Compute
\[ \lim_{N\to\infty}\int_{-N}^{N}\frac{x\sin(x)}{x^2+1}\,dx. \]
Solution
For \(N>1\), let \(C_N\) be the upper semicircle from \(N\) to \(-N\), oriented counterclockwise, and set \[ F(z)\coloneqq\frac{ze^{iz}}{z^2+1}. \]
1. The only pole of \(F\) in the upper half-plane is \(z=i\), and\[ \Res_{z=i}F(z)=\frac1{2e}. \]
Since \(z^2+1=(z-i)(z+i)\), the poles are \(i\) and \(-i\). Hence \[ \Res_{z=i}F(z) = \lim_{z\to i}\frac{ze^{iz}}{z+i} = \frac{ie^{-1}}{2i} = \frac1{2e}. \]
2. For every \(N>1\),\[ \int_{-N}^{N}\frac{xe^{ix}}{x^2+1}\,dx + \int_{C_N}F(z)\,dz = \frac{\pi i}{e}. \]
The contour consisting of the interval \([-N,N]\) followed by \(C_N\) encloses only the pole \(i\). By the residue theorem and step 1, \[ \int_{-N}^{N}\frac{xe^{ix}}{x^2+1}\,dx + \int_{C_N}F(z)\,dz = 2\pi i\Res_{z=i}F(z) = \frac{\pi i}{e}. \]
3. The semicircular contribution tends to zero:\[ \lim_{N\to\infty}\int_{C_N}F(z)\,dz=0. \]
Parametrize \(C_N\) by \(z=Ne^{i\theta}\) for \(0\le\theta\le\pi\). Since \(|e^{iz}|=e^{-N\sin\theta}\) and \(|z^2+1|\ge N^2-1\), we obtain \[ \left|\int_{C_N}F(z)\,dz\right| \le \frac{N^2}{N^2-1} \int_0^\pi e^{-N\sin\theta}\,d\theta. \] For \(0\le\theta\le\pi/2\), concavity of \(\sin\) gives \(\sin\theta\ge2\theta/\pi\). By symmetry, \[ \begin{aligned} \int_0^\pi e^{-N\sin\theta}\,d\theta &= 2\int_0^{\pi/2}e^{-N\sin\theta}\,d\theta\\ &\le 2\int_0^{\pi/2}e^{-2N\theta/\pi}\,d\theta\\ &= \frac{\pi}{N}(1-e^{-N}) \le \frac{\pi}{N}. \end{aligned} \] Therefore \[ \left|\int_{C_N}F(z)\,dz\right| \le \frac{\pi N}{N^2-1}\longrightarrow0. \]
4. Consequently,\[ \lim_{N\to\infty} \int_{-N}^{N}\frac{xe^{ix}}{x^2+1}\,dx = \frac{\pi i}{e}. \]
Let \(N\to\infty\) in the identity from step 2 and use step 3.
5. The requested limit is\[ \boxed{\frac{\pi}{e}}. \]
For real \(x\), \[ \frac{xe^{ix}}{x^2+1} = \frac{x\cos x}{x^2+1} + i\frac{x\sin x}{x^2+1}. \] Taking imaginary parts in step 4 therefore gives \[ \lim_{N\to\infty} \int_{-N}^{N}\frac{x\sin x}{x^2+1}\,dx = \frac{\pi}{e}. \]
6. Q.E.D.
Step 5 gives the required value.