1. **State the problem:** Evaluate the double integral $$\int_0^\infty \int_0^{\frac{\pi}{2}} \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln \left(\frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta}\right) d\theta \, dx$$ 2. **Understand the integrand and variables:** The integrand involves $x$ and $\theta$ with trigonometric functions and logarithms. The limits are $x$ from $0$ to $\infty$ and $\theta$ from $0$ to $\frac{\pi}{2}$. 3. **Consider the inner integral first:** Define $$I(x) = \int_0^{\frac{\pi}{2}} \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln \left(\frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta}\right) d\theta$$ 4. **Symmetry and substitution:** Note that the logarithm argument can be rewritten as $$\ln \left(\frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta}\right) = \ln(1 + x^2 \cos^2 \theta) - \ln(1 + x^2 \sin^2 \theta)$$ Split the integral accordingly: $$I(x) = \int_0^{\frac{\pi}{2}} \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln(1 + x^2 \cos^2 \theta) d\theta - \int_0^{\frac{\pi}{2}} \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln(1 + x^2 \sin^2 \theta) d\theta$$ 5. **Use substitution $\theta = \frac{\pi}{2} - \phi$ in the first integral:** Then $\sin \theta = \cos \phi$, $\cos \theta = \sin \phi$, and the limits remain $0$ to $\frac{\pi}{2}$. The first integral becomes $$\int_0^{\frac{\pi}{2}} \frac{x \cos \phi}{\sqrt{1 + x^2 \cos^2 \phi}} \ln(1 + x^2 \sin^2 \phi) d\phi$$ 6. **Rewrite $I(x)$ using this substitution:** $$I(x) = \int_0^{\frac{\pi}{2}} \frac{x \cos \theta}{\sqrt{1 + x^2 \cos^2 \theta}} \ln(1 + x^2 \sin^2 \theta) d\theta - \int_0^{\frac{\pi}{2}} \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln(1 + x^2 \sin^2 \theta) d\theta$$ 7. **Combine the integrals:** $$I(x) = \int_0^{\frac{\pi}{2}} \ln(1 + x^2 \sin^2 \theta) \left( \frac{x \cos \theta}{\sqrt{1 + x^2 \cos^2 \theta}} - \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \right) d\theta$$ 8. **Observe the integrand inside the parentheses:** This expression is antisymmetric with respect to swapping $\sin \theta$ and $\cos \theta$ over $[0, \frac{\pi}{2}]$. 9. **Integral over $\theta$ from $0$ to $\frac{\pi}{2}$ of an antisymmetric function is zero:** Because the integrand changes sign when $\theta$ is replaced by $\frac{\pi}{2} - \theta$, the integral cancels out. 10. **Therefore, for all $x$, $I(x) = 0$.** 11. **Finally, the outer integral:** $$\int_0^\infty I(x) dx = \int_0^\infty 0 \, dx = 0$$ **Answer:** $$\boxed{0}$$