1. **Stating 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. **Understanding the integral:** The integral is over $\theta$ from $0$ to $\frac{\pi}{2}$ and over $x$ from $0$ to $\infty$. The integrand involves $x$, $\sin \theta$, $\cos \theta$, and a logarithm of a ratio. 3. **Key observations:** - The logarithm argument is a ratio of terms symmetric in $\sin^2 \theta$ and $\cos^2 \theta$. - The integral is complicated but suggests symmetry or substitution might simplify it. 4. **Interchange the order of integration:** We can write $$I = \int_0^{\frac{\pi}{2}} \int_0^\infty \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln \left( \frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta} \right) dx \, d\theta$$ 5. **Substitute $t = x^2$ to simplify the logarithm:** Let $t = x^2$, so $dx = \frac{dt}{2x}$. Then $$\frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} dx = \frac{x \sin \theta}{\sqrt{1 + t \sin^2 \theta}} \frac{dt}{2x} = \frac{\sin \theta}{2 \sqrt{1 + t \sin^2 \theta}} dt$$ The integral over $x$ becomes $$\int_0^\infty \frac{\sin \theta}{2 \sqrt{1 + t \sin^2 \theta}} \ln \left( \frac{1 + t \cos^2 \theta}{1 + t \sin^2 \theta} \right) dt$$ 6. **Rewrite the integral:** $$I = \int_0^{\frac{\pi}{2}} \frac{\sin \theta}{2} \int_0^\infty \frac{\ln \left( \frac{1 + t \cos^2 \theta}{1 + t \sin^2 \theta} \right)}{\sqrt{1 + t \sin^2 \theta}} dt \, d\theta$$ 7. **Symmetry argument:** Consider the substitution $\theta \to \frac{\pi}{2} - \theta$. Then $\sin \theta \leftrightarrow \cos \theta$ and the integrand changes sign because the logarithm argument inverts. This implies the integral over $\theta$ from $0$ to $\frac{\pi}{2}$ is zero due to antisymmetry. 8. **Conclusion:** The value of the integral is $$\boxed{0}$$ This is because the integrand is antisymmetric with respect to $\theta = \frac{\pi}{4}$, causing the integral over $\theta$ to cancel out.