Let \(a>0\) and evaluate \[\displaystyle{ \int_{0}^{\infty}\frac{x\sin x}{x^2+a^2} \, dx }.\]
Solution
Since the integrand \(\frac{x \sin(x)}{x^2 + a^2}\) is even in \(x\): \[ I = \int_0^\infty \frac{x \sin(x)}{x^2 + a^2}\, dx = \frac{1}{2} \int_{-\infty}^\infty \frac{x \sin(x)}{x^2 + a^2}\, dx = \frac{1}{2} \operatorname{Im} \left( \operatorname{p.v.} \int_{-\infty}^\infty \frac{x e^{ix}}{x^2 + a^2}\, dx \right). \]
Consider the complex contour integral of \(f(z) = \frac{z e^{iz}}{z^2 + a^2}\) over the standard semi-circular contour \(\Gamma_R = [-R, R] \cup C_R\), where \(C_R = \{R e^{i\theta} : \theta \in [0, \pi]\}\) in the upper half-plane with \(R > a\).
Residue Calculation: The only singularity of \(f(z)\) in the upper half-plane is a simple pole at \(z = ia\). The residue is: \[ \operatorname{Res}(f, ia) = \lim_{z \to ia} (z - ia) \frac{z e^{iz}}{(z - ia)(z + ia)} = \frac{ia e^{i(ia)}}{2ia} = \frac{e^{-a}}{2}. \] By the Residue Theorem: \[ \int_{\Gamma_R} f(z)\, dz = 2\pi i \operatorname{Res}(f, ia) = 2\pi i \left(\frac{e^{-a}}{2}\right) = \pi i e^{-a}. \]
Arc Estimate (Jordan’s Lemma): On \(C_R\), \(z = R e^{i\theta}\), \(|z| = R\), and \(|e^{iz}| = e^{-R \sin \theta}\). For \(R > a\), \(|z^2 + a^2| \geq R^2 - a^2\). \[ \left| \int_{C_R} \frac{z e^{iz}}{z^2 + a^2}\, dz \right| \leq \frac{R}{R^2 - a^2} \int_0^\pi e^{-R \sin \theta} R\, d\theta < \frac{\pi R}{R^2 - a^2} \to 0 \quad \text{as } R \to \infty. \]
Limit: Taking \(R \to \infty\): \[ \int_{-\infty}^\infty \frac{x e^{ix}}{x^2 + a^2}\, dx = \pi i e^{-a}. \] Taking the imaginary part and dividing by 2: \[ \int_0^\infty \frac{x \sin(x)}{x^2 + a^2}\, dx = \frac{1}{2} \operatorname{Im}(\pi i e^{-a}) = \frac{\pi}{2} e^{-a}. \]