Some arguements for the unit shift on gamma(n)=(n-1)
Historically the gamma function was defined as the limit of a finite product, namely
\Gamma (z) = \lim_{k\rightarrow \infty} \frac{k! k^{z-1}}{z(z+1)\cdots (z+k-1)}
which converges for all complex z\neq 0, -1, -2,\ldots , but this still seems to have the form \Gamma (z) = f(z-1) where it seems that f(z) is somehow more simple.
Equivalently, the infiniite product form used by Euler was
\Gamma (z) = \frac{1}{z}\prod_{k=1}^{\infty} \left( 1+\frac{z}{k}\right) ^{-1}\left( 1+\frac{1}{k}\right) ^{z}
also converging for all complex z\neq 0, -1, -2,\ldots , but this
does have the look of a function that would not be any more attractive if it were shifted by a unit, and the same remarks hold for the Weierstrass product form of the gamma function, which is
\frac{1}{\Gamma (z)} = ze^{\gamma z}\prod_{k=1}^{\infty} \left( 1+\frac{z}{k}\right) ^{-1} e^{-\frac{z}{k}}
where \gamma =0.577\ldots Euler's constant. Also of note is that the Eulerian integral of the second kind, which is that definition of recent conversation:
\Gamma (z) = \int_{0}^{\infty}e^{-t}t^{z-1} \, dt ,
is notable in that it is the
Mellin transform of e^{-t} since the Mellin transform of f(t) is
\mathcal{M} \left[ f(t) \right] = \int_{0}^{\infty}f(t) t^{z-1} \, dt .
...but I think that the most complelling reason for the shift lies in that the functional equation for the Riemann zeta function, which has the nice symmetric form
\Gamma \left( \frac{s}{2}\right) \pi^{-\frac{s}{2}}\zeta (s) = \Gamma \left( \frac{1-s}{2}\right) \pi^{-\frac{1-s}{2}}\zeta (1-s)
would look like this
\left( \frac{s-2}{2}\right) ! \pi^{-\frac{s}{2}}\zeta (s) = \left( \frac{-1-s}{2}\right) ! \pi^{-\frac{1-s}{2}}\zeta (1-s)
without the shift .
