Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 13

Express each of the following integrals as a \(\Gamma\) function. By computer, evaluate numerically both the \(\Gamma\) function and the original integral. \(\int_{0}^{1} x^{2}\left(\ln \frac{1}{x}\right)^{3} d x\) Hint: Put \(x=e^{-u}\)

Short Answer

Expert verified
\(\frac{2}{27}\)

Step by step solution

01

- Substitution

Use the substitution given in the hint: Let \(x = e^{-u}\). Then, \(dx = -e^{-u} du\).
02

- Change of limits

Determine the new limits of integration for the variable \(u\). When \(x = 0\), \(u = \infty\). When \(x = 1\), \(u = 0\). Thus, the integral is now \( \int_\infty^0 x^{2} \left( \ln \frac{1}{x} \right)^{3} (-e^{-u} du)\).
03

- Simplify the integrand

Substitute \(x = e^{-u}\) into the integrand: \(x^2 = (e^{-u})^2 = e^{-2u}\) and \(\ln \frac{1}{x} = \ln e^u = u\). Now the integral becomes \( \int_ \infty ^0 e^{-2u} u^3 (-e^{-u}) du\).
04

- Integrate

Simplify the integral: \(\int_{\infty}^{0} e^{-3u} u^3 (-du) = \int_{0}^{\infty} e^{-3u} u^3 du\). This integral is in the form of a Gamma function: \(\Gamma(n) = \int_{0}^{\infty} t^{n-1} e^{-t} dt\). Rewrite the integral as \(\int_{0}^{\infty} u^{3} e^{-3u} du = \frac{1}{3^4} \int_{0}^{\infty} (3u)^3 e^{-(3u)} d(3u)\).
05

- Express in Gamma function

Recognize that the expression is now: \(\frac{1}{3^4} \int_{0}^{\infty} t^3 e^{-t} dt\), where \(t = 3u\). According to the definition of the Gamma function, this is \(\frac{1}{3^4} \Gamma(4)\) as \(\Gamma(n) = (n-1)!\).
06

- Evaluate the Gamma function

Calculate \(\Gamma(4)\): \(\Gamma(4) = 3! = 6\). Thus, the integral becomes \(\frac{1}{3^4} * 6 = \frac{6}{81} = \frac{2}{27}\). Verify this using a computational tool to ensure accuracy.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gamma function
The Gamma function, denoted as \( \Gamma(n) \), is a continuous extension of the factorial function to the real and complex plane. Specifically, for any positive integer \( n \), \( \Gamma(n) = (n-1)! \). It is defined via an improper integral: \

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Sketch or computer plot a graph of the function \(y=e^{-x^{2}}.\)

Identify each of the following integrals or expressions as one of the functions of this chapter. Check your work by evaluating both your answer and the original problem by computer. Be sure you understand your computer program's notation. $$\int_{0}^{\pi / 2}(\cos x)^{5 / 2} d x$$

Identify each of the following integrals or expressions as one of the functions of this chapter. Check your work by evaluating both your answer and the original problem by computer. Be sure you understand your computer program's notation. $$\int_{-\pi / 4}^{3 \pi / 4} \frac{d \phi}{\sqrt{1+\cos ^{2} \phi}}$$

Use Stirling's formula to evaluate \(\lim _{n \rightarrow \infty} \frac{(2 n) ! \sqrt{n}}{2^{2 n}(n !)^{2}}.\)

Express the following integrals as \(B\) functions, and then, by (7.1), in terms of \(\Gamma\) functions. When possible, use \(\Gamma\) function formulas to write an exact answer in terms of \(\pi, \sqrt{2}\), etc. Compare your answers with computer results and reconcile any discrepancies. $$\int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{\sin \theta}}$$
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free