Kummer transform of the confluent hypergeometric function of second kind I can see the kummer transformation of the confluent hypergeometric function of first kind throught the integral representation. However, I failed to see that for the second kind. More specificially, if 
$$U(a,b,z)=\frac{1}{\Gamma(a)}\int_0^\infty e^{-z t}t^{a-1}(1+t)^{b-a-1}dt,$$
then why 
$$U(a,b,z)=z^{1-b}U(1+a-b,2-b,z)?$$
Any references would be appreciated.
 A: --As always please check carefully before using--
Using the Inverse Mellin-Barnes xform DLMF 13.4.17 https://dlmf.nist.gov/13.4.E17 
$${\displaystyle U\left(a,b,z\right)dz=\frac{z^{-a}}{2\pi i}{\displaystyle \int_{-\infty}^{\infty}}\frac{\Gamma\left(a+t\right)\Gamma\left(1+a-b+t\right)\Gamma\left(-t\right)}{\Gamma\left(a\right)\Gamma\left(1+a-b\right)}z^{-t}dt}$$
Setting up 
$a'=a-b+1,b'=2-b , b=2-b',a=1+a'-b',a-b=a'-1$
$${\displaystyle U\left(a,b,z\right)=\frac{z^{-a'+1-b'}}{2\pi i}{\displaystyle \int_{-\infty}^{\infty}}\frac{\Gamma\left(1+a'-b'+t\right)\Gamma\left(a'+t\right)\Gamma\left(-t\right)}{\Gamma\left(1+a'-b'\right)\Gamma\left(a'\right)}z^{-t}dt}$$
$${\displaystyle U\left(a,b,z\right)=z^{1-b}\cdot\left(\frac{z^{-a'}}{2\pi i}{\displaystyle \int_{-\infty}^{\infty}}\frac{\Gamma\left(1+a'-b'+t\right)\Gamma\left(a'+t\right)\Gamma\left(-t\right)}{\Gamma\left(1+a'-b'\right)\Gamma\left(a'\right)}z^{-t}dt\right)}$$
and
$$U\left(a,b,z\right)=z^{1-b}U\left(1+a-b,2-b,z\right)$$
When the critical strips overlap.
Remark 1. When you wander outside the critical strip you are still calculating a function, but its a different function:)
So let's examine
“where the contour of integration separates the poles of $\Gamma\left(a+t\right)\Gamma\left(1-a-b+t\right)$ from those of $\Gamma\left(-t\right)$
1. The poles of $\Gamma\left(-t\right)$ are the integer values $0,1,2,\cdots\infty$ Which are the same in both cases.
2. The poles of $\Gamma\left(a+t\right)$ are at $t=-a,-a-1,-a-2\cdots\infty$
3. The poles of $\Gamma\left(1+a-b+t\right)$ are at $t=b-a-1,b-a-2\cdots$ 
But in the two equations, these are just interchanged. So the overlap is identical.
And the critical strip is:
$0>t>max(-a,b-a-1)$
A:   Term by term expansion. We reduce the original integral to obviously symmetric Pochhammer symbols.   
We need several facts.   
${\displaystyle \intop_{t=0}^{\infty}}e^{-z\cdot t}\cdot t^{\left(w-1\right)}=\Gamma\left(w\right)\cdot z^{-w}$
$\alpha\in\mathbb{C}$
$\left(\alpha\right)_{n}=\frac{\Gamma\left(\alpha+n\right)}{\Gamma\left(\alpha\right)}$
$\left(\begin{array}{c}
\alpha\\
n
\end{array}\right)=\frac{\Gamma\left(\alpha+1\right)}{\Gamma\left(n+1\right)\cdot\Gamma\left(\alpha-n+1\right)}  $
 Letting $m,n\in Z$ and $s\in\mathbb{C}$ and using http://mathworld.wolfram.com/BinomialCoefficient.html 
$\frac{\Gamma\left(s-m+1\right)}{\Gamma\left(s-l+1\right)}=\left(-1\right)^{l-m}\frac{\Gamma\left(l-s\right)}{\Gamma\left(m-s\right)}$
  Starting from the given integral:  
$${\displaystyle \intop_{t=0}^{\infty}}\frac{1}{\Gamma\left(a\right)}e^{-z\cdot t}\cdot t^{a-1}\cdot\left(1+t\right)^{\left(b-a-1\right)}dt$$
We select the $n^{th}$term of the binomial to work with implicit summation at the end. This assumes the the binomial summation and integral can be interchanged.
$\left[t^{n}\right]\left(1+t\right)^{\left(b-a-1\right)}=\frac{\Gamma\left(\left(b-a-1\right)+1\right)}{\Gamma\left(n+1\right)\cdot\Gamma\left(\left(b-a-1\right)-n+1\right)}=\frac{\Gamma\left(b-a\right)}{\Gamma\left(n+1\right)\cdot\Gamma\left(\left(b-a\right)-n\right)}  $ [Gamma-reciprocal]  
and the term integral is:  
$\frac{1}{\Gamma\left(a\right)}\cdot\frac{\Gamma\left(b-a\right)}{\Gamma\left(n+1\right)\cdot\Gamma\left(\left(b-a\right)-n\right)}\cdot{\displaystyle \intop_{t=0}^{\infty}}t^{\left(a-1\right)+n}\cdot e^{-zt}dt  $
$=\frac{1}{\Gamma\left(a\right)}\cdot\frac{\Gamma\left(b-a\right)}{\Gamma\left(n+1\right)\cdot\Gamma\left(\left(b-a\right)-n\right)}\cdot\Gamma\left(a+n\right)\cdot z^{-\left(a+n\right)}  $  
 Parsing  
$\left(\frac{\Gamma\left(a+n\right)}{\Gamma\left(a\right)}\right)\cdot\left(\frac{\Gamma\left(b-a\right)}{\Gamma\left(\left(b-a\right)-n\right)}\right)\cdot\frac{z^{-n}z^{-a}}{\Gamma\left(n+1\right)}  $
Using [Gamma-reciprocal]:Taking $s=b-a,m=1,l=n+1$
 $=\left(a\right)_{n}\cdot\left(\frac{\Gamma\left(b-a\right)}{\Gamma\left(\left(b-a\right)-n\right)}\right)\cdot\frac{z^{-n}z^{-a}}{\Gamma\left(n+1\right)}=\left(a\right)_{n}\cdot\left(\frac{\Gamma\left(\left(n+1\right)-\left(b-a\right)\right)}{\Gamma\left(1-\left(b-a\right)\right)}\right)\cdot\left(-1\right)^{n}\cdot\frac{z^{-n}z^{-a}}{\Gamma\left(n+1\right)}  $
$=\left(a\right)_{n}\cdot\left(a-b+1\right)_{n}\cdot\left(-1\right)^{n}\cdot\frac{z^{-n}z^{-a}}{\Gamma\left(n+1\right)}  $
Which is obviously symetric with: 
$a'=a-b+1,b'=2-b$
And the power $z^{1-b'}\cdot z^{-a}\rightarrow z^{1-b'}z^{-\left(a'-b'+1\right)}\rightarrow z^{-a'}$
