Closed form for an integral involving the incomplete Gamma function?

Question

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.

Answer 1

We may use the following integral representation for the incomplete gamma function (DLMF ref. see identity 8.6.4):

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

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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 $$\int_{-\infty}^{\infty}\frac{\left(ix\right)^{\nu_{0}}e^{-ipx}\prod_{k=1}^{n}\left(\beta_{k}+ix\right)^{\nu_{k}}}{\beta_{0}+ix}\mathrm{d}x=0.$$

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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,

$$\begin{align} I_{n}{\left(a,b;\epsilon\right)} &=\frac{2i}{\left[\Gamma{\left(a\right)}\right]^{n}}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\frac{\left(\prod_{k=1}^{n}t_{k}^{a}\right)}{\exp{\left(\sum_{k=1}^{n}t_{k}\right)}}\,f_{n}{\left(b;\epsilon\right)}\\ &=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\frac{\left(\prod_{k=1}^{n}t_{k}^{a}\right)}{\exp{\left(\sum_{k=1}^{n}t_{k}\right)}}\cdot\frac{1}{\prod_{k=1}^{n}\left(\epsilon+t_{k}\right)}\\ &=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\prod_{k=1}^{n}\left(\frac{t_{k}^{a}e^{-t_{k}}}{\epsilon+t_{k}}\right)\\ &=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\left[\int_{0}^{\infty}\mathrm{d}t\,\left(\frac{t^{a}e^{-t}}{\epsilon+t}\right)\right]^{n}\\ &=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\left[\epsilon^{a}e^{\epsilon}\,\Gamma{\left(1+a\right)}\,\Gamma{\left(-a,\epsilon\right)}\right]^{n}\\ &=2\pi a^{n}\left(1-b\epsilon\right)\epsilon^{na-1}e^{b\epsilon}\left[\Gamma{\left(-a,\epsilon\right)}\right]^{n}.\\ \end{align}$$

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Answer 2

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 $n\ge3.$