Math Problem Statement

Evaluate the integral integral, from, e, squared, to, infinity, of, start fraction, minus, 1, divided by, x, left parenthesis, natural log, x, right parenthesis, start superscript, start fraction, 5, divided by, 2, end fraction, end superscript, end fraction, d, x∫ e 2

∞ ​

x(lnx) 2 5 ​

−1 ​ dx or state that the integral diverges.

Solution

We are tasked with evaluating the improper integral:

∫e2∞−1x(ln⁡x)52 dx,\int_{e^2}^\infty \frac{-1}{x (\ln x)^{\frac{5}{2}}} \, dx,

or determining if it diverges.

Step 1: Analyze the integrand

The integrand is: −1x(ln⁡x)52.\frac{-1}{x (\ln x)^{\frac{5}{2}}}. The negative sign does not affect convergence, so we will focus on the absolute value: 1x(ln⁡x)52.\frac{1}{x (\ln x)^{\frac{5}{2}}}. We examine whether the integral converges by checking the behavior at x=∞x = \infty, since ln⁡x\ln x increases without bound as x→∞x \to \infty.

Step 2: Substitution

Set u=ln⁡xu = \ln x. Then du=1xdxdu = \frac{1}{x} dx. When x=e2x = e^2, u=2u = 2. When x→∞x \to \infty, u→∞u \to \infty. This changes the integral to: ∫e2∞−1x(ln⁡x)52 dx=−∫2∞1u52 du.\int_{e^2}^\infty \frac{-1}{x (\ln x)^{\frac{5}{2}}} \, dx = -\int_2^\infty \frac{1}{u^{\frac{5}{2}}} \, du.

Step 3: Evaluate the new integral

The integral is now: −∫2∞u−52 du.-\int_2^\infty u^{-\frac{5}{2}} \, du. The antiderivative of u−52u^{-\frac{5}{2}} is: ∫u−52 du=u−32−32=−23u−32=−231u32.\int u^{-\frac{5}{2}} \, du = \frac{u^{-\frac{3}{2}}}{-\frac{3}{2}} = -\frac{2}{3} u^{-\frac{3}{2}} = -\frac{2}{3} \frac{1}{u^{\frac{3}{2}}}.

Substituting back: [ -\int_2^\infty u^{-\frac{5}{2}} , du = \left[-\left(-\frac{2}{3} \frac{1}{u^{\frac{3}{2}}}\right)\right]_2^\infty = \left[\frac{2}{3} \frac{1}{u^{\frac{3}{2}}}\right]_2^\infty. ]

Step 4: Evaluate the limits

At u=∞u = \infty, u32→∞u^{\frac{3}{2}} \to \infty, so 1u32→0\frac{1}{u^{\frac{3}{2}}} \to 0. At u=2u = 2, 1u32=1232=122=24\frac{1}{u^{\frac{3}{2}}} = \frac{1}{2^{\frac{3}{2}}} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}. Substituting these values: 231u32∣2∞=23(0−24)=−2324=−26.\frac{2}{3} \frac{1}{u^{\frac{3}{2}}} \Big|_2^\infty = \frac{2}{3} \left(0 - \frac{\sqrt{2}}{4}\right) = -\frac{2}{3} \frac{\sqrt{2}}{4} = -\frac{\sqrt{2}}{6}.

Final Answer:

The value of the integral is: −26.-\frac{\sqrt{2}}{6}.

Would you like additional details or clarification? 😊

Related Questions:

  1. What is an improper integral, and why do we check convergence?
  2. How does the substitution u=ln⁡xu = \ln x simplify integrals involving logarithms?
  3. What are some conditions that guarantee the convergence of improper integrals?
  4. Can this result be interpreted in any real-world context?
  5. What happens if we change the exponent of (ln⁡x)(\ln x) to a value different from 52\frac{5}{2}?

Tip:

When solving improper integrals, always check the convergence of the integral before calculating its value.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Improper Integrals
Logarithmic Functions
Substitution in Integration
Convergence of Integrals

Formulas

Substitution: \(u = \ln x, \frac{du}{dx} = \frac{1}{x}\)
Integral of \(u^{-p}: \int u^{-p} du = \frac{u^{1-p}}{1-p}\) for \(p \neq 1\)

Theorems

Convergence criteria for improper integrals
Change of variables in integration

Suitable Grade Level

Grades 11-12, College

Related Recommendation

How to Solve \( \int_e^\infty \frac{1}{x (\log x)^2} \, dx \) Using Substitution

Solving the Integral \( \int_0^{\ln(2)} \frac{\ln(1 + e^x)}{e^x} \, dx \)

How to Evaluate the Integral of 1 / (x sqrt(ln x)) from e^5 to e

Evaluate the Divergent Improper Integral \( \int_e^{\infty} \frac{1}{x \ln x \left(\ln^3 x + 1\right)} dx \)

Solve the Definite Integral \( \int_1^e \frac{\ln^2(x)}{x} \, dx \)