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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.

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  • $\begingroup$ It's an exercise but would you be interested in DLMF 13.2.42? Comparing the RHS and doing substitutions seems to interchange the terms; leaving the sum intact. Of course, there is the condition "b is not an integer". Or do you really want to work through the integral? $\endgroup$ – rrogers Apr 18 at 14:28
  • $\begingroup$ A complete derivation and explanation can be found on Pages 262-265 of Special Functions & Their Applications (Dover Books on Mathematics) Kindle Edition by N. N. Lebedev (Author), Richard A. Silverman (Translator)--books.google.com/books/about/…. Actually they derive the integral from the DLMF 13.2.42 identity. $\endgroup$ – rrogers Apr 18 at 14:58
  • $\begingroup$ Yes, thanks for your comments. I have Lebedev 's book. However, as mentioned in your first comment, I am more interested in seeing it through the integral representation. I am doubting whether a change of variable works here since the gamma function on the denominator would be different. $\endgroup$ – gouwangzhangdong Apr 20 at 2:15
  • $\begingroup$ I think the duality can be shown via the Mellin (Inverse) Transform DLMF 13.10.11. Although the pullback has to down along the convergence strip and shown to be in the overlap between the two instances. Luke writes 13.10.11 as $\int((a)_{s}(1+a-b)_{s} ....$ Which can be seen by squinting :) The $z^{1-b}$ is the standard Mellin property. If you have a problem I will write it up. As for removing the restriction; I'll just wave my magic wand and say "analytic continuation" like real wizards do. Actually, if you treat your first equation as a Mellin transform on "a" it might work.(?) $\endgroup$ – rrogers Apr 21 at 21:41
  • $\begingroup$ Yes, could you please write it up? $\endgroup$ – gouwangzhangdong Apr 24 at 3:58
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  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 $z^{1-b'}\cdot z^{-a}\rightarrow z^{1-b'}z^{-\left(a'-b'+1\right)}\rightarrow z^{-a'}$

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  • $\begingroup$ Many thanks for this interesting answer. Just one tiny issue. You used the binomial expansion, which needs to require b-a is an integer. Of course, you can use the Taylor series expansion instead, but that would require t between 0 and 1. This also reminds me another question: can we see the relationship between 1F1(a,b,z) and U(a,b,z), i.e., Eq.(9.10.3), "Lebedev" simply through their integral representations given here without the differentiation equation story? $\endgroup$ – gouwangzhangdong Apr 28 at 2:49
  • $\begingroup$ Binomial expansion: en.wikipedia.org/wiki/… , It's surprising but common. In fact, it is used in several derivations in Slater "Generalized Hypergeometric Function " (for instance page 20. It might be considered the basis for int(x^a*(1+b)^b,x) type of integral representations. $\endgroup$ – rrogers Apr 28 at 22:14
  • $\begingroup$ I drifted off: No I can't see a way to do a real algebraic transform that that would reproduce 9.10.3. Personally, I would try going through the Mellin domain to demonstrate the connection algebraically. To me, it appears that complex contour integration is required. An obstacle is DLMF 13.2.6 comment "U(a,b,z), which is determined uniquely by the property" $U(a,b,z) -> z^{-a}$ as z goes to $\inf$ . $\endgroup$ – rrogers May 2 at 13:05
  • $\begingroup$ Many thanks for your help. It is really inspiring. Do you have any idea about this one: math.stackexchange.com/questions/3203040/… ? $\endgroup$ – gouwangzhangdong May 3 at 8:42
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--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:
$0>t>max(-a,b-a-1)$

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  • $\begingroup$ Ah, I see, I misunderstood your last comment. Yes, this is another way and I was aware of this method when I was reading Harry Bateman's "Higer Transcendental Functions", Volume I. Thanks for such detailed explanation, Remark 1 particularly. $\endgroup$ – gouwangzhangdong Apr 25 at 3:03

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