Evaluate \(\textstyle \int _ { 0 } ^ { \infty } { \frac { \sqrt { x } } { ( x + 1 ) ^ { 2 } } } d x\)
Solution
On the slit plane \[ \CC\sm[0,\infty), \] 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 outer radius \(R\), inner radius \(\varepsilon\), and cut along the positive real axis.
1. The only pole inside the keyhole contour is the double pole at\(z=-1\), and\[ \Res(F;-1)=-\frac{i}{2}. \]
The chosen branch is holomorphic away from the positive real axis, and the only zero of the denominator in the slit plane is \(z=-1\). Since \[ (-1)^{1/2}=e^{i\pi/2}=i \] on this branch, the residue at the double pole is \[ \begin{aligned} \Res(F;-1) &= \left. \frac{d}{dz}z^{1/2} \right|_{z=-1}\\ &= \left. \frac1{2z^{1/2}} \right|_{z=-1}\\ &= \frac1{2i} = -\frac{i}{2}. \end{aligned} \]
2. The two straight portions of the keyhole contour contribute\[ 2\int_{\varepsilon}^{R} \frac{\sqrt{x}}{(1+x)^2}\,dx. \]
On the upper bank of the cut, the argument tends to \(0\), so \[ z^{1/2}=\sqrt{x}, \] and the contribution is \[ \int_{\varepsilon}^{R} \frac{\sqrt{x}}{(1+x)^2}\,dx. \]
On the lower bank, the argument tends to \(2\pi\), so \[ z^{1/2} = \sqrt{x}e^{i\pi} = -\sqrt{x}. \] This bank is traversed from \(R\) to \(\varepsilon\), hence its contribution is \[ \int_R^{\varepsilon} \frac{-\sqrt{x}}{(1+x)^2}\,dx = \int_{\varepsilon}^{R} \frac{\sqrt{x}}{(1+x)^2}\,dx. \] Adding the two contributions gives the claim.
3. The outer circular contribution tends to zero as \(R\to\infty\).
On \(\abs{z}=R\), \[ \abs{z^{1/2}}=R^{1/2} \] and \[ \abs{1+z}\geq R-1. \] The outer arc has length at most \(2\pi R\), so \[ \abs{ \int_{\text{outer arc}}F(z)\,dz } \leq \frac{2\pi R^{3/2}}{(R-1)^2} \longrightarrow0. \]
4. The inner circular contribution tends to zero as\(\varepsilon\to0\).
On \(\abs{z}=\varepsilon\), \[ \abs{z^{1/2}}=\varepsilon^{1/2} \] and \[ \abs{1+z}\geq1-\varepsilon. \] The inner arc has length at most \(2\pi\varepsilon\), so \[ \abs{ \int_{\text{inner arc}}F(z)\,dz } \leq \frac{2\pi\varepsilon^{3/2}}{(1-\varepsilon)^2} \longrightarrow0. \]
5. The target integral is\[ \boxed{ \int_0^{\infty} \frac{\sqrt{x}}{(1+x)^2}\,dx = \frac{\pi}{2}. } \]
By the residue theorem and step 1, the keyhole contour integral is \[ 2\pi i\Res(F;-1) = 2\pi i\left(-\frac{i}{2}\right) = \pi. \] By step 2, its two straight portions contribute twice the truncated target integral. Steps 3 and 4 show that the circular contributions vanish as \(R\to\infty\) and \(\varepsilon\to0\). Hence \[ 2\int_0^{\infty} \frac{\sqrt{x}}{(1+x)^2}\,dx = \pi, \] which gives the displayed value.
6. Q.E.D.
Step 5 is the requested evaluation.