By making the transformation \(u=1 /\left(1+x^{2}ight)\) and converting to a beta-type integral on \((0,1)\), show that

\[
2 \int_{0}^{\infty} \frac{x^{d-1} d x}{\left(1+x^{2}ight)^{v+d / 2}}=B(v, d / 2)=\frac{\Gamma(v) \Gamma(d / 2)}{\Gamma(v+d / 2)}
\]

where \(B(\cdot, \cdot)\) is the beta function for strictly positive arguments.