Incomplete gamma function as a sum

Question

The incomplete gamma function is well known as: $\Gamma(n,x) = \int_x^{+\infty} t^{n-1} e^{-t} dt$. This function for positive values of its first argument $n>1$ has a "sum" representation as: $$ \Gamma(n,x) = (n-1)! e^{-x} \sum_{k=0}^{n-1} \frac{x^k}{k!} $$ with $(.)!$ being the factorial function and is valid for any value of $x$.

My question is : is there a representation as a "sum" for negative values of $n$ ? I tried to find out on the Internet but no way. A friend helped me with his own try and found out this one: $$ \Gamma(n,x) = \frac{x^n}{(-n)!} e^{-x} \sum_{k=0}^{-n+1} \frac{x^{k-1}}{(-n+k)!} $$ but I tried it on Mathematica and it does not work.

Any help ? Thank you.

Answer 1

There is a sum formula for negative $n$ but it contains the exponential integral $E_1(x)$ given in http://dlmf.nist.gov/8.4#E15: $$\Gamma(-n,x) = \frac{(-1)^n}{n!}\left( E_1(x) - e^{-x} \sum_{k=0}^{n-1} \frac{(-1)^k k!}{x^{k+1}}\right)$$

Answer 2

You can use the fundamental property $$\sum_{k=0}^n a_k=a_n+\sum_{k=0}^{n-1} a_k$$ to define the sum for negative $n$. In particular, setting $n=0$ you conclude that $$\sum_{k=0}^{-1} a_k=0.$$ Next, insert $n=-1$ in the formula to get $$\sum_{k=0}^{-2}a_k=-a_{-1},$$ and so forth. Proceed by induction.