1. **State the problem:** Evaluate the improper integral $$\int_1^\infty \frac{dx}{x\sqrt{x}}$$ and determine if it converges. 2. **Rewrite the integrand:** Note that $$\frac{1}{x\sqrt{x}} = \frac{1}{x^{3/2}}$$ because $$\sqrt{x} = x^{1/2}$$ and multiplying exponents gives $$x \cdot x^{1/2} = x^{3/2}$$. 3. **Set up the integral:** The integral becomes $$\int_1^\infty x^{-3/2} dx$$. 4. **Recall the formula for integrals of power functions:** For $$\int x^p dx = \frac{x^{p+1}}{p+1} + C$$, provided $$p \neq -1$$. 5. **Apply the formula:** Here, $$p = -\frac{3}{2}$$, so $$p+1 = -\frac{3}{2} + 1 = -\frac{1}{2}$$. 6. **Evaluate the definite integral:** $$\int_1^\infty x^{-3/2} dx = \lim_{t \to \infty} \int_1^t x^{-3/2} dx = \lim_{t \to \infty} \left[ \frac{x^{-1/2}}{-1/2} \right]_1^t = \lim_{t \to \infty} \left[ -2 x^{-1/2} \right]_1^t$$ 7. **Calculate the limit:** $$= \lim_{t \to \infty} \left( -2 t^{-1/2} + 2 \cdot 1^{-1/2} \right) = \lim_{t \to \infty} \left( -\frac{2}{\sqrt{t}} + 2 \right) = 0 + 2 = 2$$ 8. **Conclusion:** The integral converges and its value is $$2$$. This means the area under the curve $$y = \frac{1}{x\sqrt{x}}$$ from $$x=1$$ to infinity is finite and equals 2.