How was the result of the incomplete gamma function in the paper achieved?

[EngWiPy]

Homework Statement

I read in a paper that:
\Gamma\left(c,\,d\frac{x+e}{x-y}\right) = (c-1)!\,exp\left[-d\frac{y+e}{x-y}\right]\,exp[-d]\,\sum_{k=0}^{c-1}\,\sum_{l=0}^k \frac{d^k}{k!}{k\choose l}\left(\frac{y+e}{x-y}\right)^l

Homework Equations

But the incomplete gamma function defined in the book of table of integrals and series as:
\Gamma(1+n,x) = n!\,exp[-x]\,\sum_{k=0}^n \frac{x^m}{m!}

The Attempt at a Solution

Applying this we get:

\Gamma\left(c,\,d\frac{x+e}{x-y}\right) = (c-1)!\, exp\left[-d\frac{x+e}{x-y}\right]\,\sum_{k=0}^{c-1} \frac{d^k}{k!}\,\left(\frac{x+e}{x-y}\right)^k \neq (c-1)!\,exp\left[-d\frac{y+e}{x-y}\right]\,exp[-d]\,\sum_{k=0}^{c-1}\,\sum_{l=0}^k \frac{d^k}{k!}{k\choose l}\left(\frac{y+e}{x-y}\right)^l


How did the authors get their result?

Regards

 

[Avodyne]

Try simplifying the would-be identity as much as possible, and see how far you can get.

 

[EngWiPy]

Execuse me, what is the would-be identity? Can you help me a little to start?

 

[Avodyne]

The would-be identity is

(c-1)!\,\exp\!\!\left(-d\,\frac{x+e}{x-y}\right)\sum_{k=0}^{c-1} \frac{d^k}{k!}\!\left(\frac{x+e}{x-y}\right)^{\!\!k} = (c-1)!\,\exp\!\!\left(-d\,\frac{y+e}{x-y}\right)\exp(-d)\sum_{k=0}^{c-1}\sum_{l=0}^k \frac{d^k}{k!}{k\choose l}\!\!\left(\frac{y+e}{x-y}\right)^{\!\!l}

Try to simplify this, and then show that the simplified equation is true. You can start with the obvious step of canceling the (c{-}1)! on each side. Then, put all the exponentials on one side and see if you can simplify them.

 

[EngWiPy]

Ok, now I know what they did, they just changed the form of one equation, which is \left(\frac{x+e}{x-y}\right) ,to the more convenient form for mathematical manipulation \left(\frac{y+e}{x-y}+1\right).
Thank you very much Avodyne, you helped me to figure it out.
Regards