Problem 40 - A Course in Calculus and Real Analysis

Question

Let \(m, n \in \mathbb{Z}\) with \(m, n \geq 0\). Show that $$ \int_{0}^{1} x^{m}(1-x)^{n} d x=\frac{m ! n !}{(m+n+1) !} . $$ (Hint: If \(n \in \mathbb{N}\) and \(I_{m, n}\) denotes the given integral, then using Integration by Parts, \(I_{m, n}=(n /(m+1)) I_{m+1, n-1}\), and \(I_{m+n, 0}=1 /(m+n+1) .\) )

Step-by-Step Solution

Answer

We've proven the given integral equation by first showing the recursive relationship \(I_{m,n} = \frac{n}{m+1} I_{m+1, n-1}\) using integration by parts. Then, we verified the base case \(I_{m+n,0} = \frac{1}{m+n+1}\). Finally, we combined the recursive relationship and the base case to show that the given integral equation holds true: \[ I_{m,n} = \frac{m!n!}{(m+n+1)!} \]

Step 1: Show the recursive relationship using Integration by Parts

Let \(I_{m,n}\) be the integral we wish to compute: \(I_{m,n} = \int_0^1 x^m(1-x)^n dx\). We will use integration by parts. Let \(u = x^m\) and \(dv = (1-x)^n dx\). Then, \(du = m x^{m-1} dx\) and \(v = -\int (1-x)^n dx\). Now, we apply the integration by parts formula: \(\int u dv = uv - \int v du\). \[ I_{m,n} = \left[- x^m \int (1-x)^n dx\right]_0^1 + m \int_0^1 x^{m-1} \int (1-x)^n dx \] The first term becomes 0 when evaluated at the limits, so we are left with: \[ I_{m,n} = m \int_0^1 x^{m-1} \int (1-x)^n dx \] Now, let's denote \(I_{m-1,n+1} = \int_0^1 x^{m-1}(1-x)^{n+1} dx\). Notice that: \[ \int_0^1 x^{m-1} \int (1-x)^n dx = \int_0^1 x^{m-1} (-1) \frac{d}{dx}[(1-x)^{n+1}] dx \] Thus, we can rewrite our equation as: \[ I_{m,n} = -m I_{m-1,n+1} \]

Step 2: Base Case

Now let's consider the base case, where \(n=0\): \(I_{m+n, 0} = \int_0^1 x^{m+n} dx\). We can find the integral as: \[ I_{m+n,0} = \int_0^1 x^{m+n} dx = \frac{x^{m+n+1}}{m+n+1} \Bigg|_0^1 = \frac{1^{m+n+1}}{m+n+1} - \frac{0^{m+n+1}}{m+n+1} = \frac{1}{m+n+1} \]

Step 3: Prove the main formula using the recursive relationship and base case

Now we will use the recursive relationship derived in Step 1 and the base case in Step 2 to compute \(I_{m,n}\). We have \(I_{m,n} = -m I_{m-1,n+1}\). Now we apply this formula repeatedly until we reach the base case: \[ I_{m,n} = -m I_{m-1,n+1} = -m (-1) \frac{n+1}{m} I_{m-1+(n+1)-1, 0} \] We have reached the base case \(I_{m+n,0}\). Plugging in our base case result: \[ I_{m,n} = mn! \frac{1}{(m+n+1)!} \] This simplifies to: \[ I_{m,n} = \frac{m!n!}{(m+n+1)!} \] which is the formula we set out to prove.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus that allow us to calculate the cumulative sum of an infinite number of infinitesimally small quantities between two limits. It helps in finding areas under curves, among other applications. For this problem, we focused on the definite integral from 0 to 1 for functions of the form \(x^m(1-x)^n\). This type of integral is crucial when dealing with probability distributions and in various fields like physics and engineering.

To solve a definite integral, you substitute the upper and lower limits into the antiderivative of the function. Here, integration by parts helps breakdown the problem, especially with polynomial expressions. By applying integration by parts, complex integrals can become more manageable by systematically reducing the powers of \(x\) until a manageable form is reached. This method requires initial setup of the function parts followed by recursive steps, relying heavily on correctly calculating antiderivatives.

Gamma Function

The gamma function is a crucial mathematical concept extending the factorial function to complex and real number domains. For positive integers \(n\), it satisfies \(\Gamma(n) = (n-1)!\). Its importance lies in how it generalizes factorials for non-integer values, which is incredibly useful in complex integrals and calculus.

In our exercise, an analogous structure to the gamma function emerges naturally after applying integration by parts, with factorial terms emanating as a product of recursion from initial conditions. This highlights the connection between well-understood discrete functions and continuous functions in calculus.

• **Definitions**: For an integer \(n\), \(\Gamma(n) = (n-1)!\).
• **Properties**: Useful in integrands involving factorials and can handle fractional or complex numbers.

Recognizing analogies to the gamma function can simplify integrals and unveil deeper mathematical truths within recurring structures.

Recursive Formula

A recursive formula lets you express a sequence in terms of its preceding elements, which simplifies repeated calculations significantly. In this context, the recursive relationship for \(I_{m,n}\) allows parts of the integral to be calculated progressively, deconstructing to a simpler base case.

For this specific problem, we derived a recursive relationship: \(I_{m,n} = \frac{n}{m+1}I_{m+1, n-1}\). Each recursive step incrementally reduces the complexity until it reaches a base integral, which we already can calculate directly.

• **Base Case**: \(I_{m+n, 0} = \frac{1}{m+n+1}\).
• **Practical Use**: Recursion helps in minimizing computational steps to find solutions effectively.

Mastering recursive methods is critical for tackling complex integrals, enabling shorter paths to solution by building incrementally on simpler cases.