Reciprocal gamma function

Following from the infinite product definitions for the gamma function, due to Euler and Weierstrass respectively, we get the following infinite product expansion for the reciprocal gamma function: $${\displaystyle {\begin{aligned}{\frac {1}{\Gamma (z)}}&=z\prod _{n=1}^{\infty }{\frac {1+{\frac {z}{n}}}{\left(1+{\frac {1}{n}}\right)^{z}}}\\{\frac {1}{\Gamma (z)}}&=ze^{\gamma z}\prod _{n=1}^{\infty }\left(1+{\frac {z}{n}}\right)e^{-{\frac {z}{n}}}\end{aligned}}}$$

where γ = 0.577216... is the Euler–Mascheroni constant. These expansions are valid for all complex numbers z.

Taylor series expansion around 0 gives:^[1] $${\displaystyle {\frac {1}{\ \Gamma (z)\ }}=z+\gamma \ z^{2}+\left({\frac {\gamma ^{2}}{2}}-{\frac {\pi ^{2}}{12}}\right)\ z^{3}+\left({\frac {\gamma ^{3}}{6}}-{\frac {\gamma \pi ^{2}}{12}}+{\frac {\zeta (3)}{3}}\ \right)z^{4}+\cdots \ }$$ where γ is the Euler–Mascheroni constant. For n > 2, the coefficient a_n for the zⁿ term can be computed recursively as^[2][3] $${\displaystyle a_{n}={\frac {\ {a_{2}\ a_{n-1}+\sum _{j=2}^{n-1}(-1)^{j+1}\ \zeta (j)\ a_{n-j}}\ }{n-1}}={\frac {\ \gamma \ a_{n-1}-\zeta (2)\ a_{n-2}+\zeta (3)\ a_{n-3}-\cdots \ }{n-1}}}$$ where ζ is the Riemann zeta function. An integral representation for these coefficients was recently found by Fekih-Ahmed (2014):^[3] $${\displaystyle a_{n}={\frac {(-1)^{n}}{\pi n!}}\int _{0}^{\infty }e^{-t}\ \operatorname {Im} {\Bigl [}\ {\bigl (}\log(t)-i\pi {\bigr )}^{n}\ {\Bigr ]}\ \mathrm {d} t~.}$$

For small values, these give the following values:

Fekih-Ahmed (2014)^[3] also gives an approximation for a_n:

$${\displaystyle a_{n}\approx {\frac {(-1)^{n}}{\ (n-1)!\ }}\ {\sqrt {{\frac {2}{\ \pi n\ }}\ }}\ \operatorname {Im} \left({\frac {\ z_{0}^{\left(1/2-n\right)}\ e^{-nz_{0}}\ }{\sqrt {1+z_{0}\ }}}\right)\ ,}$$ where z₀ = −1/n exp(W₋₁(−n)), and W₋₁ is the negative-first branch of the Lambert W function.

The Taylor expansion around 1 has the same (but shifted) coefficients, i.e.: $${\displaystyle {\frac {1}{\Gamma (1+z)}}={\frac {1}{z\Gamma (z)}}=1+\gamma \ z+\left({\frac {\gamma ^{2}}{2}}-{\frac {\pi ^{2}}{12}}\right)\ z^{2}+\left({\frac {\gamma ^{3}}{6}}-{\frac {\gamma \pi ^{2}}{12}}+{\frac {\zeta (3)}{3}}\ \right)z^{3}+\cdots \ }$$ (the reciprocal of Gauss's pi function).

As |z| goes to infinity at a constant arg(z) we have: $${\displaystyle \ln(1/\Gamma (z))\sim -z\ln(z)+z+{\tfrac {1}{2}}\ln \left({\frac {z}{2\pi }}\right)-{\frac {1}{12z}}+{\frac {1}{360z^{3}}}-{\frac {1}{1260z^{5}}}\qquad {\text{for}}~\left|\arg(z)\right|<\pi }$$

An integral representation due to Hermann Hankel is $${\displaystyle {\frac {1}{\Gamma (z)}}={\frac {i}{2\pi }}\oint _{H}(-t)^{-z}e^{-t}\,dt,}$$ where H is the Hankel contour, that is, the path encircling 0 in the positive direction, beginning at and returning to positive infinity with respect for the branch cut along the positive real axis. According to Schmelzer & Trefethen,^[4] numerical evaluation of Hankel's integral is the basis of some of the best methods for computing the gamma function.

For positive integers n ≥ 1, there is an integral for the reciprocal factorial function given by^[5] $${\displaystyle {\frac {1}{n!}}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }e^{-nit}e^{e^{it}}\ dt.}$$

Similarly, for any real c > 0 and z ∈ C such that Re(z) > 0 we have the next integral for the reciprocal gamma function along the real axis in the form of:^[6] $${\displaystyle {\frac {1}{\Gamma (z)}}={\frac {1}{2\pi }}\int _{-\infty }^{\infty }(c+it)^{-z}e^{c+it}dt,}$$ where the particular case when z = n + 1/2 provides a corresponding relation for the reciprocal double factorial function, $${\displaystyle {\frac {1}{(2n-1)!!}}={\frac {\sqrt {\pi }}{2^{n}\cdot \Gamma \left(n+{\frac {1}{2}}\right)}}.}$$

Integration of the reciprocal gamma function along the positive real axis gives the value $${\displaystyle \int _{0}^{\infty }{\frac {1}{\Gamma (x)}}\,dx\approx 2.80777024,}$$ which is known as the Fransén–Robinson constant.^[7]

We have the following formula^{[8]: ch 9, exercise 100} $${\displaystyle \int _{0}^{\infty }{\dfrac {a^{x}}{\Gamma (x)}}\,dx=ae^{a}+a\int _{0}^{\infty }{\dfrac {e^{-ax}}{\log ^{2}(x)+\pi ^{2}}}\,dx}$$

• Bessel–Clifford function
• Inverse-gamma distribution