For \(s > 0\), the gamma function is defined by \(\Gamma(s) = \int_0^\infty e^{-t} t^{s-1}\,dt\).
Show that the gamma function is analytic in the half-plane \(\operatorname{Re}(s) > 0\), 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 \(\Gamma(1-s) = t\int_0^\infty e^{-vt}(vt)^{-s}\,dv\) for \(t > 0\).
“The formula in the previous question” refers to the question printed immediately before this one in the source (question 6): \[ \int_0^\infty \frac{x^{a-1}}{1+x^n}\,dx = \frac{\pi}{n\sin\frac{a\pi}{n}}, \qquad 0 < a < n,\ n \text{ a positive integer}. \] With \(n = 1\) and \(a = 1 - s\) for \(0 < s < 1\) it reads \(\int_0^\infty \frac{v^{-s}}{1+v}\,dv = \frac{\pi}{\sin \pi s}\), which the hint turns into the reflection formula.