Question

Express the following integrals as \(\Gamma\) functions and evaluate them using a table of \(\Gamma\) functions. $$ \int_{0}^{1} \sqrt[3]{\ln x} d x $$

Short answer

\( \Gamma \left( \frac{4}{3} \right) = \frac{1}{3} \Gamma \left( \frac{1}{3} \right) \)

Step-by-step solution

Identify the integral form

Recognize that the integral can be transformed to match the form of a Gamma function. The given integral is \[ \ \ \ \ \ \ \int_{0}^{1} \sqrt[3]{\ln x} d x. \]

Make a substitution

Use the substitution \( \ln x = -t \) and differentiate: \( dx = -e^{-t} dt \). Changing limits of integration: when \( x = 1, t = 0 \) and when \( x = 0, t = \infty \). Thus, the integral becomes: \[ -\int_{\infty}^{0} \sqrt[3]{-t} e^{-t} (-dt) = \int_{0}^{\infty} t^{1/3} e^{-t} dt. \]

Recognize the Gamma function

Notice that the integral \( \int_{0}^{\infty} t^{a-1} e^{-t} dt \) is the definition of the Gamma function: \[ \Gamma(a) = \int_{0}^{\infty} t^{a-1} e^{-t} dt. \] Here, \( t^{1/3} \) implies \( a - 1 = \frac{1}{3} \), so \( a = \frac{4}{3} \). Thus, the integral can be written as \[ \Gamma\left( \frac{4}{3} \right). \]

Evaluate using the Gamma function table

Using a table of Gamma functions, \( \Gamma \left( \frac{4}{3} \right) \) can be evaluated or expressed in terms of known values. \( \Gamma \left( \frac{4}{3} \right) = \frac{1}{3} \Gamma \left( \frac{1}{3} \right) \).

Key concepts

Integral Transformation

The goal of integral transformation is to reshape an integral into a more recognizable form. In this exercise, we start with the integral: \ \( \int_{0}^{1} \sqrt[3]{\ln x} \ d x \ \)
To transform this, we look for a substitution that simplifies the form. Here, we use the substitution \ \( \ln x = -t\ \ \), which transforms the variable of integration.
This substitution changes the integral's limits:

• When \( x = 1 \), \( t = 0 \)
• When \( x = 0 \), \( t = \infty \)

By transforming these limits and substituting into the integral, we get a new integral that can be recognized as a Gamma function form.

Substitution Method

The substitution method is a powerful tool to simplify integrals. For our example, starting with \ \( \int_{0}^{1} \sqrt[3]{\ln x} d x \),\ \ we use \ \( \ln x = -t\ \).
Differentiate this to find \ \( d(\ln x) = \-dt\ \) and express \ \( dx = -e^{-t} dt \).
After substitution, the given integral becomes: \ \( -\int_{\infty}^{0} \sqrt[3]{-t} e^{-t} (-dt) \ \).
This further simplifies to: \ \( \int_{0}^{\infty} t^{1/3} e^{-t} dt \).
Recognizing this new form is crucial as it matches the definition of the Gamma function.

Evaluation Using Gamma Function Tables

To evaluate the transformed integral, we match it with the Gamma function. The Gamma function is defined as: \ \( \Gamma(a) = \int_{0}^{\infty} t^{a-1} e^{-t} dt\ \).
In our case, \ \( t^{1/3} \) suggests \ \( a-1 = \frac{1}{3}\ \), so \ \( a = \frac{4}{3}\ \).
This means our integral \ \( \int_{0}^{\infty} t^{1/3} e^{-t} dt \) is \ \( \Gamma \left( \frac{4}{3}\ \right) \).
Using Gamma function tables, \ \( \Gamma \left( \frac{4}{3}\ \right) = \frac{1}{3} \Gamma \left( \frac{1}{3}\ \right) \).
Gamma function tables provide these values, making the final calculation straightforward.