Question

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. $$\int_{0}^{1} \frac{x^{4} d x}{\sqrt{1-x^{2}}}$$

Short answer

The integral evaluates to \( \frac{1}{3} \).

Step-by-step solution

Identify the Integral as a B Function

The given integral is \(\int_{0}^{1} \frac{x^4 \, dx}{\sqrt{1-x^2}}\). Recognize that it can be transformed to the form of a Beta function integral.

Use Substitution to Transform Integral

Use the substitution \( x = \sin{u} \), \( dx = \cos{u} \, du \). The integral becomes \(\int_{0}^{\frac{\pi}{2}} \frac{\sin^4{u} \, \cos{u} \, du}{\cos{u}} = \int_{0}^{\frac{\pi}{2}} \sin^4{u} \, du \).

Express it in Beta Function Form

Recognize that the Beta function is given by \( B(p,q) = 2 \int_{0}^{\frac{\pi}{2}} \sin^{2p-1}{u} \cos^{2q-1}{u} \, du \). Rewrite the integral in this form recognizing \( \sin^4{u} = \sin^{2\cdot2}{u} \).

Identify Parameters p and q

Compare with the standard Beta function form. Here \( 2p-1 = 5 \) and \( 2q-1 = 1 \), so \( p = \frac{5+1}{2} = 3 \) and \( q = \frac{1+1}{2} = 1 \). Thus, \( B(3,1) \).

Relate the Beta Function to Gamma Functions

Use the relation \( B(p, q) = \frac{\Gamma(p) \Gamma(q)}{\Gamma(p+q)} \). Thus, \( B(3,1) = \frac{\Gamma(3) \Gamma(1)}{\Gamma(4)} \).

Evaluate Gamma Functions

Use known values: \( \Gamma(3) = 2! = 2 \), \( \Gamma(1) = 1 \), and \( \Gamma(4) = 3! = 6 \). Thus, \( \frac{2 \cdot 1}{6} = \frac{1}{3} \).

Express Integral in Terms of Known Constants

Thus, the integral's value, in exact terms, is \( \frac{1}{3} \). Compare with computer results for verification.

Key concepts

Gamma function

The Gamma function, denoted by \(\beta{(x)}\), is a fundamental and widely used special function in mathematics. It extends the concept of factorial functions to complex and real number arguments. The function is defined for positive integers as \(\beta{(n+1)} = n!\) and for real and complex numbers using an integral: \[ \beta{(x)} = \int_{0}^{\infty}t^{x-1}e^{-t}dt \].
The Gamma function has several notable properties:

• Recurrence: \(\beta{(x+1)} = x\beta{(x)}\)
• Reflection: \(\beta{(1-x)} \beta{(x)} = \frac{\pi}{sin(\pi \beta)}\)
• Special Values: For integer values, \(\beta{(n+1)} = n!\)

These properties make the Gamma function useful in various fields, including probability and statistics, complex analysis, and physics.

substitution method

The substitution method is a common technique used to evaluate integrals by simplifying the integral into a more manageable form. This is achieved by substituting parts of the integral with new variables. In our problem, we used a trigonometric substitution:
We let \( x = \sin{u} \). This automatically means that \( dx = \cos{u} \ du \), leading to the transformed integral:
\[ \int_{0}^{1} \frac{x^4 \ dx}{\sqrt{1-x^2}} = \int_{0}^{\frac{\pi}{2}} \sin^4{u} \ du \]
The substitution \( x = \sin{u} \) helped simplify the integral by leveraging the periodic properties of the trigonometric functions.

integral transform

Integral transforms are mathematical tools involving integral equations to convert functions into a different domain, typically to simplify problem-solving. They have applications in various fields, such as physics, engineering, and signal processing. An example is transforming a differential equation into an algebraic one to facilitate its solution. The Beta function itself arises from an integral transform, expressing it in terms of the Gamma function:
\(\beta(p, q) = 2 \int_{0}^{\frac{\pi}{2}} \sin^{2p-1}(u) \cos^{2q-1}(u) \ du\)
This coherent interpretation helps to deal with more intricate integrals by converting them into forms that are easier to handle or already well-defined.

trigonometric substitution

Trigonometric substitution is a specific type of substitution used to evaluate integrals containing square roots of quadratic expressions. This method leverages trigonometric identities to simplify such integrals. In our problem, \( x = \sin{u} \) was chosen because it allows conversion of the integral of a square root into a simpler form. Key trigonometric identities used in such substitutions include:

• For \( \sqrt{a^2 - x^2} \ : x = \sin(u) \)
• For \( \sqrt{a^2 + x^2}\ : x = \tan(u)\)
• For \(\sqrt{x^2 - a^2}\ : x = \sec(u)\)

Using these identities, complex integral forms are transformed into simpler ones, often reducing them to well-known integrals that can be evaluated using standard formulas or properties of trigonometric functions.