Evaluate \(\textstyle \int _ { 0 } ^ { \infty } { \frac { \sin x } { x ( x ^ { 2 } + 1 ) } } d x\)
Solution
Set \[ F(z)=\frac{e^{iz}}{z(z^2+1)}. \] For \(0<\varepsilon<1<R\), use the upper-half-plane contour consisting of \([-R,-\varepsilon]\), a clockwise upper semicircle of radius \(\varepsilon\) about \(0\), \([\varepsilon,R]\), and the large upper semicircle of radius \(R\).
1. The only pole inside the indented contour is \(z=i\), and\[ \Res(F;i)=-\frac{e^{-1}}2. \]
The poles of \(F\) are \(0\) and \(\pm i\). The indentation excludes \(0\), and only \(i\) lies in the upper half-plane. Since the pole at \(i\) is simple, \[ \begin{aligned} \Res(F;i) &= \frac{e^{ii}}{i(i+i)}\\ &= \frac{e^{-1}}{-2}\\ &= -\frac{e^{-1}}2. \end{aligned} \]
2. The small indented semicircle contributes\[ -\pi i \]in the limit \(\varepsilon\to0\).
Near \(0\), \[ F(z) = \frac1z +O(1). \] Indeed, \(e^{iz}=1+O(z)\) and \((1+z^2)^{-1}=1+O(z^2)\). On the clockwise upper semicircle \[ z=\varepsilon e^{i\theta}, \qquad \pi\geq\theta\geq0, \] one has \[ \int\frac{dz}{z} = \int_\pi^0 i\,d\theta = -\pi i. \] The integral of the bounded remainder tends to zero with the arc length.
3. The integral over the large upper semicircle tends to zero as\(R\to\infty\).
For \(z\) in the upper half-plane, \[ \abs{e^{iz}}\leq1. \] On \(\abs{z}=R\), \[ \abs{z^2+1}\geq R^2-1. \] Hence \[ \abs{F(z)} \leq \frac1{R(R^2-1)}. \] The arc length is \(\pi R\), so the ML estimate gives \[ \abs{ \int_{\text{large arc}}F(z)\,dz } \leq \frac{\pi}{R^2-1} \longrightarrow0. \]
4. The symmetric indented real-axis integrals satisfy\[ \lim_{\substack{R\to\infty\\ \varepsilon\to0^+}} \left( \int_{-R}^{-\varepsilon}F(x)\,dx + \int_{\varepsilon}^{R}F(x)\,dx \right) = \pi i(1-e^{-1}). \]
By the residue theorem and step 1, the full contour integral is \[ 2\pi i\Res(F;i) = -\pi i e^{-1}. \] Thus the two real-axis segments equal this residue contribution minus the small and large arc integrals. By steps 2 and 3, their limit is \[ -\pi i e^{-1} -(-\pi i) = \pi i(1-e^{-1}). \]
5. The whole-line sine integral is\[ \int_{-\infty}^{\infty} \frac{\sin x}{x(x^2+1)}\,dx = \pi(1-e^{-1}). \]
The imaginary part of \(F(x)\) is \[ \frac{\sin x}{x(x^2+1)}. \] This function extends continuously across \(x=0\), since \[ \frac{\sin x}{x}\longrightarrow1. \] Taking imaginary parts in step 4 therefore gives the value \(\pi(1-e^{-1})\) for the ordinary improper whole-line integral.
6. The requested value is\[ \boxed{ \int_0^{\infty} \frac{\sin x}{x(x^2+1)}\,dx = \frac{\pi}{2}(1-e^{-1}). } \]
The integrand in step 5 is even, so its whole-line integral is twice the half-line integral.
7. Q.E.D.
Step 6 is the requested evaluation.