Evaluate \[ \int_{-\infty}^{\infty}\frac{\sin x}{x(1+x^2)}\,dx. \]
Solution
For \(0\le a\le1\), define \[ I(a)\coloneqq \int_{-\infty}^{\infty} \frac{\sin(ax)}{x(1+x^2)}\,dx, \] where the integrand is interpreted at \(x=0\) by its removable value \(a\).
1. The function \(I\) is differentiable and\[ I'(a) = \int_{-\infty}^{\infty} \frac{\cos(ax)}{1+x^2}\,dx. \]
For fixed \(x\), \[ \frac{\partial}{\partial a} \frac{\sin(ax)}{x(1+x^2)} = \frac{\cos(ax)}{1+x^2}. \] Moreover, \[ \left|\frac{\cos(ax)}{1+x^2}\right| \le \frac1{1+x^2}, \] and the right-hand side is integrable on \(\RR\). Differentiation under the integral sign follows from dominated convergence.
2. For every \(a\ge0\),\[ \int_{-\infty}^{\infty} \frac{\cos(ax)}{1+x^2}\,dx = \pi e^{-a}. \]
For \(a>0\), integrate \[ F_a(z)=\frac{e^{iaz}}{1+z^2} \] over the positively oriented upper semicircle of radius \(R>1\). The only enclosed pole is \(z=i\), with residue \[ \operatorname{Res}(F_a;i) = \frac{e^{ia i}}{2i} = \frac{e^{-a}}{2i}. \] On the semicircular arc, \[ \abs{e^{iaz}}=e^{-a\operatorname{Im}z}\le1, \] while \(\abs{1+z^2}\ge R^2-1\). Hence the arc integral has absolute value at most \[ \frac{\pi R}{R^2-1}\longrightarrow0. \] The residue theorem therefore gives \[ \int_{-\infty}^{\infty}\frac{e^{iax}}{1+x^2}\,dx = 2\pi i\frac{e^{-a}}{2i} = \pi e^{-a}. \] Its imaginary part is zero because \(\sin(ax)/(1+x^2)\) is odd, so the real part is the displayed cosine integral.
For \(a=0\), the identity is \[ \int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\pi. \]
3. For \(0\le a\le1\),\[ I'(a)=\pi e^{-a}. \]
Combine steps 1 and 2.
4. One has\[ I(0)=0. \]
The numerator \(\sin(0\cdot x)\) is identically zero.
5. The requested integral is\[ \boxed{\pi(1-e^{-1})}. \]
By steps <1>3–<1>4 and the fundamental theorem of calculus, \[ \begin{aligned} I(1) &=I(0)+\int_0^1 I'(a)\,da\\ &=\pi\int_0^1 e^{-a}\,da\\ &=\pi(1-e^{-1}). \end{aligned} \] By definition, \(I(1)\) is the integral in the problem.
6. Q.E.D.
Step 5 gives the requested value.