Let \(a>0\). Evaluate
\[ \int _ { 0 } ^ { \infty } { \frac { x \sin x } { x ^ { 2 } + a ^ { 2 } } } d x . \]
Some retained UGA/Tie appearances omit a condition on the parameter \(a\). The Azoff appearance explicitly states \(a>0\), and a later UGA appearance states the corresponding whole-line identity for all \(a>0\). The problem above records that hypothesis explicitly.
Solution
Set \[ F(z)=\frac{ze^{iz}}{z^2+a^2}. \] For \(R>a\), let \(C_R\) be the positively oriented contour consisting of \([-R,R]\) and the upper semicircle \[ \Gamma_R=\{Re^{i\theta}:0\leq\theta\leq\pi\}. \]
1. The only pole of \(F\) inside \(C_R\) is \(z=ia\), and\[ \Res(F;ia)=\frac{e^{-a}}2. \]
Since \[ z^2+a^2=(z-ia)(z+ia), \] the poles are \(\pm ia\). Because \(a>0\), only \(ia\) lies in the upper half-plane. Its residue is \[ \begin{aligned} \Res(F;ia) &= \lim_{z\to ia} \frac{ze^{iz}}{z+ia}\\ &= \frac{ia\,e^{-a}}{2ia}\\ &= \frac{e^{-a}}2. \end{aligned} \]
2. The integral over the semicircular arc tends to zero:\[ \int_{\Gamma_R}F(z)\,dz\longrightarrow0 \]as \(R\to\infty\).
On \(z=Re^{i\theta}\), \[ \abs{e^{iz}}=e^{-R\sin\theta}. \] Also, for \(R>a\), \[ \abs{z^2+a^2} \geq R^2-a^2, \] so \[ \abs{F(Re^{i\theta})} \leq \frac{R}{R^2-a^2}e^{-R\sin\theta}. \] Parametrizing the arc gives \[ \abs{ \int_{\Gamma_R}F(z)\,dz } \leq \frac{R^2}{R^2-a^2} \int_0^\pi e^{-R\sin\theta}\,d\theta. \]
For \(0\leq\theta\leq\pi/2\), \[ \sin\theta\geq\frac{2\theta}{\pi}. \] By symmetry, \[ \begin{aligned} \int_0^\pi e^{-R\sin\theta}\,d\theta &= 2\int_0^{\pi/2}e^{-R\sin\theta}\,d\theta\\ &\leq 2\int_0^{\pi/2}e^{-2R\theta/\pi}\,d\theta\\ &= \frac{\pi}{R}(1-e^{-R})\\ &\leq \frac{\pi}{R}. \end{aligned} \] Therefore \[ \abs{ \int_{\Gamma_R}F(z)\,dz } \leq \frac{\pi R}{R^2-a^2} \longrightarrow0. \]
3. One has\[ \lim_{R\to\infty} \int_{-R}^{R} \frac{x e^{ix}}{x^2+a^2}\,dx = \pi i e^{-a}. \]
By the residue theorem and step 1, \[ \int_{C_R}F(z)\,dz = 2\pi i\Res(F;ia) = \pi i e^{-a}. \] Thus \[ \int_{-R}^{R}\frac{x e^{ix}}{x^2+a^2}\,dx + \int_{\Gamma_R}F(z)\,dz = \pi i e^{-a}. \] Letting \(R\to\infty\) and using step 2 gives the displayed limit.
4. The symmetric sine integrals satisfy\[ \lim_{R\to\infty} \int_{-R}^{R} \frac{x\sin x}{x^2+a^2}\,dx = \pi e^{-a}. \]
Taking imaginary parts in step 3 gives \[ \lim_{R\to\infty} \int_{-R}^{R} \frac{x\sin x}{x^2+a^2}\,dx = \pi e^{-a}. \]
5. The requested value is\[ \boxed{ \int_0^{\infty} \frac{x\sin x}{x^2+a^2}\,dx = \frac{\pi}{2}e^{-a}. } \]
The function \[ x\longmapsto\frac{x\sin x}{x^2+a^2} \] is even. Hence for every \(R>0\), \[ \int_{-R}^{R} \frac{x\sin x}{x^2+a^2}\,dx = 2\int_0^R \frac{x\sin x}{x^2+a^2}\,dx. \] Step 4 therefore proves that the half-line improper integral converges and has the displayed value.
6. Q.E.D.
Step 5 is the requested value.