1. ## Gamma and factorial

Is this expression true?
$(n+s)! \Gamma (s+1)= s! \Gamma (n+s+1)$
i was looking at the proof of $J_{-n} (x)= (-1)^n J_n (x)$ and i deduced that expression must be true for the proof provided to be sound, the proof was not from a rigourous source so it may have cut corners & have botched up somewhere.

2. Originally Posted by phycdude
Is this expression true?
$(n+s)! \Gamma (s+1)= s! \Gamma (n+s+1)$.
For this to make sense requires that $s$ and $n$ be natural numbers. In which case the gammas may be replaced by factorials and both sides are equal to $s!(n+s)!$

CB

3. Originally Posted by phycdude
Is this expression true?
$(n+s)! \Gamma (s+1)= s! \Gamma (n+s+1)$
What makes sense if s is non-integer and n is positive integer is the following: $\Gamma(n+s+1)=(n+s)\Gamma(n+s)=(n+s)(n+s-1)\Gamma(n+s-1)$ $=\cdots=(n+s)\cdots (s+1)\Gamma(s+1)$. On the other hand, it is not uncommon to define $s!=\Gamma(s+1)$ for non-integer $s$, in which case your expression is trivially true.

4. Originally Posted by Laurent
What makes sense if s is non-integer and n is positive integer is the following: $\Gamma(n+s+1)=(n+s)\Gamma(n+s)=(n+s)(n+s-1)\Gamma(n+s-1)$ $=\cdots=(n+s)\cdots (s+1)\Gamma(s+1)$. On the other hand, it is not uncommon to define $s!=\Gamma(s+1)$ for non-integer $s$, in which case your expression is trivially true.
thanks , makes lots of sense now