Question

a) Evaluate \(\int_{0}^{1} x^{n} \log x d x\) by differentiating both sides of the equation \(\int_{0}^{1} x^{n} d x=\) \(\frac{1}{n+1}\) with respect to \(n(n>-1)\). b) Evaluate \(\int_{0}^{\infty} x^{n} e^{-a x} d x\) by repeated differentiation of \(\int_{0}^{\infty} e^{-a x} d x(a>0)\). c) Evaluate \(\int_{0}^{\infty} \frac{d y}{\left(x^{2}+y^{2}\right)^{n}}\) by repeated differentiation of \(\int_{0}^{\infty} \frac{d y}{x^{2}+y^{2}}\). [In (b) and (c) the improper integrals are of a type to which Leibnitz's Rule is applicable, as is shown in Chapter 6 . The result of (a) can be explicitly verified.]

Short answer

a) \(\int_{0}^{1} e^{nx} dx\) b) \(\int_{0}^{1} e^{-nx} dx\) c) \(\int_{0}^{1} x^{n} \log x \, dx\) d) \(\int_{0}^{1} x^{n} e^{-x} \, dx\) Answer: c) \(\int_{0}^{1} x^{n} \log x \, dx\)

Step-by-step solution

Differentiate the identity under the integral sign

First, we differentiate both sides of the given equation with respect to n. \(\frac{d}{dn} \int_0^1 x^n \, dx = \frac{d}{dn}(\frac{1}{n+1})\)

Apply the Differentiation to the Integral

Applying differentiation inside the integral, we get: \(\int_0^1 \frac{d}{dn}(x^n) \, dx = -\frac{1}{(n+1)^2}\)

Differentiate the integrand

Now, we differentiate the integrand with respect to n: \(\int_0^1 nx^{n-1} \log(x) \, dx = -\frac{1}{(n+1)^2}\)

Solve the Integral

The integral for part a is now given by: \(\int_{0}^{1} x^{n} \log x \, dx = -\frac{1}{(n+1)^2}\) #Part b#

Differentiate the identity under the integral sign

First, we differentiate both sides of the given equation with respect to n repeatedly: \(\frac{d^n}{dn^n} \int_0^{\infty} e^{-ax} \, dx = \frac{d^n}{dn^n}(\frac{1}{a})\)

Apply the differentiation to the integral

Applying differentiation inside the integral multiple times, we get: \(\int_0^{\infty} \frac{d^n}{dn^n}(x^n e^{-ax}) \, dx = \frac{d^n}{dn^n}(\frac{1}{a})\)

Differentiate the integrand

Now, we differentiate the integrand repeatedly with respect to n: \(\int_0^{\infty} x^n e^{-ax} \, dx = \frac{n!}{a^{n+1}}\) #Part c#

Differentiate the identity under the integral sign

First, we differentiate both sides of the given equation with respect to n repeatedly: \(\frac{d^n}{dn^n} \int_0^{\infty} \frac{1}{x^2+y^2} \, dy = \frac{d^n}{dn^n}(\frac{1}{x^2})\)

Apply the differentiation to the integral

Applying differentiation inside the integral multiple times, we get: \(\int_0^{\infty} \frac{d^n}{dn^n}\left(\frac{1}{(x^2+y^2)^n}\right) \, dy = \frac{d^n}{dn^n}(\frac{1}{x^2})\)

Differentiate the integrand

Now, we differentiate the integrand repeatedly with respect to n: \(\int_0^{\infty} \frac{1}{(x^2+y^2)^n} \, dy = \frac{\pi}{2x^2} \frac{\Gamma(n-\frac{1}{2})}{\Gamma(n)}\) Where \(\Gamma(\cdot)\) is the gamma function.

Key concepts

Leibniz's Rule

Leibniz's Rule is a powerful technique used in calculus for differentiating integrals. It is especially useful when integrating a function that depends on a parameter. This rule allows us to differentiate under the integral sign, which can simplify complicated integrals.

In many cases, we deal with integrals where the integrand changes with a parameter. Suppose we have an integral in the form:

• \( F(n) = \int_{a}^{b} f(x, n) \, dx \)

Here, \( n \) is a parameter that affects the function \( f(x, n) \).

Leibniz's Rule states that we can differentiate \( F(n) \) with respect to \( n \) as follows:

• \( \frac{d}{dn} \int_{a}^{b} f(x, n) \, dx = \int_{a}^{b} \frac{\partial}{\partial n} f(x, n) \, dx \)

This transformation to move the derivative inside the integral can greatly simplify the evaluation. It is particularly useful for functions where a closed-form antiderivative is difficult to find.

This rule underlies the steps taken in the original solution for handling the integrals involving differentiation with respect to \( n \). It also makes it easier to handle problems with improper integrals by providing a systematic approach to derive solutions through differentiation.

Improper Integrals

Improper integrals arise when the bounds of integration are infinite or when the integrand becomes infinite within the range of integration. They often appear in calculus and analysis, especially when dealing with physical applications involving infinite domains.

Consider an integral with an infinite bound:

• \( \int_{0}^{\infty} f(x) \, dx \)

This integral is improper because the upper limit is \( \infty \).

To evaluate such an integral, one typically takes the limit:

• \( \int_{a}^{b} f(x) \, dx = \lim_{t \to b} \int_{a}^{t} f(x) \, dx \)

This approach ensures the integrals are computed over finite limits before taking the limit to infinity.

In addition, when the integrand has singularities, such as points where the function becomes infinite, the integral needs to be defined using limits as well. For example:

• \( \int_{0}^{1} \frac{1}{x^p} \, dx \quad (p > 1) \)

Integrals like these are assessed by finding a limit near the point of singularity.

The original exercise exploits improper integrals within the context of applications of Leibniz's Rule. By using differentiation and improper integrals, more complex expressions can be simplified to a form involving basic functions.

Gamma Function

The gamma function, denoted as \( \Gamma(n) \), is a fundamental concept in advanced mathematics, especially in calculus and analysis. It extends the factorial function to complex numbers. For positive integers, the gamma function is defined as:

• \( \Gamma(n) = (n-1)! \)

For complex and real numbers other than integers, it is defined by an improper integral:

• \( \Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} \, dx \quad (\text{for} \ n > 0) \)

Notice how it resembles the integrals in the exercise. This link shows the versatility of the Gamma function. It's highly useful for solving different types of improper integrals due to its extensive properties.

Another key property of \( \Gamma(n) \) is the recursion formula:

• \( \Gamma(n+1) = n \Gamma(n) \)

This recursive approach mirrors the factorial operation but extends it, providing solutions to a multitude of mathematical issues. It also facilitates solving problems that involve parameter-dependent integrals. This characteristic is clearly seen in the solution for part c, where the gamma function handles the complexity emerging from the parametric integration setup.

Overall, the gamma function serves as a bridge between simple factorial operations and more complex analyses in continuous domains.