To evaluate the given integral:

$\int_1^{+\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx$

we will proceed with a substitution method. Let us analyze the problem step-by-step:

Step 1: Substitution

Let $u = \sqrt{x}$. Then, $x = u^2$ and $dx = 2u \, du$. Substituting these into the integral:

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

The integral becomes: $\int_1^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx = \int_1^{\infty} \frac{e^{-u}}{u} \cdot 2u \, du.$

The $u$'s cancel, simplifying to: $2 \int_1^{\infty} e^{-u} \, du.$

Step 2: Evaluate the Integral

The integral of $e^{-u}$ is straightforward: $\int e^{-u} \, du = -e^{-u}.$

Applying the limits from $u = 1$ to $u = \infty$: $\int_1^{\infty} e^{-u} \, du = \left[ -e^{-u} \right]_1^{\infty}.$

At $u = \infty$, $e^{-u} \to 0$. At $u = 1$, $e^{-1} = \frac{1}{e}$. Thus: $\int_1^{\infty} e^{-u} \, du = 0 - \left(-\frac{1}{e}\right) = \frac{1}{e}.$

Step 3: Multiply by 2

Returning to the original substitution, the integral evaluates to: $2 \int_1^{\infty} e^{-u} \, du = 2 \cdot \frac{1}{e} = \frac{2}{e}.$

Final Answer:

$\int_1^{+\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx = \frac{2}{e}.$

Would you like more details on substitution methods or exponential integral properties? Let me know!

Related Questions:

1. What is the significance of exponential decay in integral evaluations?
2. How does substitution simplify integral problems involving square roots?
3. Could this problem be solved using a numerical approach instead of substitution?
4. How does the convergence of improper integrals like this one work?
5. What are other applications of the exponential function in integrals?

Tip:

Always check for substitution opportunities in integrals with nested functions or compositions involving square roots, logarithms, or exponentials—they often simplify complex expressions!