Gamma function ($\Gamma$)

$\Gamma (z)={\int }_0^{\infty }{t}^{z-1}{e}^{-t}dt$ for $\Re z>0$, and $\Gamma (z+1)=z\Gamma (z)$ gives meromorphic continuation.

The Gamma function recurrence and $\Gamma (z)\sim 1/z$ at zero imply a simple pole at every $z=-n$, $n\ge 0$, with residue $(-1{)}^n/n!$.

For $z$ away from the poles, the gamma function satisfiesIts finite product is $m!(m+1{)}^z/[z(z+1)\cdots (z+m)]$. Splitting even and odd factors gives the gamma duplication formula.

The Gamma function recurrence gives $\Gamma (s)=\Gamma (s+j+1)/(s(s+1)\cdots (s+j))$. Its numerator equals one at $s=-j$, so the residue there is $(-1{)}^j/j!$ for every nonnegative integer $j$.

For $a,s>0$, the change of variable $u=at$ in the defining integral for the Gamma function givesIn particular, $\Gamma (1/2)=\sqrt{\pi }$ implies

For $t>0$, contour rotation through angle $-\pi /6$ givesTaking real parts yields

Integration by parts gives

For $\mathrm{Re}s>1$, expansion of $(e^x-1{)}^{-1}$ as a geometric series and termwise integration give

The Euler--Mascheroni constant is

The reciprocal gamma function has the entire-product representation

The digamma function is the logarithmic derivativeLogarithmically differentiating the Weierstrass product gives

The trigamma function is ${\psi }^{\prime }(z)$. For real $z>0$,

The digamma function is strictly increasing on the positive real axis, whileIt therefore has exactly one positive zero, and that zero lies in $(1,2)$.

It follows by making the logarithm of the quotient entire and periodic; growth forces its periodic remainder to be constant.

For $\mathrm{Re}p,\mathrm{Re}q>0$, Euler's beta function is

For $c=1-a-b$ with positive real parts $\mathrm{Re}a,\mathrm{Re}b,\mathrm{Re}c$, the integral converges at its two finite singularities and at infinity, and equalsTo prove it, use Schwinger parameterization with exponents $1-a$ and $1-b$. Integrating the resulting planar Gaussian integral gives $\pi /(\lambda +\rho )$ times $\exp [-\lambda \rho /(\lambda +\rho )]$. Set $q=\lambda +\rho$ and $x=\lambda /q$; the $q$ integral is a gamma function and the $x$ integral is $B(a,b)$, giving the stated ratio. This domain justifies the interchanges; elsewhere the answer is interpreted by analytic continuation. The string measure $d^{2}z=2dxdy$ multiplies the answer by two.

The sum-and-ratio change of variables in a product of gamma integrals gives

For $s,t>0$, put $r=s+t$ and $u=t/(s+t)$. Then $(s,t)=(r(1-u),ru)$ and $dsdt=rdrdu$, separating radial gamma and ratio beta integrals.

The Beta function obeysThese identities follow immediately from the beta--gamma identity and the Gamma function recurrence.

For $0<a<2$,Differentiating at $a=1$ and expanding the secant gives

The functional equation and $\Gamma (1)=1$ give $\Gamma (z)\sim 1/z$ as $z\to 0$ away from the negative real axis.

The beta--gamma identity gives $B(z,z)=\Gamma (z{)}^2/\Gamma (2z)\sim 2/z$ as $z\to 0$ through the right half-plane.