The given improper integral is:

$\int_{1}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx.$

To solve this, let's follow these steps:

Step 1: Substitution

Use the substitution $u = \sqrt{x}$, which implies:

$x = u^2 \quad \text{and} \quad dx = 2u \, du.$

Changing the limits:

• When $x = 1$, $u = \sqrt{1} = 1$.
• When $x = \infty$, $u = \infty$.

Substituting into the integral:

$\int_{1}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx = \int_{1}^{\infty} \frac{e^{-u}}{u} \cdot 2u \, du = 2\int_{1}^{\infty} e^{-u} \, du.$

Step 2: Solve the Integral

Now, solve the simplified integral:

$2\int_{1}^{\infty} e^{-u} \, du.$

Evaluating this:

$\int_{1}^{\infty} e^{-u} \, du = \left[-e^{-u}\right]_{1}^{\infty}.$

Calculating the limits:

$\left[-e^{-u}\right]_{1}^{\infty} = 0 - (-e^{-1}) = e^{-1}.$

Therefore:

$2\int_{1}^{\infty} e^{-u} \, du = 2 \cdot e^{-1} = \frac{2}{e}.$

Final Answer

$\int_{1}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx = \frac{2}{e}.$

Would you like further explanation or have any questions? Here are some related questions:

1. How do you handle improper integrals with different types of convergence?
2. What are the criteria for an improper integral to converge?
3. How can substitution methods simplify complex integrals?
4. What are some common techniques for evaluating improper integrals involving exponential functions?
5. How do you determine if an improper integral diverges?

Tip: Always check if the integral converges before evaluating it, especially when dealing with infinite limits or undefined regions.