Evaluate \(\textstyle \int _ { 0 } ^ { \infty } { \frac { \sqrt { x } } { 1 + x ^ { 2 } } } d x\)
Solution
On \[ \CC\sm[0,\infty), \] choose the branch \[ z^{1/2}=r^{1/2}e^{i\theta/2}, \qquad z=re^{i\theta}, \qquad 0<\theta<2\pi. \] Set \[ F(z)=\frac{z^{1/2}}{1+z^2}. \] For \(0<\varepsilon<1<R\), integrate \(F\) around the positively oriented keyhole contour with cut along the positive real axis.
1. The poles inside the contour are \(i\) and \(-i\), with\[ \Res(F;i) = \frac{1-i}{2\sqrt2}, \qquad \Res(F;-i) = -\frac{1+i}{2\sqrt2}. \]
The poles are the simple zeros of \(1+z^2\). On the chosen branch, \[ i^{1/2}=e^{i\pi/4} = \frac{1+i}{\sqrt2}, \] while \(-i\) has argument \(3\pi/2\), so \[ (-i)^{1/2} = e^{3\pi i/4} = \frac{-1+i}{\sqrt2}. \] Therefore \[ \begin{aligned} \Res(F;i) &= \frac{i^{1/2}}{2i} = \frac{1-i}{2\sqrt2}, \\ \Res(F;-i) &= \frac{(-i)^{1/2}}{-2i} = -\frac{1+i}{2\sqrt2}. \end{aligned} \]
2. The sum of the enclosed residues is\[ -\frac{i}{\sqrt2}. \]
By step 1, \[ \frac{1-i}{2\sqrt2} -\frac{1+i}{2\sqrt2} = -\frac{i}{\sqrt2}. \]
3. The two straight portions of the keyhole contour contribute\[ 2\int_{\varepsilon}^{R} \frac{\sqrt{x}}{1+x^2}\,dx. \]
On the upper bank, \[ z^{1/2}=\sqrt{x}, \] so the contribution is the truncated target integral.
On the lower bank, \[ z^{1/2} = \sqrt{x}e^{i\pi} = -\sqrt{x}, \] and the orientation is from \(R\) to \(\varepsilon\). Thus the lower-bank contribution is again \[ \int_{\varepsilon}^{R} \frac{\sqrt{x}}{1+x^2}\,dx. \]
4. The outer circular contribution tends to zero as \(R\to\infty\).
On \(\abs{z}=R\), \[ \abs{z^{1/2}}=R^{1/2}, \qquad \abs{1+z^2}\geq R^2-1. \] The outer arc has length at most \(2\pi R\), so \[ \abs{ \int_{\text{outer arc}}F(z)\,dz } \leq \frac{2\pi R^{3/2}}{R^2-1} \longrightarrow0. \]
5. The inner circular contribution tends to zero as\(\varepsilon\to0\).
On \(\abs{z}=\varepsilon\), \[ \abs{z^{1/2}}=\varepsilon^{1/2}, \qquad \abs{1+z^2}\geq1-\varepsilon^2. \] The inner arc has length at most \(2\pi\varepsilon\), so \[ \abs{ \int_{\text{inner arc}}F(z)\,dz } \leq \frac{2\pi\varepsilon^{3/2}}{1-\varepsilon^2} \longrightarrow0. \]
6. The requested value is\[ \boxed{ \int_0^{\infty} \frac{\sqrt{x}}{1+x^2}\,dx = \frac{\pi}{\sqrt2}. } \]
By the residue theorem and step 2, the keyhole contour integral is \[ 2\pi i \left(-\frac{i}{\sqrt2}\right) = \sqrt2\,\pi. \] By step 3, the two straight portions contribute twice the truncated target integral. Steps 4 and 5 show that both circular contributions vanish. Hence \[ 2\int_0^{\infty} \frac{\sqrt{x}}{1+x^2}\,dx = \sqrt2\,\pi, \] which yields the displayed value.
7. Q.E.D.
Step 6 is the requested evaluation.