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
Let \[ \Gamma(s)=\int_0^\infty e^{-t}t^{s-1}\,dt, \qquad \Re s>0, \] where \(t^{s-1}=e^{(s-1)\log t}\) for the real logarithm on \((0,\infty)\).
Fix a compact set \(K\subset\{\Re s>0\}\). Choose numbers \(0<\sigma_0\le \Re s\le \sigma_1\) for \(s\in K\). On \((0,1]\), \[ |e^{-t}t^{s-1}\log t| \le t^{\sigma_0-1}|\log t|, \] and on \([1,\infty)\), \[ |e^{-t}t^{s-1}\log t| \le e^{-t}t^{\sigma_1-1}\log t. \] Both majorants are integrable. Hence differentiation under the integral sign is valid uniformly on compact subsets, so \(\Gamma\) is holomorphic on \(\Re s>0\) and \[ \Gamma'(s)=\int_0^\infty e^{-t}t^{s-1}\log t\,dt. \]
Now suppose \(0<\Re s<1\). Using the change of variables \(u=vt\) in the second factor, \[ \Gamma(1-s)=t\int_0^\infty e^{-vt}(vt)^{-s}\,dv. \] Therefore Tonelli/Fubini gives \[ \begin{aligned} \Gamma(s)\Gamma(1-s) &=\int_0^\infty\int_0^\infty e^{-t(1+v)}v^{-s}\,dt\,dv \\ &=\int_0^\infty\frac{v^{-s}}{1+v}\,dv. \end{aligned} \] Equivalently, replacing \(s\) by \(1-s\), \[ \Gamma(s)\Gamma(1-s) =\int_0^\infty \frac{x^{s-1}}{1+x}\,dx. \]
Evaluate this integral by a keyhole contour around the positive real axis for \[ F(z)=\frac{z^{s-1}}{1+z}, \qquad 0<\arg z<2\pi. \] The small and large circular arcs vanish because \(0<\Re s<1\). The upper bank contributes \(I(s)\) and the lower bank contributes \(-e^{2\pi i(s-1)}I(s)\), where \[ I(s)=\int_0^\infty\frac{x^{s-1}}{1+x}\,dx. \] The only enclosed pole is at \(z=-1\), with residue \[ (-1)^{s-1}=e^{i\pi(s-1)}. \] Thus \[ (1-e^{2\pi i(s-1)})I(s)=2\pi i e^{i\pi(s-1)}. \] Since \[ 1-e^{2\pi i(s-1)}=-2i e^{i\pi(s-1)}\sin(\pi(s-1)) =2i e^{i\pi(s-1)}\sin(\pi s), \] we obtain \[ I(s)=\frac{\pi}{\sin\pi s}. \] Hence, for \(0<\Re s<1\), \[ \boxed{\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}}. \] After the usual meromorphic continuation of \(\Gamma\), the same identity extends to every \(s\notin\mathbb Z\) by the identity theorem.