Problem 47 - A Course in Calculus and Real Analysis

Question

Let \(p, q \in \mathbb{R}\) with \(p>0\) and \(q>0 .\) Show that $$ \beta(p, q)=2 \int_{0^{+}}^{(\pi / 2)^{-}}(\sin u)^{2 p-1}(\cos u)^{2 q-1} d u $$ and in particular, \(\beta\left(\frac{1}{2}, \frac{1}{2}\right)=\pi .\) (Hint: Substitute \(\left.t:=\sin ^{2} u .\right)\)

Step-by-Step Solution

Answer

We have shown that the Beta function can be expressed as the given integral: \[ \beta(p, q) = 2 \int_0^{\frac{\pi}{2}} (\sin u)^{2p-1} (\cos u)^{2q-1} du \] We also demonstrated that \(\beta\left(\frac{1}{2}, \frac{1}{2}\right) = \pi\). Using the hint substituting \(t:=\sin^2u\) helped us transform the integral and show the equivalence.

Step 1: Review of the Beta function

The Beta function is defined as: \[ \beta(p, q)=\int_0^{1} t^{p-1}(1-t)^{q-1} dt \] for real numbers \(p > 0\) and \(q > 0\).

Step 1: Perform the substitution

We'll use the hint provided to substitute \(t\) with \(\sin^2 u\). We have: \[ t = \sin^2u \\ dt = 2\sin u \cos u \space du \] When \(u = 0\), we have \(t = 0\). When \(u = \frac{\pi}{2}\), we have \(t = 1\). Now let's rewrite the given expression for \(\beta(p, q)\) with the new variables. We get: \[ \beta(p, q) = 2\int_0^{\frac{\pi}{2}} (\sin u)^{2p - 1} (\cos u)^{2q - 1} du \]

Step 2: Show the equivalence

Now, let's show the equivalence between the original expression for \(\beta(p, q)\) and the rewritten expression: \[ \beta(p, q) = \int_0^{1} t^{p-1}(1-t)^{q-1} dt \\ = 2\int_0^{\frac{\pi}{2}} (\sin^2 u)^{p-1} (\cos^2 u)^{q-1} (2\sin u \cos u) du \\ = 2 \int_0^{\frac{\pi}{2}} (\sin u)^{2p-1} (\cos u)^{2q-1} du \] Therefore, the Beta function expression is equivalent to the given expression.

Step 3: Evaluate for specific case

Finally, let's evaluate the specific case \(\beta\left(\frac{1}{2}, \frac{1}{2}\right)\). Using the expression involving trigonometric functions, we have: \[ \beta\left(\frac{1}{2}, \frac{1}{2}\right) = 2 \int_0^{\frac{\pi}{2}} (\sin u)^{2(\frac{1}{2})-1} (\cos u)^{2(\frac{1}{2})-1} du \\ = 2 \int_0^{\frac{\pi}{2}} \sqrt{\sin u \cos u}\hspace {0.5 cm} du \] Using the substitution \(x = \sin u\), we get: \[ = 2 \int_0^{\frac{\pi}{2}} \sqrt{x(1 - x)} dx \\ \] Now, noting that the integral in the above expression is an integral representation of the Beta function with \(p = \frac{1}{2}\) and \(q = \frac{1}{2}\), we’ll find: \[ \beta\left(\frac{1}{2}, \frac{1}{2}\right) = 2 \cdot \frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2}+\frac{1}{2})} \\ \] Recalling that \(\Gamma(\frac{1}{2}) = \sqrt{\pi}\) and \(\Gamma(1) = 1\), we get: \[ \beta\left(\frac{1}{2}, \frac{1}{2}\right) = 2\cdot\frac{\sqrt{\pi}\sqrt{\pi}}{1} \\ = \pi \] Hence, we have proven that \(\beta\left(\frac{1}{2}, \frac{1}{2}\right) = \pi\).

Key Concepts

Integration by substitution

Integration by substitution is a powerful technique used to simplify integrals by changing variables. It involves substituting a part of the original integral with a new variable, making the integral easier to solve. Here's how it works in our scenario:

• We start with the integral of the Beta function in terms of the trigonometric functions sine and cosine.
• By substituting \(t = \sin^2 u\), we transform the variables to a more familiar form.
• This substitution involves some calculus: the derivative of \(t\) gives us \(dt = 2\sin u\cos u\ du\).
• The limits of integration also change; when \(u = 0\), \(t\) is 0, and when \(u = \frac{\pi}{2}\), \(t\) becomes 1.

By substituting, we simplify the integral to match the standard definition of the Beta function, which helps in solving complex integrals efficiently. This method is frequently used in calculus to navigate tricky integrations where a direct approach would be cumbersome.

Trigonometric identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the involved variables. They are essential in transforming and simplifying expressions in calculus, especially during integration. In the context of our Beta function problem, we use trigonometric identities to assist in the process:

• The identity \(\sin^2 u + \cos^2 u = 1\) simplifies expressions involving powers of sine and cosine.
• Using the identity \(\sin 2u = 2\sin u \cos u\), we derive \(dt = 2\sin u \cos u\ du\).
• This step confirms the correct transformation of variables needed for substitution.

These identities enable transformations that maintain the integrity of the equation while facilitating easier handling of the integral. It's like finding a shortcut in a maze: the journey gets simpler, but it reaches the same destination.

Gamma function

The Gamma function is a more extensive concept related to factorials, and it generalizes the factorial function to complex numbers. For positive integers, it holds that \(\Gamma(n) = (n-1)!\), but it also works with non-integers. This function connects deeply with the Beta function and is crucial in evaluating it:

• The Beta function can be expressed in terms of the Gamma function as \(B(p, q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}\).
• In our specific case, \(\beta\left(\frac{1}{2}, \frac{1}{2}\right)\), we use known values: \(\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}\) and \(\Gamma(1) = 1\).
• This allows us to conclude that \(\beta\left(\frac{1}{2}, \frac{1}{2}\right) = \pi\), proving our given relation.

The Gamma function thus bridges the gap between the Beta function and familiar constants like \(\pi\), showing its importance in higher-level calculus. Understanding this connection helps us see how various calculus concepts interlink, highlighting the elegance of mathematical analysis.