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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.

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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: $$\Gamma(-n,x) = \frac{(-1)^n}{n!}\left( E_1(x) - e^{-x} \sum_{k=0}^{n-1} \frac{(-1)^k k!}{x^{k+1}}\right)$$

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  • $\begingroup$ Simply excellent ! Thank you ! $\endgroup$ – قيس بن فرج Sep 3 '13 at 14:33
  • $\begingroup$ @gammatester What about if n starts from 0, (i.e. 0, -1, -2, ....). $\endgroup$ – sky-light Jun 19 '14 at 15:12
  • $\begingroup$ @barznjy: There is nothing special about the case $n=0,\;$ the sum is just empty: $\Gamma(0,x) = E_1(x).\;$ But it is also listed separately in dlmf.nist.gov/8.4.E4 $\endgroup$ – gammatester Jun 23 '14 at 7:16
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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.

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  • $\begingroup$ Granted, the resulting definition looks somewhat odd. It might have looked better had $\sum_{k=0}^n a_k$ been defined to exclude the final term $a_n$. $\endgroup$ – Harald Hanche-Olsen Sep 3 '13 at 13:39
  • $\begingroup$ I am just now trying integration by parts. Definying $\Gamma(-n,x) = I_{n+1}$ for $n>0$, I get: $$ I_{n+1} = \frac{e^{-x}}{n x^n} - \frac{1}{n} I_{n}, $$ and I am just trying to find a good form to represent the final result. $\endgroup$ – قيس بن فرج Sep 3 '13 at 14:08

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