Compute \[ I(b)=\int_0^{\pi/2}(\tan t)^{ib}\,dt \] for \(b\in\mathbb R\).
Hint: a simple substitution reduces the integral to one treated in the homework and lectures.
Solution
For \(0<t<\pi/2\), \(\tan t>0\), so we use the real logarithm in \[ (\tan t)^{ib}=e^{ib\log(\tan t)}. \] Set \(x=\tan t\). Then \(dt=dx/(1+x^2)\) and \[ I(b)=\int_0^\infty \frac{x^{ib}}{1+x^2}\,dx. \] Now let \(u=x^2\). This gives \[ I(b)=\frac12\int_0^\infty \frac{u^{a-1}}{1+u}\,du, \qquad a=\frac{1+ib}{2}. \] Since \(\Re a=1/2\), Euler’s beta integral gives \[ \int_0^\infty \frac{u^{a-1}}{1+u}\,du =B(a,1-a)=\Gamma(a)\Gamma(1-a) =\frac{\pi}{\sin(\pi a)}. \] But \[ \sin\!\left(\frac{\pi}{2}+i\frac{\pi b}{2}\right) =\cosh\!\left(\frac{\pi b}{2}\right). \] Therefore \[ \boxed{ I(b)=\frac{\pi}{2\cosh(\pi b/2)} }. \]