(a) The Beta function is defined by

$$\mathrm{B}(p, q)=\int_{0}^{1} x^{p-1}(1-x)^{q-1} d x$$

Show that

$$\mathrm{B}(p, q)=\int_{1}^{\infty} x^{-p-q}(x-1)^{q-1} d x$$

(b) The function $J(p, q)$ is defined by

$$J(p, q)=\int_{\gamma} t^{p-1}(1-t)^{q-1} d t$$

where the integrand has a branch cut along the positive real axis. Just above the cut, $\arg t=0$. For $t>1$ just above the cut, arg $(1-t)=-\pi$. The contour $\gamma$ runs from $t=\infty e^{2 \pi i}$, round the origin in the negative sense, to $t=\infty$ (i.e. the contour is a reflection of the usual Hankel contour). What restriction must be placed on $p$ and $q$ for the integral to converge?

By evaluating $J(p, q)$ in two ways, show that

$$\left(1-e^{2 \pi i p}\right) \mathrm{B}(p, q)+\left(e^{-\pi i(q-1)}-e^{\pi i(2 p+q-1)}\right) \mathrm{B}(1-p-q, q)=0,$$

where $p$ and $q$ are any non-integer complex numbers.

Using the identity

$$B(p, q)=\frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)}$$

deduce that

$$\Gamma(p) \Gamma(1-p) \sin (\pi p)=\Gamma(p+q) \Gamma(1-p-q) \sin [\pi(1-p-q)]$$

and hence that

$$\pi=\Gamma(q) \Gamma(1-q) \sin [\pi(1-q)]$$