu_{0}}\prod_{k=1}^{n}\left(\beta_{0}+\beta_{k}\right)^{\nu_{k}},$$
  and
  $$\int_{-\infty}^{\infty}\frac{\left(ix\right)^{\nu_{0}}e^{-ipx}\prod_{k=1}^{n}\left(\beta_{k}+ix\right)^{\nu_{k}}}{\beta_{0}+ix}\mathrm{d}x=0.$$


The following partial fraction decomposition is easily verified:
$$\frac{-bx-i}{x^{2}+\epsilon^{2}}=\frac{1}{2i\epsilon}\left[\frac{\left(1-b\epsilon\right)}{\epsilon-ix}+\frac{\left(1+b\epsilon\right)}{\epsilon+ix}\right].$$
Then, assuming $0<\Re{\left(t_{k}\right)}\land b<n$,
$$\begin{align}
f_{n}{\left(b;\epsilon\right)}
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{\left(-bx-i\right)e^{-i\left(n-b\right)x}}{\left(x^{2}+\epsilon^{2}\right)\prod_{k=1}^{n}\left(t_{k}+ix\right)}\\
&=\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{1}{2i\epsilon}\left[\frac{\left(1-b\epsilon\right)}{\epsilon-ix}+\frac{\left(1+b\epsilon\right)}{\epsilon+ix}\right]\frac{e^{-i\left(n-b\right)x}}{\prod_{k=1}^{n}\left(t_{k}+ix\right)}\\
&=\frac{1-b\epsilon}{2i\epsilon}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{e^{-i\left(n-b\right)x}}{\left(\epsilon-ix\right)\prod_{k=1}^{n}\left(t_{k}+ix\right)}\\
&~~~~~+\frac{1+b\epsilon}{2i\epsilon}\int_{-\infty}^{\infty}\mathrm{d}x\,\frac{e^{-i\left(n-b\right)x}}{\left(\epsilon+ix\right)\prod_{k=1}^{n}\left(t_{k}+ix\right)}\\
&=\frac{1-b\epsilon}{2i\epsilon}\cdot\frac{2\pi e^{-\left(n-b\right)\epsilon}}{\prod_{k=1}^{n}\left(\epsilon+t_{k}\right)}.\\
\end{align}$$
Thus,
$$\begin{align}
I_{n}{\left(a,b;\epsilon\right)}
&=\frac{2i}{\left[\Gamma{\left(a\right)}\right]^{n}}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\frac{\left(\prod_{k=1}^{n}t_{k}^{a}\right)}{\exp{\left(\sum_{k=1}^{n}t_{k}\right)}}\,f_{n}{\left(b;\epsilon\right)}\\
&=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\frac{\left(\prod_{k=1}^{n}t_{k}^{a}\right)}{\exp{\left(\sum_{k=1}^{n}t_{k}\right)}}\cdot\frac{1}{\prod_{k=1}^{n}\left(\epsilon+t_{k}\right)}\\
&=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\prod_{k=1}^{n}\left(\frac{t_{k}^{a}e^{-t_{k}}}{\epsilon+t_{k}}\right)\\
&=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\left[\int_{0}^{\infty}\mathrm{d}t\,\left(\frac{t^{a}e^{-t}}{\epsilon+t}\right)\right]^{n}\\
&=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\left[\epsilon^{a}e^{\epsilon}\,\Gamma{\left(1+a\right)}\,\Gamma{\left(-a,\epsilon\right)}\right]^{n}\\
&=2\pi a^{n}\left(1-b\epsilon\right)\epsilon^{na-1}e^{b\epsilon}\left[\Gamma{\left(-a,\epsilon\right)}\right]^{n}.\\
\end{align}$$

