TITLE: Closed form for an integral involving the incomplete Gamma function?
QUESTION [13 upvotes]: 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.

REPLY [2 votes]: This is a partial answer, linked with the calculations via multiple convolution.
$\color{brown}{\textbf{Representation of the integral.}}$
Is known that

$$s^p\Gamma(-p,s) = E_{p+1}(s),$$
$$\dfrac1{2\pi i} \int\limits_{-i\infty}^{+i\infty}F(s)\,e^{ts}\,\text ds 
= \mathcal L^{-1}_{s\mapsto t}(F(s)),$$
$$\mathcal L^{-1}_{s\mapsto t}(e^{s} E_{p+1}(s)) = \dfrac1{(t+1)^{p+1}},$$

where $\;E_{p+1}(s)\;$ is the exponential integral $\;E\;$ and $\;\mathcal L^{-1}_{s\mapsto t}\;$ is the inverse Laplace transform from $\;s\;$ to $\;t.$
Let $\;-iu =s,\quad K=n+t,\;$ then
$$I(\alpha,n+t,n) = \dfrac{4\pi}{2\pi i}\int\limits_{-i\infty}^{+i\infty}\dfrac{(n+t)s-1}{s^2}\big(\alpha s^\alpha\Gamma(-\alpha, s)\big)^ne^{(n+t)s}\text ds$$
$$ = \dfrac{4\pi\alpha^n}{2\pi i}\int\limits_{-i\infty}^{+i\infty}\dfrac{(n+t)s-1}{s^2}\big(e^s\,E_{\alpha+1}(s)\big)^n\,e^{ts}\,\text ds
=4\pi\alpha^n\mathcal L^{-1}_{s\,\mapsto t}\left(\dfrac{(n+t)s-1}{s^2}\big(e^s\,E_{\alpha+1}(s)\big)^n\right),$$
$$=4\pi\alpha^n\left((n+t)\int\limits_0^t G_n(\alpha,\lambda)\,\text d\lambda
-\int\limits_0^t\int\limits_0^\lambda G_n(\alpha,\mu)\,\text d\mu\,\text d\lambda \right)$$
$$=4\pi\alpha^n\left((n+t)\int\limits_0^t G_n(\alpha,\lambda)\,\text d\lambda
-\int\limits_0^t\int\limits_0^\lambda G_n(\alpha,\mu)\,\text d\mu\,\text d(n+\lambda) \right)$$
$$\;\overset{\text{IBP}}{=\!=}\; 4\pi\alpha^n \int\limits_0^t(n+\lambda)G_n(\alpha,\lambda)\,\text d\lambda,$$
$$I(\alpha,n+t,n)=4\pi\alpha^n\int\limits_0^t(n+\lambda)
G_n(\alpha,\lambda)\,\text d\lambda,\tag1$$
where
$$G_n(\alpha,t) = \underbrace{\dfrac1{(t+1)^{\alpha+1}}*\dfrac1{(t+1)^{\alpha+1}}*\dots*\dfrac1{(t+1)^{\alpha+1}}}_n\tag2$$
is the multiple convolution.
$\color{brown}{\mathbf{Case\; n = 1.}}$
\begin{cases}
G_1(\alpha,t)=\dfrac1{(t+1)^{\alpha+1}}\\
\big(G_1(\alpha,t)\big)'_t = -\dfrac1{(\alpha+1)(t+1)^{\alpha+2}}\\
\int\limits_0^t G_1(\alpha,t)\,\text dt
=\dfrac1\alpha \left(1 - \dfrac1{(t+1)^{\alpha}}\right).\tag3
\end{cases}
$$I(\alpha,t+1,1) = 4\pi\alpha\int\limits_0^t \dfrac{\text d\lambda} {(\lambda+1)^{\alpha}} 
= \dfrac{4\pi\alpha}{\alpha-1}\left(1-\dfrac1{(t+1)^{\alpha-1}}\right),$$
$$I(\alpha,K,1) = \dfrac{4\pi\alpha}{\alpha-1}\left(1-\dfrac1{K^{\alpha-1}}\right).\tag4$$
$\color{brown}{\mathbf{Case\; n = 2.}}$
$$G_2(\alpha,t) = G_1(\alpha,t) * \dfrac1{(t+1)^{\alpha+1}}
=\int\limits_0^t G_1(\alpha,u)\,\dfrac{\text du}{(t-u+1)^{\alpha+1}},\tag5$$
$$I(\alpha,t+2,2) = 4\pi\alpha^2\int\limits_0^t \int\limits_0^\lambda \dfrac{(\lambda+2) G_1(\alpha,\mu)}{(\lambda-\mu+1)^{\alpha+1}}\,\text d\mu\,\text d\lambda$$
$$= 4\pi\alpha^2\int\limits_0^t \int\limits_0^\mu \dfrac{\big((\lambda-\mu+1) + (\mu+1)\big) G_1(\alpha,\mu)}{(\lambda-\mu+1)^{\alpha+1}}\,\text d\lambda\,\text d\mu$$
$$= 4\pi\alpha^2\int\limits_0^t \int\limits_0^\mu \left(
\dfrac{G_1(\alpha,\mu)}{(\lambda-\mu+1)^{\alpha}} 
+\dfrac{G_1(\alpha-1,\mu)}{(\lambda-\mu+1)^{\alpha+1}} 
\right)\,\text d\lambda\,\text d\mu$$
$$= -4\pi\alpha^2\int\limits_0^t \left(
\dfrac{G_1(\alpha,\mu)}{(\alpha-1)(\lambda-\mu+1)^{\alpha-1}} 
+\dfrac{G_1(\alpha-1,\mu)}{\alpha(\lambda-\mu+1)^{\alpha}} 
\right)\bigg|_0^\mu\,\text d\mu$$
$$= \dfrac{4\pi\alpha}{\alpha-1}\int\limits_0^t
\big(\alpha G_1(\alpha,\mu)(1-G_1(\alpha-2,-\mu))
+(\alpha-1)G_1(\alpha-1,\mu)(1-G_1(\alpha-1,-\mu))\big)
\,\text d\mu$$
$$= \dfrac{4\pi\alpha}{\alpha-1}\int\limits_0^t
(1-G_1(\alpha-2,-\mu))\,\text d G_1(\alpha-1,\mu)$$
$$- 4\pi\alpha \int\limits_0^t G_1(\alpha-1,\mu)(1-G_1(\alpha-1,-\mu)))
\,\text d\mu$$
$$= \dfrac{4\pi\alpha}{\alpha-1}
(1-G_1(\alpha-2,-\mu))\,G_1(\alpha-1,\mu)\bigg|_0^t
- 4\pi\alpha\int\limits_0^t G_1(\alpha-1,\mu) G_1(\alpha-1,-\mu) \,\text d\mu$$
$$- 4\pi\alpha (1-G_1(\alpha-2,-t)))
+ 4\pi\alpha \int\limits_0^t G_1(\alpha-1,\mu)G_1(\alpha-1,-\mu)\,\text d\mu$$
$$= \dfrac{4\pi}{\alpha-1}(1-G_1(\alpha-2,-t))\big(\alpha G_1(\alpha-1,t)
- \alpha+1)\big),$$
$$I(\alpha,K,2) = \dfrac{4\pi}{\alpha-1}(1-G_1(\alpha-2,K-2))\big(\alpha G_1(\alpha-1,K-2)
- \alpha+1)\big),\tag6$$
$$G_2(\alpha,t)=\int\limits_0^t\dfrac{\text du}{(u+1)^{\alpha+1}(t-u+1)^{\alpha+1}}
=\int\limits_1^{t+1}\dfrac{\text du}{u^{\alpha+1}(t+2-u)^{\alpha+1}}\\[4pt]
=\dfrac1{(t+2)^{2\alpha+1}}\int\limits_1^{t+1}\dfrac{\text d\left(\frac u{t+2}\right)}{\left(\frac u{t+2}\right)^{\alpha+1}\left(1-\frac u{t+2}\right)^{\alpha+1}} 
=\dfrac1{(t+2)^{2\alpha+1}}\int\limits_{\large\frac1{t+2}}^{\large\frac{t+1}{t+2}} s^{-\alpha-1}(1-s)^{-\alpha-1}\text ds,$$
$$G_2(\alpha,t) = \dfrac{\text B_{\large\frac{t+1}{t+2}}(-\alpha,-\alpha)
-\text B_{\large\frac1{t+2}}(-\alpha,-\alpha)}{(t+2)^{2\alpha+1}}.\tag7$$
Obtained results show that algebraic closed form of the given integral can exist even if the convolution has not such form.
However, I have not any closed forms for $n\ge3.$