Question

Express the following integrals as \(B\) functions, hence in terms of \(\Gamma\) functions, and evaluate using a table of \(\Gamma\) functions. $$ \int_{0}^{1} x^{2}\left(1-x^{2}\right)^{3 / 2} d x $$

Short answer

The integral's value is \[ \frac{4}{3} \].

Step-by-step solution

Identify the Beta Function Form

Recognize that the integral can be expressed in the form of a Beta function. The Beta function is defined as \[ B(x, y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt \]. Identify corresponding terms. Here, let's set \[ u = x^2 \]. So, the integrand becomes \int_0^1 u^{1/2 - 1} (1-u)^{3/2} du.

Replace with Variables for Beta Function

From Step 1, define \[ a = 0.5 \] and \[ b = 2 \]. Thus, the integral becomes the Beta function: \[ B\left(\frac{1}{2}, 2\right) \].

Use Relationship Between Beta and Gamma Functions

Recall that the Beta function can be expressed in terms of the Gamma function using: \[ B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)} \]. So, \[ B\left(\frac{1}{2}, 2\right) = \frac{\Gamma\left(\frac{1}{2}\right) \Gamma(2)}{\Gamma\left(\frac{1}{2} + 2\right)} \].

Evaluate Using Gamma Function

Use known values: \[ \Gamma(\frac{1}{2}) = \sqrt{\pi} \] and \[ \Gamma(2) = 1! = 1 \]. Then, \[ \Gamma\left(\frac{5}{2}\right) = \frac{3\sqrt{\pi}}{4} \]. So, \[ B\left(\frac{1}{2}, 2\right) = \frac{\sqrt{\pi} \cdot 1}{\frac{3\sqrt{\pi}}{4}} = \frac{4}{3} \], hence the integral is \[ \int_{0}^{1} x^{2}(1-x^{2})^{3 / 2} dx = \frac{4}{3} \].

Key concepts

Beta Function

The Beta function is a special function in mathematics that is closely related to the Gamma function. It is denoted as \( B(x, y) \) and defined by the integral:
\B(x, y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt
In simpler terms, the Beta function describes an integral where the integrand involves a product of a power of \( t \) and a power of \( (1-t) \). For example, in our exercise, we recognized that the given integral:
\( \int_{0}^{1} x^2 (1-x^2)^{3/2} dx \),
can be rewritten to fit the Beta function form by making a substitution.
Specifically, setting \( u = x^2 \), transforms our integral's limits and integrand to fit the Beta function template. This technique is essential in converting complex integrals to a form where special functions like Beta and Gamma can be used for evaluation.
Understanding how to identify and transform integrals into the Beta function form is crucial in higher mathematics and various applied fields such as statistics.

Gamma Function

The Gamma function, denoted as \( \Gamma(x) \), extends the concept of factorials to real and complex numbers. It is defined by the integral:
\Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} dt
This function plays a significant role in many areas of mathematics. One of the fascinating properties of the Gamma function is its relationship with the Beta function:
\B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}
This relationship allows us to express the Beta function in terms of Gamma functions, making it easier to evaluate complex integrals.
In our exercise, we used known values of the Gamma function such as \( \Gamma(1/2) = \sqrt{\pi} \) and \( \Gamma(2) = 1 \) to solve the Beta function expression.
Additionally, we used the property that for any positive integer n, \( \Gamma(n) = (n-1)! \). These known values and properties simplify the process of evaluating integrals involving special functions.

Integral Evaluation

Evaluating integrals is a fundamental skill in calculus and mathematical analysis. The integral provided in the original exercise gives us an opportunity to explore advanced techniques involving the Beta and Gamma functions.
Let’s recap the steps we used to evaluate our integral \( \int_{0}^{1} x^{2}(1-x^{2})^{3 / 2} dx \):

• Identify that the integral can fit the Beta function form using a substitution \( u = x^2 \).
• Rewrite the integral in terms of \( u \) and recognize the corresponding parameters of the Beta function.
• Express the Beta function in terms of Gamma functions using \( B(x, y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x + y)} \).
• Use known values of Gamma functions to simplify and evaluate the integral.

The process demonstrates how powerful properties of special functions can simplify complex integral evaluations.
Remember to practice these techniques frequently. Recognizing when and how to use special functions like Beta and Gamma will strengthen your mathematical problem-solving skills.