Let's analyze the **convergence** of the improper integral: \[ I = \int_0^\infty \frac{e^{-x}}{\sqrt{x}}\, dx \] We are told to **split it at \( x=1 \)**: \[ I = \int_0^1 \frac{e^{-x}}{\sqrt{x}}\, dx + \int_1^\infty \frac{e^{-x}}{\sqrt{x}}\, dx \] We need to check convergence at both \( x=0 \) and as \( x \to \infty \). --- ## 1. Near \( x = 0 \): As \( x \to 0 \), \( e^{-x} \to 1 \). So: \[ \frac{e^{-x}}{\sqrt{x}} \sim \frac{1}{\sqrt{x}} \] Recall: \[ \int_0^1 x^{-p}\,dx \] converges iff \( p < 1 \) (i.e., the power is less than 1). Here, \( p = 1/2 < 1 \), so \[ \int_0^1 \frac{1}{\sqrt{x}}\,dx = \left[ 2\sqrt{x} \right]_0^1 = 2 \] is **convergent**. Since \( \frac{e^{-x}}{\sqrt{x}} < \frac{1}{\sqrt{x}} \) for \( x > 0 \), by the **comparison test**, \( \int_0^1 \frac{e^{-x}}{\sqrt{x}}\,dx \) also converges. --- ## 2. As \( x \to \infty \): As \( x \to \infty \), \( e^{-x} \) decays **much faster** than any power of \( x \), and \( \sqrt{x} \) grows slowly. Let’s check the comparison with \( e^{-x} \): \[ \frac{e^{-x}}{\sqrt{x}} < e^{-x} \] The integral \[ \int_1^\infty e^{-x}\,dx = \left[ -e^{-x} \right]_1^\infty = e^{-1} \] is **convergent**. So, by the **comparison test**, \( \int_1^\infty \frac{e^{-x}}{\sqrt{x}}\,dx \) also **converges**. --- ## **Conclusion** Both integrals converge, so \[ \boxed{ \int_0^\infty \frac{e^{-x}}{\sqrt{x}}\, dx \quad \text{is convergent.} } \] **Summary of reasoning:** - Near \( x = 0 \): behaves like \( x^{-1/2} \), which is integrable at 0. - As \( x \to \infty \): decays faster than \( e^{-x} \), which is integrable at infinity. - Therefore, the original improper integral converges by the comparison test.