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The Gamma function extends factorials to a continuous argument. Its defining Gamma integral iswhere $t^{z-1}=\exp ((z-1)\log t)$ uses the real natural logarithm for $t>0$. The Gamma function recurrence gives its meromorphic continuation to the complex plane, with simple poles at $0,-1,-2,\dots$ and no zeros. See Gamma function residue at a nonpositive integer and Gamma function has no zeros for proof outlines.^[ref]

For nonnegative integers $n$, the factorial identity has a shift of one:Thus $\Gamma (n)=(n-1)!$ for positive integers $n$. Half-integer arguments supply a different familiar constant:The first value follows from the Gaussian integral; the second follows from the Gamma function recurrence.^[ref]

The plot shows only positive real arguments; the Gamma function also accepts complex numbers away from its poles. Matching the factorials alone does not select a unique interpolation, even if the Gamma function recurrence is imposed: $f(x)=\Gamma (x)\exp (\sin (2\pi x))$ has the same positive-integer values and the same recurrence relation. The Bohr-Mollerup theorem supplies the extra condition that selects the Gamma function.

The useful idea is to replace a discrete product of numbers by an integral whose exponent varies continuously. The Gamma integral then evaluates whole families of integrals at once, provides the normalizing constant for the Gamma distribution in probability theory, and appears in physics, for example in the Bose integral. Its large-positive-argument growth reconnects this continuous description to factorials through Stirling's formula.^[ref]

Historically, Leonhard Euler studied the interpolation of factorials in his paper E19, written in 1729 and published in 1738. His integral over $(0,1)$ becomes the Gamma integral after a change of variables, with the same one-step shift in the argument.^[ref]

For $\mathrm{Re}z>0$, absolute convergence permits the two-dimensional gamma integralPut $u=rs$, $v=r(1-s)$, with $r>0$, $0<s<1$ and Jacobian determinant $r$. Separation of the integrals gives the beta--gamma identityThe beta integral is finite in this half-plane. Thus a zero of $\Gamma (z)$ would imply $\Gamma (z/2)=0$, and iteration would imply $\Gamma (z/2^m)=0$ for every positive integer $m$. But integration by parts gives $w\Gamma (w)=\Gamma (1+w)$, and continuity near one makes $\Gamma (1+w)\to 1$ as $w\to 0$. It cannot vanish for sufficiently small $w$, a contradiction. Therefore the Gamma function has no zeros in the right half-plane.

For any $z$ outside the nonpositive integers, choose an integer $n$ with $\mathrm{Re}(z+n)>0$. The Gamma function recurrence givesAt the excluded integers gamma has poles, with nonzero residues by the same recurrence. Hence the meromorphic Gamma function has no zeros anywhere; at its poles it has no finite value to which a nonzero-value assertion could apply.