To solve the integral $I = \int_{0}^{\infty} \frac{x \sin x}{x^2 + 1} \, dx$ using the residue theorem, we first extend the integral to the complex plane and then apply the residue theorem.

Steps:

1. Consider the Complex Function: Define the complex function $f(z) = \frac{z e^{iz}}{z^2 + 1}$, where $z$ is a complex variable.

2. Contour Integration: To apply the residue theorem, consider the contour $C$ that consists of the real axis from $-R$ to $R$ and a semicircle of radius $R$ in the upper half-plane, centered at the origin.

3. Residues of the Function: The poles of $f(z)$ are at $z = \pm i$. Only the pole at $z = i$ lies within the upper half-plane. The residue at this pole is given by: $\text{Res}(f, i) = \lim_{z \to i} (z - i) \frac{z e^{iz}}{z^2 + 1} = \lim_{z \to i} \frac{z e^{iz}}{(z - i)(z + i)} = \frac{i e^{-1}}{2i} = \frac{e^{-1}}{2}$

4. Integral Along the Real Axis: The contour integral over $C$ can be written as: $\int_C f(z) \, dz = \int_{-R}^{R} \frac{x e^{ix}}{x^2 + 1} \, dx + \int_{\text{semicircle}} f(z) \, dz$ As $R \to \infty$, the integral over the semicircle tends to zero.

   Therefore: $\int_{-\infty}^{\infty} \frac{x e^{ix}}{x^2 + 1} \, dx = 2\pi i \cdot \text{Res}(f, i) = 2\pi i \cdot \frac{e^{-1}}{2} = \pi i e^{-1}$

5. Taking the Real Part: The original integral $I$ is the real part of the complex integral: $I = \text{Re} \left( \int_{0}^{\infty} \frac{x \sin x}{x^2 + 1} \, dx \right) = \text{Re} \left( \frac{\pi i e^{-1}}{2} \right)$ Thus: $I = \frac{\pi e^{-1}}{2}$

The final answer is: $I = \frac{\pi e^{-1}}{2}$

Would you like more details on any part of this solution?

Related Questions:

1. How can contour integration be used in evaluating other improper integrals?
2. What are the key differences between using the residue theorem and other techniques like Laplace transforms?
3. What are the steps to find the residues of more complex functions?
4. How does Jordan's lemma assist in evaluating integrals over semicircular contours?
5. Can you provide examples of other integrals solvable using the residue theorem?

Tip:

When using the residue theorem for improper integrals, always ensure the contour in the complex plane is chosen to simplify calculations, often by reducing the problem to evaluating the integral over the real axis.