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. **Analyze the integrand:** The integrand is $$f(x,\theta) = \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln \left( \frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta} \right)$$ 3. **Symmetry insight:** Note that swapping $\sin \theta$ and $\cos \theta$ in the logarithm argument inverts the fraction inside the logarithm: $$\ln \left( \frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta} \right) = - \ln \left( \frac{1 + x^2 \sin^2 \theta}{1 + x^2 \cos^2 \theta} \right)$$ 4. **Consider the substitution $\theta \to \frac{\pi}{2} - \theta$:** Under this substitution, - $\sin \theta \to \cos \theta$ - $\cos \theta \to \sin \theta$ The integral over $\theta$ from $0$ to $\frac{\pi}{2}$ can be split and analyzed using this symmetry. 5. **Define the inner integral:** $$I(x) = \int_0^{\frac{\pi}{2}} \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln \left( \frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta} \right) d\theta$$ 6. **Apply substitution $\phi = \frac{\pi}{2} - \theta$ in $I(x)$:** $$I(x) = \int_0^{\frac{\pi}{2}} \frac{x \cos \phi}{\sqrt{1 + x^2 \cos^2 \phi}} \ln \left( \frac{1 + x^2 \sin^2 \phi}{1 + x^2 \cos^2 \phi} \right) d\phi$$ 7. **Add the two expressions for $I(x)$:** $$2I(x) = \int_0^{\frac{\pi}{2}} x \left( \frac{\sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} - \frac{\cos \theta}{\sqrt{1 + x^2 \cos^2 \theta}} \right) \ln \left( \frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta} \right) d\theta$$ 8. **Note the integrand is antisymmetric under $\theta \to \frac{\pi}{2} - \theta$:** This implies $$I(x) = -I(x) \implies I(x) = 0$$ 9. **Therefore, the entire double integral is zero:** $$\int_0^\infty I(x) dx = \int_0^\infty 0 \, dx = 0$$ **Final answer:** $$\boxed{0}$$