Evaluate \[ \int_0^\infty\frac{\sqrt{x}}{1+x^2}\,dx. \]
Solution
Let \[ F(z)\coloneqq\frac{z^{1/2}}{1+z^2}, \] where \[ z^{1/2} = \exp\left(\frac12(\log\abs z+i\arg z)\right), \qquad 0<\arg z<2\pi. \] Thus the branch cut is the positive real axis.
1. On the upper bank of the positive real axis,\[ F(x)=\frac{\sqrt{x}}{1+x^2}, \]while on the lower bank,\[ F(x)=-\frac{\sqrt{x}}{1+x^2}. \]
On the upper bank, \(\arg x=0\), so \[ x^{1/2}=\sqrt x. \] On the lower bank, the branch has \(\arg x=2\pi\), so \[ x^{1/2} = \sqrt x\,e^{i\pi} = -\sqrt x. \]
2. On a keyhole contour with outer radius \(R\) and inner radius\(\varepsilon\), the two circular-arc integrals tend to zero as\[ R\to\infty \qquad\text{and}\qquad \varepsilon\to0. \]
On \(\abs z=R\) with \(R>2\), \[ \abs{F(z)} \leq \frac{R^{1/2}}{R^2-1}, \] while the arc length is at most \(2\pi R\). Hence the outer-arc integral has absolute value \[ O(R^{-1/2}) \longrightarrow0. \]
On \(\abs z=\varepsilon<1/2\), \[ \abs{F(z)} \leq \frac{\varepsilon^{1/2}}{1-\varepsilon^2}, \] and the arc length is at most \(2\pi\varepsilon\). Hence the inner-arc integral is \[ O(\varepsilon^{3/2}) \longrightarrow0. \]
3. The poles inside the keyhole contour are \(i\) and \(-i\), with\[ \Res_{z=i}F = \frac{e^{i\pi/4}}{2i} \]and\[ \Res_{z=-i}F = -\frac{e^{3i\pi/4}}{2i}. \]
Both poles are simple. At \(i\), the chosen branch has \[ i^{1/2}=e^{i\pi/4}, \] so \[ \Res_{z=i}F = \frac{i^{1/2}}{2i} = \frac{e^{i\pi/4}}{2i}. \] At \(-i\), the chosen argument is \(3\pi/2\), hence \[ (-i)^{1/2}=e^{3i\pi/4}. \] Since the derivative of \(1+z^2\) at \(-i\) is \(-2i\), \[ \Res_{z=-i}F = \frac{e^{3i\pi/4}}{-2i}. \]
4. The sum of the residues is\[ \Res_{z=i}F+\Res_{z=-i}F = -\frac{i}{\sqrt2}. \]
Using \[ e^{i\pi/4} = \frac{1+i}{\sqrt2} \qquad\text{and}\qquad e^{3i\pi/4} = \frac{-1+i}{\sqrt2}, \] one obtains \[ \begin{aligned} \Res_{z=i}F+\Res_{z=-i}F &= \frac{ e^{i\pi/4}-e^{3i\pi/4} }{2i}\\ &= \frac{\sqrt2}{2i}\\ &= -\frac{i}{\sqrt2}. \end{aligned} \]
5. If\[ I\coloneqq \int_0^\infty\frac{\sqrt{x}}{1+x^2}\,dx, \]then\[ 2I=\sqrt2\,\pi. \]
The upper bank is traversed from \(\varepsilon\) to \(R\) and contributes \[ \int_\varepsilon^R \frac{\sqrt{x}}{1+x^2}\,dx. \] By step 1, the lower-bank integrand has the opposite sign, but that bank is traversed from \(R\) back to \(\varepsilon\). It therefore contributes the same quantity. By step 2, the circular arcs vanish in the limit.
The residue theorem and step 4 consequently give \[ 2I = 2\pi i \left( -\frac{i}{\sqrt2} \right) = \sqrt2\,\pi. \]
6. Therefore\[ \boxed{ \int_0^\infty\frac{\sqrt{x}}{1+x^2}\,dx = \frac{\pi}{\sqrt2} }. \]
Divide the identity in step 5 by \(2\).
7. Q.E.D.
Step 6 gives the requested value.