By considering the integral

$$\int_{C}\left(\frac{t}{1-t}\right)^{i} d t$$

where $C$ is a large circle centred on the origin, show that

$$B(1+i, 1-i)=\pi \operatorname{cosech} \pi$$

where

$$B(p, q)=\int_{0}^{1} t^{p-1}(1-t)^{q-1} d t, \operatorname{Re}(p)>0, \operatorname{Re}(q)>0$$

By using $B(p, q)=\frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}$, deduce that $\Gamma(i) \Gamma(-i)=\pi \operatorname{cosech} \pi$.