For \(s>0\), the gamma function is defined by \(\displaystyle{\Gamma(s)=\int_0^{\infty} e^{-t}t^{s-1} dt}\).
Show that the gamma function is analytic in the half-plane \(\Re (s)>0\), and is still given there by the integral formula above.
Apply the formula in the previous question to show that \[\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin \pi s}.\]
Hint: You may need \(\displaystyle{\Gamma(1-s)=t \int_0^{\infty}e^{-vt}(vt)^{-s} dv}\) for \(t>0\).
Solution
For \(\Re s>0\) define \[ \Gamma(s)=\int_0^\infty e^{-t}t^{s-1}\,dt. \] On every compact subset of the half-plane \(\Re s>0\), both the integrand and its \(s\)-derivative \[ e^{-t}t^{s-1}\log t \] are dominated by an integrable function: near \(0\) use \(t^{\sigma-1}(1+|\log t|)\) with \(\sigma>0\), and near infinity use \(e^{-t}t^M(1+\log t)\). Hence differentiation under the integral sign is valid, so \(\Gamma\) is holomorphic on \(\Re s>0\).
For \(0<\Re s<1\), the change of variables \(u=vt\) in the second factor gives \[ \Gamma(1-s) =t\int_0^\infty e^{-vt}(vt)^{-s}\,dv. \] Multiplying by the integral for \(\Gamma(s)\) and applying Fubini, \[ \Gamma(s)\Gamma(1-s) =\int_0^\infty\int_0^\infty e^{-t(1+v)}v^{-s}\,dt\,dv =\int_0^\infty\frac{v^{-s}}{1+v}\,dv. \] Replacing \(s\) by \(1-s\) yields \[ \Gamma(s)\Gamma(1-s) =\int_0^\infty\frac{x^{s-1}}{1+x}\,dx. \]
Evaluate the last integral by a keyhole contour for \(z^{s-1}/(1+z)\) with branch \(0<\arg z<2\pi\). The two sides of the cut differ by the factor \(e^{2\pi i(s-1)}\), and the only enclosed pole is at \(z=-1\). The small and large circular arcs vanish because \(0<\Re s<1\). Thus \[ (1-e^{2\pi i(s-1)}) \int_0^\infty\frac{x^{s-1}}{1+x}\,dx =2\pi i e^{i\pi(s-1)}. \] Since \[ 1-e^{2\pi i(s-1)} =2i e^{i\pi(s-1)}\sin(\pi s), \] we obtain \[ \boxed{\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}}, \qquad 0<\Re s<1. \] After the usual meromorphic continuation of \(\Gamma\), the identity extends to all \(s\notin\mathbb Z\).