Calculate the following integral:
\[\int_0^\infty \frac{x^{1/2}}{4 + x^2} \, dx.\]
Solution
The integral equals \(\boxed{\pi/2}\).
1. The integral converges at both endpoints.
For \(0<x\leq1\), the positive integrand is at most \(x^{1/2}/4\), which is integrable near zero. For \(x\geq1\), it is at most \(x^{-3/2}\), which is integrable at infinity. Denote the resulting finite integral by \(I\).
2. An indented upper semicircle gives a residue identity.
Choose the logarithm branch with \(-\pi/2<\arg z<3\pi/2\), and define \(z^{1/2}=\exp(\operatorname{Log}z/2)\) [@SS03]. Its cut is the nonpositive imaginary axis; hence the branch is holomorphic on a neighborhood of the closed upper half-plane away from zero. For \(x>0\), it gives \(x^{1/2}=\sqrt x\) and \((-x)^{1/2}=i\sqrt x\).
For \(0<\varepsilon<1<2<R\), integrate \(F(z)=z^{1/2}/(z^2+4)\) along \([-R,-\varepsilon]\), a clockwise upper semicircle of radius \(\varepsilon\), \([\varepsilon,R]\), and the counterclockwise upper semicircle of radius \(R\). This positively oriented contour encloses only the pole \(2i\). Since \((2i)^{1/2}=1+i\), \[ \operatorname{Res}_{2i}F=\frac{1+i}{4i}. \] The substitution \(z=-x\) on the negative real segment shows that the two straight segments contribute \[ (1+i)\int_\varepsilon^R\frac{\sqrt x}{x^2+4}\,dx. \] The residue theorem equates this contribution plus the two arc integrals to \(\pi(1+i)/2\) [@SS03].
3. The arc integrals vanish, determining \(I\).
On the outer semicircle, \(|z^{1/2}|=\sqrt R\) and \(|z^2+4|\geq R^2-4\). Its integral therefore has modulus at most \(\pi R^{3/2}/(R^2-4)\), which tends to zero. On the inner semicircle the corresponding bound is \(\pi\varepsilon^{3/2}/(4-\varepsilon^2)\), also tending to zero. These estimates use the arc lengths \(\pi R\) and \(\pi\varepsilon\) and do not depend on orientation.
Letting \(R\to\infty\) and \(\varepsilon\downarrow0\) in step 2 gives \[ (1+i)I=\frac{\pi}{2}(1+i). \] The endpoint limits exist by step 1, and \(1+i\ne0\). Division yields \(I=\pi/2\).