For real \(s>0\), define \[ \Gamma(s)=\int_0^\infty e^{-t}t^{s-1}\,dt. \] Show that the same integral defines an analytic function in the half-plane \(\Re(s)>0\).
Using \[ \int_0^\infty {x^{a-1}\over1+x^n}\,dx ={\pi\over n\sin(a\pi/n)}, \qquad 0<a<n, \] show that, for \(0<\Re(s)<1\), \[ \Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin \pi s} .\]
Hint: You may need \(\displaystyle{\Gamma(1-s)=t \int_0^{\infty}e^{-vt}(vt)^{-s} dv}\) for \(t>0\).
Solution
Fix a compact set \(K\subset\{s:\Re s>0\}\) and put \[ \sigma_0=\min_{s\in K}\Re s>0, \qquad \sigma_1=\max_{s\in K}\Re s. \] For \(0<t\le1\) and \(s\in K\), \[ |e^{-t}t^{s-1}|\le t^{\sigma_0-1}, \] while for \(t\ge1\), \[ |e^{-t}t^{s-1}|\le e^{-t}t^{\sigma_1-1}. \] Both majorants are integrable. The same argument, with an additional factor \(|\log t|^m\), applies after any number \(m\) of differentiations in \(s\). Therefore differentiation under the integral sign is valid locally uniformly, and \[ \Gamma^{(m)}(s)=\int_0^\infty e^{-t}t^{s-1}(\log t)^m\,dt. \] In particular, \(\Gamma\) is holomorphic on \(\Re s>0\).
Now assume \(0<\Re s<1\). Using the substitution \(u=vt\) in the second gamma factor, \[ \Gamma(1-s) =t\int_0^\infty e^{-vt}(vt)^{-s}\,dv. \] Hence, by absolute convergence and Fubini, \[ \begin{aligned} \Gamma(s)\Gamma(1-s) &=\int_0^\infty\int_0^\infty e^{-t}t^{s-1}\,t e^{-vt}(vt)^{-s}\,dv\,dt\\ &=\int_0^\infty v^{-s} \left(\int_0^\infty e^{-(1+v)t}\,dt\right)dv\\ &=\int_0^\infty {v^{-s}\over1+v}\,dv. \end{aligned} \] For \[ M(a)=\int_0^\infty {v^{a-1}\over1+v}\,dv, \qquad 0<\Re a<1, \] the same compact-majorant argument used above shows that \(M\) is holomorphic on the strip \(0<\Re a<1\). The function \(\pi/\sin(\pi a)\) is holomorphic there as well. The formula in the statement with \(n=1\) gives \[ M(a)={\pi\over\sin(\pi a)} \] for every real \(a\in(0,1)\), so the identity theorem extends this equality to the whole strip. Taking \(a=1-s\) therefore gives \[ \int_0^\infty {v^{-s}\over1+v}\,dv ={\pi\over\sin(\pi(1-s))} ={\pi\over\sin(\pi s)}. \] Thus \[ \Gamma(s)\Gamma(1-s)={\pi\over\sin\pi s}, \qquad 0<\Re s<1. \]