definite integralsgamma functionmellin-transform
\begin{align}
& \int_{0}^{\infty}y^{\nu-1}e^{-y}\int_{0}^{\infty}e^{-x}(\alpha x+y)^{\beta}dxdy\\
=& \int_{0}^{\infty}y^{\nu-1}e^{-y}(\alpha^{\beta}e^{\frac{y}{\alpha}}\Gamma(\beta+1,\frac{y}{\alpha}))dy,\quad (3)
\end{align}
where $\Gamma(\cdot,\cdot)$ represents the upper incomplete gamma function, i.e., $\Gamma(s,x)=\int_{x}^{\infty}t^{s-1}e^{-t}dt$.
If I can find a compact form of the above integral (3) ?
Thanks a lot, Liu.
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,
$$\begin{align} I_{n}{\left(a,b;\epsilon\right)} &=\frac{2i}{\left[\Gamma{\left(a\right)}\right]^{n}}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\frac{\left(\prod_{k=1}^{n}t_{k}^{a}\right)}{\exp{\left(\sum_{k=1}^{n}t_{k}\right)}}\,f_{n}{\left(b;\epsilon\right)}\\ &=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\frac{\left(\prod_{k=1}^{n}t_{k}^{a}\right)}{\exp{\left(\sum_{k=1}^{n}t_{k}\right)}}\cdot\frac{1}{\prod_{k=1}^{n}\left(\epsilon+t_{k}\right)}\\ &=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\int_{[0,\infty)^{n}}\mathrm{d}^{n}\mathbf{t}\,\prod_{k=1}^{n}\left(\frac{t_{k}^{a}e^{-t_{k}}}{\epsilon+t_{k}}\right)\\ &=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\left[\int_{0}^{\infty}\mathrm{d}t\,\left(\frac{t^{a}e^{-t}}{\epsilon+t}\right)\right]^{n}\\ &=\frac{2\pi\left(1-b\epsilon\right)e^{-\left(n-b\right)\epsilon}}{\left[\Gamma{\left(a\right)}\right]^{n}\epsilon}\left[\epsilon^{a}e^{\epsilon}\,\Gamma{\left(1+a\right)}\,\Gamma{\left(-a,\epsilon\right)}\right]^{n}\\ &=2\pi a^{n}\left(1-b\epsilon\right)\epsilon^{na-1}e^{b\epsilon}\left[\Gamma{\left(-a,\epsilon\right)}\right]^{n}.\\ \end{align}$$
For $\operatorname{Re}(\alpha) < 1$ and $\beta > 0$ the function $$ f \colon \mathbb{R}^+ \to \mathbb{C}\, , \, f(x) = \int \limits_\beta^\infty \mathrm{e}^{- x \theta} \frac{1}{\theta} \left(\frac{\theta}{\beta}-1\right)^{-\alpha} \, \mathrm{d} \theta \, , $$ is well-defined and differentiable. Using the change of variables $\theta = \beta \left(1 + \frac{\phi}{\beta x}\right)$ we find for $x > 0$ $$ f'(x) = - \int \limits_\beta^\infty \mathrm{e}^{- x \theta} \left(\frac{\theta}{\beta}-1\right)^{-\alpha} \, \mathrm{d} \theta = - \beta^{\alpha} x^{\alpha-1} \mathrm{e}^{-\beta x} \int \limits_0^\infty \phi^{-\alpha} \mathrm{e}^{- \phi} \, \mathrm{d} \phi = - \beta^{\alpha} x^{\alpha-1} \mathrm{e}^{-\beta x} \, \Gamma(1-\alpha) \, .$$ Since $\lim_{x \to \infty} f(x) = 0$ , we obtain \begin{align} f(x) &= - \int \limits_x^\infty f'(y) \, \mathrm{d} y = \Gamma(1-\alpha) \beta^{\alpha} \int \limits_x^\infty y^{\alpha-1} \mathrm{e}^{- \beta y} \, \mathrm{d} y = \Gamma(1-\alpha) \int \limits_{\beta x}^\infty z^{\alpha-1} \mathrm{e}^{- z} \, \mathrm{d} z \\ &= \Gamma(1-\alpha) \Gamma(\alpha, \beta x) \, \end{align} for $x > 0$ .
The Laplace transform of the gamma function does not exist, since $\Gamma$ grows faster than any exponential function.
Best Answer
Given real parameters $\left(\beta,\nu,a\right)\in\mathbb{R}^{3}$, suppose
$$0<\beta+\nu+1\land0<\nu\land\left[0<a\lor\left(a=0\land0<\beta+\nu\right)\right].$$
These are the most general conditions for which the following integral is convergent and real-valued:
$$\mathcal{I}{\left(\beta,\nu,a\right)}=\int_{0}^{\infty}\mathrm{d}y\,y^{\nu-1}e^{-y}\int_{0}^{\infty}\mathrm{d}x\,\left(ax+y\right)^{\beta}e^{-x}.$$
Changing the order of integration and rescaling the integrals in the right way, the double integral in question turns out to be separable. We obtain
$$\begin{align} \mathcal{I}{\left(\beta,\nu,a\right)} &=\int_{0}^{\infty}\mathrm{d}y\,y^{\nu-1}e^{-y}\int_{0}^{\infty}\mathrm{d}x\,\left(ax+y\right)^{\beta}e^{-x}\\ &=\int_{0}^{\infty}\mathrm{d}y\int_{0}^{\infty}\mathrm{d}x\,\left(ax+y\right)^{\beta}e^{-x}y^{\nu-1}e^{-y}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}y\,\left(ax+y\right)^{\beta}y^{\nu-1}e^{-x-y}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}t\,x\,\left(ax+xt\right)^{\beta}(xt)^{\nu-1}e^{-x-xt};~~~\small{\left[y=xt\right]}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{\infty}\mathrm{d}t\,\left(a+t\right)^{\beta}x^{\beta+\nu}t^{\nu-1}e^{-x(1+t)}\\ &=\int_{0}^{\infty}\mathrm{d}t\int_{0}^{\infty}\mathrm{d}x\,t^{\nu-1}\left(a+t\right)^{\beta}x^{\beta+\nu}e^{-(1+t)x}\\ &=\int_{0}^{\infty}\mathrm{d}t\,\frac{t^{\nu-1}\left(a+t\right)^{\beta}}{\left(1+t\right)^{\beta+\nu+1}}\int_{0}^{\infty}\mathrm{d}u\,u^{\beta+\nu}e^{-u};~~~\small{\left[x=\frac{u}{1+t}\right]}\\ &=\Gamma{\left(\beta+\nu+1\right)}\int_{0}^{\infty}\mathrm{d}t\,\frac{t^{\nu-1}\left(a+t\right)^{\beta}}{\left(1+t\right)^{\beta+\nu+1}};~~~\small{\left[-1<\beta+\nu\right]}\\ &=\nu^{-1}\Gamma{\left(\beta+\nu+1\right)}\,{_2F_1}{\left(-\beta,1;\nu+1;1-a\right)},\\ \end{align}$$
where in the last line above we've made use of this integral representation variant for the Gauss hypergeometric function.