Special functions

It shows the Gamma integral is not an arbitrary generalization but the unique smooth continuation that reproduces the multiplicative structure of factorials, which is why it is the right object to plug non-integers into.

Step 1 (Base case). $\Gamma(1) = \int_0^\infty t^0 e^{-t}\,dt = \int_0^\infty e^{-t}\,dt = \big[-e^{-t}\big]_0^\infty = 0-(-1) = 1 = 0!$, so the formula holds for $n=0$.

Step 2 (Recurrence by integration by parts). For $\operatorname{Re}(z) > 0$, integrate $\Gamma(z)=\displaystyle\int_0^\infty t^{z-1}e^{-t}\,dt$ by parts with $u=t^z$, $dv=e^{-t}\,dt$: $\Gamma(z+1) = \int_0^\infty t^z e^{-t}\,dt = \big[-t^z e^{-t}\big]_0^\infty + z\int_0^\infty t^{z-1}e^{-t}\,dt = 0 + z\,\Gamma(z)$, using that the boundary term vanishes at both ends. This proves $\Gamma(z+1) = z\,\Gamma(z)$.

Step 3 (Induction). Assume $\Gamma(k+1) = k!$ for some non-negative integer $k$. By Step 2 with $z=k+1$, $\Gamma(k+2) = (k+1)\,\Gamma(k+1) = (k+1)\cdot k! = (k+1)!$.

Step 4 (Conclude). Since the base case $n=0$ holds (Step 1) and the inductive step carries the identity from $n=k$ to $n=k+1$ (Step 3), $\Gamma(n+1)=n!$ holds for every non-negative integer $n$ by mathematical induction.

It links a purely algebraic-looking object (a ratio of two Gamma values) to trigonometry, and it is exactly the identity that makes the poles of $\Gamma$ at non-positive integers line up with the zeros of $\sin(\pi z)$ at all integers — the residue calculus from contour integration is what produces the sine on the right-hand side.

Step 1 (Reduce to the Beta function). Using $B(x,y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ with $y=1-z$, $\Gamma(z)\Gamma(1-z) = B(z,1-z) = \int_0^1 t^{z-1}(1-t)^{-z}\,dt$. Substituting $u = t/(1-t)$ turns this into $\displaystyle\int_0^\infty \dfrac{u^{z-1}}{1+u}\,du$ for $0<\operatorname{Re}(z)<1$.

Step 2 (Set up a keyhole contour). To evaluate $I=\int_0^\infty \dfrac{u^{z-1}}{1+u}\,du$, use the branch of $w^{z-1}$ cut along the positive real axis and integrate $\dfrac{w^{z-1}}{1+w}$ around a keyhole contour: outward just above the cut, a large circle of radius $R$, back just below the cut, and a small circle of radius $\varepsilon$ around $0$.

Step 3 (Residue at the only pole). The only singularity of $\dfrac{w^{z-1}}{1+w}$ inside the contour is the simple pole at $w=-1$, where (using the branch with argument $\pi$ there) $\mathrm{Res} = (-1)^{z-1} = e^{i\pi(z-1)}$, so the residue theorem gives the keyhole integral as $2\pi i\, e^{i\pi(z-1)}$.

Step 4 (Compare the two sides of the cut and let $R\to\infty$, $\varepsilon\to 0$). The large and small circles vanish for $0<\operatorname{Re}(z)<1$, and the two straight segments differ by the phase picked up by $w^{z-1}$ crossing the cut, giving $I\big(1 - e^{2\pi i(z-1)}\big) = 2\pi i\, e^{i\pi(z-1)}$. Solving for $I$ and simplifying the exponentials using $\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$ gives $I = \dfrac{\pi}{\sin(\pi z)}$, so combining with Step 1, $\Gamma(z)\Gamma(1-z) = \dfrac{\pi}{\sin(\pi z)}$.