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 ${\displaystyle a_{n}}$ :

${\displaystyle a_{n}\approx {\frac {(-1)^{n}}{\ (n-1)!\ }}\ {\sqrt {{\frac {2}{\ \pi n\ }}\ }}\ \operatorname {Im} \left({\frac {\ z_{0}^{\left(1/2-n\right)}\ e^{-nz_{0}}\ }{\sqrt {1+z_{0}\ }}}\right)\ ,}$ 

where ${\displaystyle z_{0}=-{\frac {1}{n}}\exp \!{\Bigl (}W_{-1}(-n){\Bigr )}\ ,}$  and ${\displaystyle W_{-1}}$  is the minus-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' 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 ${\displaystyle n\geq 1}$ , 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 ${\displaystyle c>0}$  and ${\displaystyle z\in \mathbb {C} }$  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 ${\displaystyle z=n+1/2}$  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.

We have the following formula (^[7] chapter 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

• Mette Lund, An integral for the reciprocal Gamma function
• Milton Abramowitz & Irene A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables
• Eric W. Weisstein, Gamma Function, MathWorld