Question

Compute \(\int_{0}^{1} \frac{x^{m} d x}{\sqrt{1-x^{2}}}\) when \(m\) is (a) even, (b) odd \((m>0)\).

Short answer

In summary, the integral \(\int_{0}^{1} \frac{x^{m} dx}{\sqrt{1-x^{2}}}\) can be computed with different methods depending on whether \(m\) is even or odd. When \(m\) is even, the integral is: \(\int_{0}^{1} \frac{x^{2n} dx}{\sqrt{1-x^{2}}} = \frac{1}{2}\frac{\Gamma(n+1)\Gamma(\frac{1}{2})}{\Gamma(n+\frac{3}{2})}\) When \(m\) is odd, the integral is: \(\int_{0}^{1} \frac{x^{2n+1} dx}{\sqrt{1-x^{2}}} = \frac{1}{2}\frac{\Gamma(n+1)\Gamma(\frac{3}{2})}{\Gamma(n+\frac{5}{2})}\) where \(m=2n\) for even values of \(m\) and \(m=2n+1\) for odd values of \(m\), with \(n\) being a non-negative integer.

Step-by-step solution

Identify whether \(m\) is even or odd

Depending on whether \(m\) is even or odd, different techniques may be used to evaluate the integral. To solve the exercise, we will consider two cases: (a) even and (b) odd, and provide appropriate methods for each case.

Step 2a: Compute the integral when \(m\) is even

If \(m\) is even, then \(m=2n\) for some non-negative integer \(n\). In this case, the integral can be expressed as \(\int_{0}^{1} \frac{x^{2n} d x}{\sqrt{1-x^{2}}}\). We can apply substitution method with \(x=\sin(t)\), which yields \(dx=\cos(t) dt\) and \(x^2 = \sin^2(t)\). The limits of integration are changed from \(0\) to \(1\), to \(0\) to \(\frac{\pi}{2}\). Substituting these variables into the integral, we get: \(\int_{0}^{\frac{\pi}{2}} \frac{\sin^{2n}(t) \cos{t} dt}{\cos{t}}\) Now, we can cancel the \(\cos{t}\) in the numerator and denominator to get: \(\int_{0}^{\frac{\pi}{2}} \sin^{2n}(t) dt\) Now, we compute the integral using the Beta function: \(B(p, q) = \int_{0}^{\pi/2} (\sin t)^{2p-1}(\cos t)^{2q-1} dt\) We have \(2n=2(2p-1)\), thus \(p=n\). Since we need the integral from \(0\) to \(\frac{\pi}{2}\), we can rewrite the integral as: \(\int_{0}^{\frac{\pi}{2}} \sin^{2n}(t) dt= \frac{1}{2}B(n+1, \frac{1}{2})\) Now we can use the properties of the Gamma function to further simplify: \(B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\) \(\frac{1}{2}B(n+1, \frac{1}{2}) = \frac{1}{2}\frac{\Gamma(n+1)\Gamma(\frac{1}{2})}{\Gamma(n+\frac{3}{2})}\) Thus, the integral when \(m\) is even is: \(\int_{0}^{1} \frac{x^{2n} d x}{\sqrt{1-x^{2}}} = \frac{1}{2}\frac{\Gamma(n+1)\Gamma(\frac{1}{2})}{\Gamma(n+\frac{3}{2})}\)

Step 2b: Compute the integral when \(m\) is odd

If \(m\) is odd, then \(m=2n+1\) for some non-negative integer \(n\). In this case, the integral can be expressed as \(\int_{0}^{1} \frac{x^{2n+1} d x}{\sqrt{1-x^{2}}}\). By performing similar substitution as in the even case, we get: \(\int_{0}^{\frac{\pi}{2}} \sin^{2n+1}(t) \cos{t}dt\) We use the formula for the Beta function: \(B(p, q) = \int_{0}^{\pi/2} (\sin t)^{2p-1}(\cos t)^{2q-3} dt\) We have \(2n+1=2(2p-1)\), thus \(p=n+1\). And since we need the integral from \(0\) to \(\frac{\pi}{2}\), we can rewrite the integral as: \(\int_{0}^{\frac{\pi}{2}} \sin^{2n+1}(t) \cos{t} dt= \frac{1}{2}B(n+1, \frac{3}{2})\) Now we can use the properties of the Gamma function to further simplify: \(B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\) \(\frac{1}{2}B(n+1, \frac{3}{2}) = \frac{1}{2}\frac{\Gamma(n+1)\Gamma(\frac{3}{2})}{\Gamma(n+\frac{5}{2})}\) Thus, the integral when \(m\) is odd is: \(\int_{0}^{1} \frac{x^{2n+1} d x}{\sqrt{1-x^{2}}} = \frac{1}{2}\frac{\Gamma(n+1)\Gamma(\frac{3}{2})}{\Gamma(n+\frac{5}{2})}\)

Key concepts

Beta Function

The Beta function, denoted as B(p, q), is an integral that plays a key role in calculus, particularly in evaluating complex integrals that arise in probability, statistics, and other mathematical areas. It is defined by the integral:
\[ B(p, q) = \text{int}_{0}^{1} t^{p-1} (1 - t)^{q-1} dt \]
In the context of our exercise, the Beta function appears when we use the substitution method to evaluate trigonometric integrals. Specifically, we look at the integral of the form:
\[ B(p, q) = \text{int}_{0}^{\pi/2} (\sin t)^{2p-1}(\cos t)^{2q-1} dt \]
which is a modified version tailored for the trigonometric substitution applied. The relationship between the Beta function and the integral in the provided exercise serves to simplify the evaluation process, transforming a rather complex integral into a more manageable form that can be expressed in terms of known functions.
Moreover, the Beta function connects to the Gamma function, allowing further simplification. The identity:
\[ B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \]
shows how the Beta function can be decomposed into Gamma function terms, providing a pathway to a solution for our integral when applying the appropriate conditions for even and odd m values.

Gamma Function

The Gamma function, represented as \(\Gamma(n)\), is a crucial element in various branches of mathematics and science, such as probability and physics. It is often introduced as an extension of the factorial function to complex and real number arguments, defined by the integral:
\[ \Gamma(n) = \text{int}_{0}^{\infty} x^{n-1} e^{-x} dx \]
for real numbers n greater than zero. In the context of our exercise, the Gamma function allows us to express the Beta function in terms of more elementary components when dealing with the integral of powers of \(x\) over a square root of \(1 - x^2\).
When substituting the Beta function into our problem's solution, we use the identity connecting it to the Gamma function. For example, with even m:
\[ \frac{1}{2}B(n+1, \frac{1}{2}) = \frac{1}{2}\frac{\Gamma(n+1)\Gamma(\frac{1}{2})}{\Gamma(n+\frac{3}{2})} \]
This remarkable interrelation transforms the problem into a form where we can directly apply values to find the solution without needing to compute the full integral from scratch. It streamlines the process of solving integrals by utilizing the properties of these higher-level functions.

Integration Techniques

Integration techniques are the various methods used to evaluate integrals, which are key components in integral calculus. The exercise provided illustrates the importance of selecting the correct method based on the nature of the function being integrated.
For cases when m is even, the substitution \(x = \sin(t)\) is used, taking advantage of trigonometric identities and the inherent symmetry of the function over the interval [0,1]. The substitution simplifies the integral by changing the variable from x to a trigonometric function, which is then more easily integrated using the Beta function as demonstrated.
For odd values of m, the integral requires a slight adjustment but still follows a similar approach. In both cases, the ability to choose and apply the right technique—trigonometric substitution in this instance—greatly simplifies the problem.
Understanding these integration techniques, such as substitution, parts, trigonometric integration, and integration of rational functions, gives one the tools needed to tackle a wide range of integrals, including those found in homework exercises or more advanced mathematical problems.