# The Incomplete Gamma Function

1. Mar 5, 2009

### S_David

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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

Last edited: Mar 5, 2009
2. Mar 5, 2009

### Avodyne

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

3. Mar 5, 2009

### S_David

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

4. Mar 5, 2009

### 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.

5. Mar 6, 2009

### S_David

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