November 25th
Today I learned the reflection formula for the gamma function. Quickly, recall that, provided $\op{Re}(s) \gt -1,$\[\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}\,dt.\]
We begin by claiming that\[\Gamma(s)=\lim_{n\to\infty}\frac{n^s}s\prod_{k=1}^n\frac k{s+k}.\]The main idea to prove this statement is to note that\[\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}\,dt=\lim_{n\to\infty}\int_0^\infty t^{s-1}\left(1-\frac tn\right)^n\,dt\]by definition of $\exp.$ However, we can be even more greedy with our limit by exchanging the upper bound of the integral to\[\Gamma(s)=\lim_{n\to\infty}\int_0^nt^{s-1}\left(1-\frac tn\right)^n\,dt\]I guess there is some uniform convergence complications we need to exchange the limit and the integral, but I don't care enough. We can evaluate this integral for $n\in\ZZ$ (which is good enough to approach $\exp$) by using repeated integration by parts. Indeed,\[\int_0^nt^{s-1}\left(1-\frac tn\right)^n\,dt=\frac{t^s}s\left(1-\frac tn\right)^n\bigg|_0^n-\int_0^n\frac{t^s}s\cdot n\left(1-\frac tn\right)^{n-1}\cdot-\frac1n\,dt.\]The main term vanishes, leaving us with\[\int_0^nt^{s-1}\left(1-\frac tn\right)^n\,dt=\frac n{ns}\int_0^nt^s\left(1-\frac tn\right)^{n-1}\,dt.\]Repeating this to a total of $n$ times gives us\[\int_0^nt^{s-1}\left(1-\frac tn\right)^n\,dt=\frac n{ns}\cdot\frac{n-1}{n(s+1)}\cdot\frac{n-2}{n(s+2)}\cdots\frac1{n(s+n-1)}\int_0^nt^{s+n-1}\,dt.\]The integral evaluates to $\frac1{s+n}n^{s+n},$ which removes the $n^n$ in the denominator and leaves us with\[\int_0^nt^{s-1}\left(1-\frac tn\right)^n\,dt=\frac ns\cdot\frac{n-1}{s+1}\cdot\frac{n-2}{s+2}\cdots\frac1{s+n-1}\cdot\frac{n^s}{s+n}=\frac{n^s}s\prod_{k=1}^n\frac k{s+k}.\]From this we get that\[\Gamma(s)=\lim_{n\to\infty}\frac{n^s}s\prod_{k=1}^n\frac k{s+k}.\]As an aside, this provides an analytic continuation of $\Gamma$ to all complex numbers except nonpositive integers; some care has to be taken to make this rigorous, but I think it works out.
We are now ready to show the reflection formula. The main idea is that the above formula has some symmetry that we can use. Explicitly, we see\[\Gamma(-s)=\lim_{n\to\infty}\frac{n^{-s}}{-s}\prod_{k=1}^n\frac k{-s+k}.\]Multiplying this by $\Gamma(s),$ now, does some lovely things, most notably cancelling the pesky $n^s$ term, giving\[\Gamma(s)\Gamma(-s)=\lim_{n\to\infty}\frac1{-s^2}\prod_{k=1}^n\frac{k^2}{(s+k)(-s+k)}.\]We may now remove the $-\frac1{s^2}$ term, no longer dependent on $n.$ The inner product comes out to be\[\prod_{k=1}^\infty\frac{k^2}{k^2-s^2}=\left(\prod_{k=1}^\infty\left(1-\frac{s^2}{k^2}\right)\right)^{-1}.\]Quickly, the product on the right-hand side has roots at all nonzero integers, so Euler would say the Weierstrass product makes it $\frac{\sin(\pi s)}{\pi s}.$ Plugging back in, we see\[\Gamma(s)\Gamma(-s)=-\frac1{s^2}\cdot\frac{\pi s}{\sin(\pi s)}.\]This rearranges to\[\Gamma(s)\Gamma(1-s)=\Gamma(s)\cdot-s\Gamma(-s)=\frac\pi{\sin(\pi s)},\]which is our reflection formula.
Just for fun, we do a quick application of this: we compute $\left(\frac12\right)!.$ Plugging in $s=\frac12$ to the reflection formula implies\[\Gamma\left(\frac12\right)^2=\frac\pi{\sin\left(\frac\pi2\right)}=\pi.\]This implies $\Gamma\left(\frac12\right)=\pm\sqrt\pi.$ As for which sign it is, note that the integral representation of $\Gamma$ given at the beginning works at $s=\frac12,$ and the integral is always positive. It follows\[\left(\frac12\right)!=\Gamma\left(\frac32\right)=\frac12\Gamma\left(\frac12\right)=\frac{\sqrt\pi}2,\]which is what we wanted.