Integral involving power, exponential and confluent hypergeometric function I am seeking a solution for the following integral:
\begin{equation}
\int\frac{e^{-\beta t}}{t}\,U(a,b,t) \,\mathrm{d}t
\end{equation}
where $0<\beta<1$, $a<1$, $b<2$, and $U(a,b,t)$ is the confluent hypergeometric function of the 2nd kind. I have already solved this integral for $a=0,-1,-2,\dots$


Motivation for asking this question:
I am trying to compute the following Cauchy principal valued integral
  \begin{equation}
\lim_{\varepsilon\to 0^{+}}\int_{\varepsilon}^{\infty}\frac{1}{t}\left(\frac{e^{-\beta_{1} t}}{\Gamma(\alpha_{1})}U(1-\alpha_{1},2-\alpha_{3},\beta_{3}t)
-
\frac{e^{-\beta_{2} t}}{\Gamma(\alpha_{2})}U(1-\alpha_{2},2-\alpha_{3},\beta_{3}t)
\right)\mathrm{d}t,
\end{equation}
  where $\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}>0$, $\alpha_{3}=\alpha_{1}+\alpha_{2}$, and $\beta_{3}=\beta_{1}+\beta_{2}$. I have already solved this integral for the following increasingly general special cases:
$\qquad(1).\ \ $ $\alpha_{1}=\alpha_{2}$ and $\beta_{1}=\beta_{2}$ (trivial case).
$\qquad(2).\ \ $ $\alpha_{1}=\alpha_{2}=1$.
$\qquad(3).\ \ $ $\alpha_{1}\in\mathbb{N}^{+}$ and $\alpha_{2}\in\mathbb{N}^{+}$.
My ultimate goal is to find the solution for $\alpha_{1},\alpha_{2}\in\mathbb{R}^{+}$.

 A: If you consider this not to a relevant answer, just tell me and I will delete it.  Of course, constructive remarks are always welcome.
Be aware that $U()$ has a lot of special cases (computational branch points) that I have ignored in favour of what I consider typical. If you have special cases of interest that makes this blow up, I will try to look into them.
An equation is:${\displaystyle \int_{0}^{\infty}}t^{s-1}e^{-\alpha t}U\left(a,b,\lambda\cdot t\right)dt$
Using Mellin transform interpretation of 
http://dlmf.nist.gov/13.10#E7
${\displaystyle \int_{0}^{\infty}}t^{s-1}e^{-z\cdot t}U\left(a,c,t\right)dt=\Gamma\left(s\right)\cdot\Gamma\left(s-c+1\right)\cdot z^{-s}\cdot\,_{2}F_{1}\left(a,s;a+s-c+1;1-\frac{1}{z}\right)$
Requiring $\mathbb{\mathcal{R}}\left(z\right)>0,\mathcal{R}\left(s\right)>max\left(\mathbb{\mathcal{R}}\left(c\right),0\right)$
We can examine the terms of the right hand side and make sure the power series expansion matches up after inversion on a term by term basis. Failure of the inverse Mellin transform implies that when $\mathcal{R}\left(s\right)\leq0$ either the closure of the contour around the left hand poles fails because of essential or logarithmic singularity ; or because of the problem of the obvious pole/zero cancellation. But let's just move the goal posts a little using http://dlmf.nist.gov/15.5.E15 and some substitution
$_{2}F_{1}\left(a,s;a+s-c+1;1-\frac{1}{z}\right)=\left(1-\frac{b}{c}\right)\cdot\left(1-\frac{1}{z}\right)\cdot\,_{2}F_{1}\left(a,s+1;a+s-c+2;1-\frac{1}{z}\right)-\frac{1}{z}\cdot\,_{2}F_{1}\left(a,s+1;a+s-c+1;1-\frac{1}{z}\right)$
$\Gamma\left(s\right)\cdot\Gamma\left(s-c+1\right)\Rightarrow\frac{\Gamma\left(s+1\right)}{s}\cdot\Gamma\left(s-c+1\right)$
Which pulls out $\frac{1}{s}$ as a simple pole and the other terms are well behaved. In fact we can recognize the alteration as almost a simple integration.
Now I can probably pull this back to two integrals of $\int e^{-z\cdot t}U(..)$ and do a term by term check; but given the complexity of the series expansion $U()$ you probably won't like it and I would put it in a link. An alternative is rearrange the original to $U()=M()\cdot...+M()\cdot...$ using http://dlmf.nist.gov/13.2.E42. But that has the restriction on c not be an integer.
Yet another alternative is using “Cauchy principal value” which, as I recall, deals with this sort of thing for the Casimir effect. I think I can find the usage explanation and usage in an MAA article.
