Calculate \[ \int_0^\infty {\sqrt x \over 1 + x^2} \, dx .\]
Solution
Use the standard beta-integral identity \[ \int_0^\infty\frac{x^{a-1}}{1+x^n}\,dx =\frac\pi n\csc\frac{\pi a}{n}, \qquad 0<a<n. \] Here \(a=3/2\) and \(n=2\), because \(\sqrt x=x^{1/2}=x^{a-1}\). Therefore \[ \int_0^\infty\frac{\sqrt x}{1+x^2}\,dx =\frac\pi2\csc\frac{3\pi}{4} =\frac\pi2\cdot\sqrt2 =\boxed{\frac{\pi}{\sqrt2}}. \]
Equivalently, the beta-integral identity itself follows from a keyhole contour for \(z^{a-1}/(1+z^n)\).