Math Problem Statement

Find the value

Find the value of the integral \( \int_1^{+\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx \).

Solution

To evaluate the given integral:

1+exxdx\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=xu = \sqrt{x}. Then, x=u2x = u^2 and dx=2ududx = 2u \, du. Substituting these into the integral:

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

The integral becomes: 1exxdx=1euu2udu.\int_1^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx = \int_1^{\infty} \frac{e^{-u}}{u} \cdot 2u \, du.

The uu's cancel, simplifying to: 21eudu.2 \int_1^{\infty} e^{-u} \, du.


Step 2: Evaluate the Integral

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

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

At u=u = \infty, eu0e^{-u} \to 0. At u=1u = 1, e1=1ee^{-1} = \frac{1}{e}. Thus: 1eudu=0(1e)=1e.\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: 21eudu=21e=2e.2 \int_1^{\infty} e^{-u} \, du = 2 \cdot \frac{1}{e} = \frac{2}{e}.


Final Answer:

1+exxdx=2e.\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!

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Math Problem Analysis

Mathematical Concepts

Substitution
Exponential Functions
Improper Integrals
Integration Techniques

Formulas

Substitution formula: u = sqrt(x), dx = 2u du
Integral of e^{-u}: \int e^{-u} du = -e^{-u}

Theorems

Substitution method for integrals
Convergence of improper integrals

Suitable Grade Level

Grades 11-12