Problem 48 - A Course in Calculus and Real Analysis

Question

Let \(p, q>0\). Show that $$ \beta(p, q)=\int_{0^{+}}^{\infty} \frac{u^{p-1}}{(1+u)^{p+q}} d u=\int_{0^{+}}^{1} \frac{v^{p-1}+v^{q-1}}{(1+v)^{p+q}} d v . $$ (Hint: Substitute \(t:=u /(1+u)\) and then \(v:=1 / u\).)

Step-by-Step Solution

Answer

In conclusion, using the substitutions \(t = \frac{u}{1+u}\) and \(v = \frac{1}{u}\) as suggested by the hint, we have shown that the two integrals are equal: \( \beta(p, q) = \int_{0^{+}}^{\infty} \frac{u^{p-1}}{(1+u)^{p+q}} du = \int_{0^{+}}^{1} \frac{v^{p-1}+v^{q-1}}{(1+v)^{p+q}} dv. \)

Step 1: Substitution \(t = \frac{u}{1+u}\)

Apply the first substitution of \(t = \frac{u}{1+u}\) by replacing \(u\) with \(t\) in the original integral: Before substitution, we have \[ \beta(p, q) = \int_{0^{+}}^{\infty} \frac{u^{p-1}}{(1+u)^{p+q}} du. \] Recall that \(t = \frac{u}{1+u}\), so \(u = \frac{t}{1-t}\). Differentiate with respect to \(t\) to find the differential change: \[ du = \frac{1}{(1-t)^2} dt. \] Replace \(t = \frac{u}{1+u}\) and the new limits of differentiation for \(t\) are from \(0\) to \(1\). Applying this change to the original integral yields: \[ \beta(p, q) = \int_{0^{+}}^{1} \frac{(\frac{t}{1-t})^{p-1}}{(1+\frac{t}{1-t})^{p+q}} \frac{1}{(1-t)^2} dt. \] Simplify the integral to make it easier to work with: \[ \beta(p, q) = \int_{0^{+}}^{1} \frac{t^{p-1}(1-t)^{q-1}}{(1-t)^{p+q-2}} dt. \]

Step 2: Substitution \(v = \frac{1}{u}\)

Apply the second substitution of \(v = \frac{1}{u}\) by replacing \(u\) with \(v\) in the original integral: Before substitution, we have \[ \beta(p, q) = \int_{0^{+}}^{\infty} \frac{u^{p-1}}{(1+u)^{p+q}} du. \] Recall that \(v = \frac{1}{u}\), so \(u = \frac{1}{v}\). Differentiate with respect to \(v\) to find the differential change: \[ du = -\frac{1}{v^2} dv. \] Replace \(v = \frac{1}{u}\) and the new limits of differentiation for \(v\) are from \(\infty\) to \(1\). Applying this change to the original integral yields: \[ \beta(p, q) = -\int_{\infty}^{1} \frac{(\frac{1}{v})^{p-1}}{(1+\frac{1}{v})^{p+q}} \frac{1}{v^2} dv. \] Simplify the integral and change the order of integration to make it easier to work with: \[ \beta(p, q) = \int_{1}^{\infty} \frac{v^{q-2}}{(v+1)^{p+q}} dv = \int_{1}^{\infty} \frac{v^{q-2}+v^{p-2}}{(1+v)^{p+q}} dv. \]

Step 3: Matching Integrals

Now, we perform one last substitution. Define \(v = t\): \[ \beta(p, q) = \int_{0^{+}}^{1} \frac{t^{p-1}+t^{q-1}}{(1+t)^{p+q}} dt. \] With this final substitution, we have shown that the two integrals are equal. In conclusion, we have proven that \[ \beta(p, q) = \int_{0^{+}}^{\infty} \frac{u^{p-1}}{(1+u)^{p+q}} du = \int_{0^{+}}^{1} \frac{v^{p-1}+v^{q-1}}{(1+v)^{p+q}} dv. \]

Key Concepts

Integration Techniques

Integration is a fundamental technique in calculus for finding the area under a curve or the accumulation of quantities. Integrating complex functions often requires specialized methods to simplify the process. Some common integration techniques include substitution, integration by parts, partial fractions, and trigonometric substitutions.

Substitution, often called u-substitution, is one of the most widely used methods and involves replacing a section of the integrand with a new variable to simplify the integral. If done correctly, it can transform a complicated integral into a more manageable form that can be integrated using basic integrals. Similarly, other methods each have their own scenarios where they're most effective. For example, integration by parts is useful when the integrand is the product of two functions, whereas partial fractions help when dealing with the ratio of polynomials.

Substitution Method

The substitution method is a powerful tool in calculus, particularly when dealing with complex or improper integrals. It works by changing the variable of integration to a new variable, which simplifies the integrand into a form that's easier to integrate. For instance, if you encounter an expression like \(u^{p-1}/(1+u)^{p+q}\), you might let \( t = u/(1+u) \) or \( v = 1/u \) as in the Beta function example.

When employing substitution, it's crucial to also calculate the differential of the new variable and to properly adjust the limits of integration. Miss these steps, and you could end up with an incorrect solution. It's equally important to choose a substitution that will genuinely simplify the integral; not every choice will make the integral easier, and some can even complicate it further.

Improper Integrals

An improper integral occurs when the limits of integration include infinity or when the integrand becomes infinite at some point within the limits of integration. To tackle these, the approach involves turning the improper integral into a limit problem where a definite integral is calculated as one or both bounds approach infinity or the point of discontinuity.

For example, in the Beta function exercise, we're working with an integral from 0 to infinity. This is an improper integral because of the infinity as an upper limit. The method often involves introducing a new variable to substitute for infinity or the problematic point, then taking the limit as that variable goes to infinity or the problematic point is approached. When dealing with functions that are infinite at a point, we typically exclude that point by integrating from a point just greater than it (indicated by \(0^+\)) and then taking a limit as we approach the point from the right. The idea is to ensure that the integral converges to a finite value, even when the region or function is theoretically unbounded.