# Double integral $\int_{z=u}^{+\infty}\int_{t=u}^{+\infty}\frac{e^{-Az}}{z+B}\frac{te^{-tD}}{t-zC}\,dtdz$

I am doing research, and while calculating a closed form expression, I got a form of integration like the following:

$$\int_{z=u}^{+\infty}\int_{t=u}^{+\infty}\frac{e^{-Az}}{z+B}\frac{te^{-tD}}{t-zC}dtdz$$ where $A$, $B$, $C$, $D$ and $u$ are positive reals.

I don't know if there is way to get the closed form of it or we have to rely on some approximations.

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\begin{align} \int_{z=u}^{+\infty}\int_{t=u}^{+\infty}\frac{te^{-tD}}{t-zC} \mathbb{d}t\mathbb{d}z &= \int_{z=u}^{+\infty} \left( \frac{e^{-Az}}{z+B} \int_{t=u}^{+\infty}\frac{te^{-tD}}{t-zC}\mathbb{d}t \right)\mathbb{d}z \\ &=\int_{z=u}^{+\infty} \left( \frac{e^{-Az}}{z+B} \underbrace{\int_{t=u}^{+\infty}\frac{te^{-tD}}{t-zC}\mathbb{d}t}_{J_{D,C}(u,z)} \right)\mathbb{d}z \\ \end{align} with \begin{align} J_{D,C}(u,z) &= \int_{t=u}^{+\infty}\frac{te^{-tD}}{t-zC}\mathbb{d}t \\ &= \int_{t=u}^{+\infty}\frac{t-zC+zC}{t-zC}e^{-tD}\mathbb{d}t \\ &= \int_{t=u}^{+\infty}e^{-tD}\mathbb{d}t + zC\int_{t=u}^{+\infty}\frac{e^{-tD}}{t-zC}\mathbb{d}t\\ &= \left[\frac{e^{-tD}}{-D}\right]_u^{+\infty} + zC I_{D,zC}(u)\\ \end{align}
where $$I_{a,b}(u) = \int\limits_{u}^{\infty} \frac{\ \exp\left(-a t\right)}{t-b} \mathrm{d}t \stackrel{s=\frac{t-b}{u-b}}= (u-b)e^{-ab}\int\limits_{1}^{\infty} \frac{\ e^{-a(u-b)s}}{s} \mathrm{d}s =(u-b)e^{-ab}E_1(a(u-b))$$ $=>$ $$J_{D,C}(u,z) =\frac{e^{-uD}}{D} + zC(u-zC)e^{-zCD}E_1(D(u-zC))$$ Now :
\begin{align} \int_{z=u}^{+\infty}\int_{t=u}^{+\infty}\frac{te^{-tD}}{t-zC} \mathbb{d}t\mathbb{d}z &=\frac{e^{-uD}}{D}\int_{z=u}^{+\infty} \frac{e^{-Az}}{z+B} \mathbb{d}z + \int_{z=u}^{+\infty} \frac{zC(u-zC)e^{-zCD}E_1(D(u-zC))}{z+B} e^{-Az}\mathbb{d}z\\ &=\frac{e^{-uD}}{D}I_{A,-B} + \underbrace{\int_{z=u}^{+\infty} \frac{zC(u-zC)e^{-zCD}E_1(D(u-zC))}{z+B} e^{-Az}\mathbb{d}z}_{(1)}\\ \end{align} Let $g(z)=(u-zc)De^{D(u-zC)}E_1(D(u-zC))$
since $\frac{z}{z+B}= 1-\frac{B}{z+B}$ then : \begin{align} (1) &= \frac{C}{D}e^{-uD}\int_{z=u}^{+\infty}{g(z)} e^{-Az}\mathbb{d}z-\frac{BC}{D}e^{-uD}\int_{z=u}^{+\infty} \frac{g(z)}{z+B} e^{-Az}\mathbb{d}z \\ \end{align}
Consider the substitution $s=z-u$ \begin{align}(1) &= \frac{C}{D}e^{-uD}\int_{0}^{+\infty}{g(s+u)} e^{-As}\mathbb{d}s-\frac{BC}{D}e^{-uD}\int_{0}^{+\infty} \frac{g(s+u)}{z+B} e^{-As}\mathbb{d}s\\ &= \mathcal{L}\{g(s+u)\} (A) -\frac{BC}{D}e^{-uD}\mathcal{L}^2\{g(s+u)e^{-As}\} (B) \\ \end{align} where $\mathcal{L}\{f(t)\}$ The Laplace transform and
$\mathcal{L}^2\{f(t)\}$ the second iterate of Laplace transform wich the same as the Stieltjes Transform given by: $\mathcal{S}\{f(x)\}(y)= \int_{0}^{+\infty} \frac{f(x)}{x+y} \mathbb{d}x$