Evaluating The Limit Of A Logarithmic Integral Expression

This article delves into the intricate evaluation of the limit:

$$\lim_{n\to \infty}\log_{n}\left(\int_{0}^{1}(1-x^{3})^{n}dx\right)$$

This fascinating limit, originating from the MIT BEE examination, presents a compelling challenge that necessitates a multifaceted approach, drawing upon concepts from calculus and limit theory. We will explore various techniques, providing a comprehensive understanding of the solution. By the end of this discussion, you will have a solid grasp of how to tackle such problems and appreciate the beauty of mathematical problem-solving.

Initial Exploration and Challenges

When confronted with this limit, a natural first step is to understand the behavior of the integral $\int_{0}^{1}(1-x{3})^{n}dx$ as n approaches infinity. The integrand, $(1-x^{3})^{n}$, is a function that decays rapidly as n increases, especially for x away from 0. This suggests that the main contribution to the integral comes from the region near x = 0. However, directly evaluating this integral is not straightforward, and we need to employ suitable approximation techniques. Trigonometric substitutions, as the user initially attempted, might seem promising, but they often lead to complex expressions that are difficult to handle. Instead, we will explore alternative methods, such as substitution and the use of the Gamma function, to simplify the integral.

The Significance of the Problem

Problems of this nature are significant in mathematical analysis because they test our ability to combine different concepts and techniques. Evaluating limits involving integrals often requires a deep understanding of asymptotic behavior, special functions, and approximation methods. Moreover, such problems frequently appear in competitive examinations, highlighting their importance in assessing mathematical proficiency. The challenge lies not only in finding the correct answer but also in developing a rigorous and clear solution. In the following sections, we will dissect the problem step by step, providing detailed explanations and justifications for each step.

A Strategic Approach to Solving the Limit

To effectively tackle this limit, we need a strategic plan. Our approach can be broken down into several key steps:

1. Simplify the Integral: The primary challenge is to find a manageable expression for the integral $\int_{0}^{1}(1-x{3})^{n}dx$. We'll achieve this by using a suitable substitution to transform the integral into a form that can be related to the Gamma function.
2. Relate to the Gamma Function: The Gamma function is a generalization of the factorial function to complex numbers and appears frequently in integral evaluations. We'll use the properties of the Gamma function to express our integral in terms of Gamma functions.
3. Apply Asymptotic Approximations: As n approaches infinity, we'll need to use asymptotic approximations to simplify the expressions involving Gamma functions. Stirling's approximation is a powerful tool for this purpose.
4. Evaluate the Logarithmic Limit: Finally, we'll substitute the simplified expression back into the original limit and evaluate it using logarithmic properties and limit rules.

By following this structured approach, we can systematically solve the problem and gain a deeper understanding of the underlying mathematical concepts. Each step is crucial, and we'll provide detailed explanations to ensure clarity and rigor.

Step 1: Simplifying the Integral Using Substitution

The first critical step in evaluating the limit is to simplify the integral:

$$\int_{0}^{1}(1-x^{3})^{n}dx$$

A suitable substitution can help transform this integral into a more manageable form. Let's consider the substitution:

$$u = x^{3}$$

This implies that:

$$x = u^{1/3}$$

and

$$dx = \frac{1}{3}u^{-2/3}du$$

When x = 0, u = 0, and when x = 1, u = 1. Thus, the limits of integration remain the same. Substituting these into the integral, we get:

$$\int_{0}^{1}(1-x^{3})^{n}dx = \int_{0}^{1}(1-u)^{n}\frac{1}{3}u^{-2/3}du = \frac{1}{3}\int_{0}^{1}(1-u)^{n}u^{-2/3}du$$

This substitution is crucial because it transforms the integral into a form that can be related to the Beta function, which is closely connected to the Gamma function. The Beta function is defined as:

$$B(x, y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$$

Comparing this with our transformed integral, we can see a clear resemblance. This connection will be exploited in the next step.

Importance of Choosing the Right Substitution

The choice of substitution is often the key to solving complex integrals. In this case, substituting u = x³ was strategic because it simplified the integrand into a form that involves a power of (1-u) and a power of u, which aligns perfectly with the structure of the Beta function. Other substitutions might not lead to such a simplification, highlighting the importance of experience and intuition in selecting the appropriate technique. By recognizing the form of the Beta function, we can proceed to express the integral in terms of Gamma functions, paving the way for further simplification.

Step 2: Relating the Integral to the Gamma Function

Now that we have simplified the integral using substitution, the next step is to express it in terms of the Gamma function. Recall that we transformed the integral into:

$$\frac{1}{3}\int_{0}^{1}(1-u)^{n}u^{-2/3}du$$

This integral closely resembles the Beta function, which is defined as:

$$B(x, y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$$

The Beta function can be expressed in terms of the Gamma function as:

$$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

Comparing our integral with the Beta function, we can identify:

$$x - 1 = -\frac{2}{3} \implies x = \frac{1}{3}$$

$$y - 1 = n \implies y = n + 1$$

Thus, our integral can be written in terms of the Beta function as:

$$\frac{1}{3}\int_{0}^{1}(1-u)^{n}u^{-2/3}du = \frac{1}{3}B\left(\frac{1}{3}, n+1\right)$$

Using the relationship between the Beta and Gamma functions, we get:

$$\frac{1}{3}B\left(\frac{1}{3}, n+1\right) = \frac{1}{3}\frac{\Gamma(\frac{1}{3})\Gamma(n+1)}{\Gamma(n+\frac{4}{3})}$$

We know that Γ(n+1) = n!, so we have:

$$\frac{1}{3}\frac{\Gamma(\frac{1}{3})n!}{\Gamma(n+\frac{4}{3})}$$

This expression is a significant step forward. We have successfully transformed the integral into a form involving the Gamma function and the factorial function. The next step will involve using asymptotic approximations to simplify this expression as n approaches infinity.

The Role of Special Functions

Special functions like the Gamma and Beta functions play a crucial role in evaluating integrals and solving various mathematical problems. The Gamma function, in particular, is a versatile tool that appears in many areas of mathematics and physics. By recognizing the connection between our integral and these special functions, we were able to make significant progress in simplifying the expression. This highlights the importance of being familiar with the properties and applications of these functions in mathematical problem-solving.

Step 3: Applying Asymptotic Approximations (Stirling's Approximation)

With the integral now expressed in terms of the Gamma function, we need to tackle the limit as n approaches infinity. This is where asymptotic approximations become crucial. Specifically, we will use Stirling's approximation, a powerful tool for approximating the Gamma function (and factorials) for large values of n. Stirling's approximation states that:

$$\Gamma(z) \approx \sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^{z} \text{ as } |z| \to \infty$$

For our purposes, we can also write Stirling's approximation for the factorial function as:

$$n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}$$

We have the expression:

$$\frac{1}{3}\frac{\Gamma(\frac{1}{3})n!}{\Gamma(n+\frac{4}{3})}$$

Applying Stirling's approximation to n! and Γ(n + 4/3), we get:

$$n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}$$

$$\Gamma\left(n+\frac{4}{3}\right) \approx \sqrt{\frac{2\pi}{n+\frac{4}{3}}}\left(\frac{n+\frac{4}{3}}{e}\right)^{n+\frac{4}{3}}$$

Substituting these approximations into our expression, we have:

$$\frac{1}{3}\frac{\Gamma(\frac{1}{3})\sqrt{2\pi n}(\frac{n}{e})^{n}}{\sqrt{\frac{2\pi}{n+\frac{4}{3}}}(\frac{n+\frac{4}{3}}{e})^{n+\frac{4}{3}}}$$

Now, let's simplify this expression. We can rewrite it as:

$$\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\sqrt{\frac{n(n+\frac{4}{3})}{1}}\frac{(\frac{n}{e})^{n}}{(\frac{n+\frac{4}{3}}{e})^{n+\frac{4}{3}}}$$

$$\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\sqrt{n\left(n+\frac{4}{3}\right)}\frac{n^{n}e^{n+\frac{4}{3}}}{e^{n}\left(n+\frac{4}{3}\right)^{n+\frac{4}{3}}}$$

$$\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\sqrt{n^{2}+\frac{4}{3}n}\frac{n^{n}e^{\frac{4}{3}}}{\left(n+\frac{4}{3}\right)^{n+\frac{4}{3}}}$$

This simplified form is much easier to handle when evaluating the limit as n approaches infinity. The next step is to substitute this expression back into the original logarithmic limit and evaluate it.

The Power of Asymptotic Methods

Asymptotic methods, such as Stirling's approximation, are indispensable tools in analyzing the behavior of functions as their arguments tend to extreme values. In this case, Stirling's approximation allowed us to approximate the Gamma function and the factorial function, which are otherwise difficult to handle directly. These approximations are not exact, but they become increasingly accurate as n becomes large, making them invaluable for evaluating limits. Understanding and applying asymptotic methods is a crucial skill in advanced mathematical analysis.

Step 4: Evaluating the Logarithmic Limit

Now that we have approximated the integral, we can substitute the result back into the original limit and evaluate it. Recall the original limit:

$$\lim_{n\to \infty}\log_{n}\left(\int_{0}^{1}(1-x^{3})^{n}dx\right)$$

We approximated the integral as:

$$\int_{0}^{1}(1-x^{3})^{n}dx \approx \frac{1}{3}\Gamma\left(\frac{1}{3}\right)\sqrt{n^{2}+\frac{4}{3}n}\frac{n^{n}e^{\frac{4}{3}}}{\left(n+\frac{4}{3}\right)^{n+\frac{4}{3}}}$$

Substituting this into the limit, we get:

$$\lim_{n\to \infty}\log_{n}\left(\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\sqrt{n^{2}+\frac{4}{3}n}\frac{n^{n}e^{\frac{4}{3}}}{\left(n+\frac{4}{3}\right)^{n+\frac{4}{3}}}\right)$$

To simplify this limit, we can use logarithmic properties. Recall that \log_b(xy) = log_b(x) + log_b(y) and log_b(x^r) = rlog_b(x). Applying these properties, we have:

$$\lim_{n\to \infty}\left[\log_{n}\left(\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\right) + \log_{n}\left(\sqrt{n^{2}+\frac{4}{3}n}\right) + \log_{n}\left(\frac{n^{n}e^{\frac{4}{3}}}{\left(n+\frac{4}{3}\right)^{n+\frac{4}{3}}}\right)\right]$$

Now, let's analyze each term separately. The first term is:

$$\lim_{n\to \infty}\log_{n}\left(\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\right) = 0$$

Because the argument of the logarithm is a constant, and the logarithm to the base n of a constant goes to 0 as n goes to infinity.

The second term is:

$$\lim_{n\to \infty}\log_{n}\left(\sqrt{n^{2}+\frac{4}{3}n}\right) = \lim_{n\to \infty}\frac{1}{2}\log_{n}\left(n^{2}+\frac{4}{3}n\right) = \lim_{n\to \infty}\frac{1}{2}\log_{n}\left[n^{2}\left(1+\frac{4}{3n}\right)\right]$$

$$= \lim_{n\to \infty}\frac{1}{2}\left[\log_{n}(n^{2}) + \log_{n}\left(1+\frac{4}{3n}\right)\right] = \lim_{n\to \infty}\frac{1}{2}\left[2 + \log_{n}\left(1+\frac{4}{3n}\right)\right] = 1$$

Because log_n(1 + 4/(3n)) approaches 0 as n approaches infinity.

The third term is the most complex:

$$\lim_{n\to \infty}\log_{n}\left(\frac{n^{n}e^{\frac{4}{3}}}{\left(n+\frac{4}{3}\right)^{n+\frac{4}{3}}}\right) = \lim_{n\to \infty}\left[\log_{n}(n^{n}e^{\frac{4}{3}}) - \log_{n}\left(\left(n+\frac{4}{3}\right)^{n+\frac{4}{3}}\right)\right]$$

$$= \lim_{n\to \infty}\left[n\log_{n}(n) + \frac{4}{3}\log_{n}(e) - \left(n+\frac{4}{3}\right)\log_{n}\left(n+\frac{4}{3}\right)\right]$$

$$= \lim_{n\to \infty}\left[n + \frac{4}{3}\log_{n}(e) - \left(n+\frac{4}{3}\right)\log_{n}\left(n\left(1+\frac{4}{3n}\right)\right)\right]$$

$$= \lim_{n\to \infty}\left[n + \frac{4}{3}\log_{n}(e) - \left(n+\frac{4}{3}\right)\left(\log_{n}(n) + \log_{n}\left(1+\frac{4}{3n}\right)\right)\right]$$

$$= \lim_{n\to \infty}\left[n + \frac{4}{3}\log_{n}(e) - \left(n+\frac{4}{3}\right)\left(1 + \log_{n}\left(1+\frac{4}{3n}\right)\right)\right]$$

$$= \lim_{n\to \infty}\left[n + \frac{4}{3}\log_{n}(e) - n - n\log_{n}\left(1+\frac{4}{3n}\right) - \frac{4}{3} - \frac{4}{3}\log_{n}\left(1+\frac{4}{3n}\right)\right]$$

$$= \lim_{n\to \infty}\left[\frac{4}{3}\log_{n}(e) - n\log_{n}\left(1+\frac{4}{3n}\right) - \frac{4}{3} - \frac{4}{3}\log_{n}\left(1+\frac{4}{3n}\right)\right]$$

As n approaches infinity, log_n(1 + 4/(3n)) approaches 0. Using the approximation log_n(1 + x) ≈ x/ln(n) for small x, we have:

$$\lim_{n\to \infty} n\log_{n}\left(1+\frac{4}{3n}\right) = \lim_{n\to \infty} n\frac{\frac{4}{3n}}{\ln(n)} = \lim_{n\to \infty} \frac{\frac{4}{3}}{\ln(n)} = 0$$

Thus, the third term simplifies to:

$$\lim_{n\to \infty}\left[\frac{4}{3}\log_{n}(e) - 0 - \frac{4}{3} - 0\right] = -1$$

Combining all the terms, we have:

$$\lim_{n\to \infty}\left[0 + 1 + (-1)\right] = 0$$

Therefore, the final answer is 0.

Importance of Logarithmic Properties and Limit Rules

Evaluating the logarithmic limit required a careful application of logarithmic properties and limit rules. We used the properties of logarithms to break down the complex expression into simpler terms, and we used limit rules to evaluate each term separately. The strategic use of approximations, such as Stirling's approximation and the logarithmic approximation, was also crucial. This step highlights the importance of mastering these fundamental tools in calculus and limit theory.

Conclusion

The limit we set out to evaluate, $\lim_{n\to \infty}\log_{n}\left(\int_{0}^{1}(1-x{3})^{n}dx\right)$, is a complex problem that demands a comprehensive understanding of calculus, limit theory, and special functions. By systematically breaking down the problem into smaller steps, we were able to find a solution. We used substitution to simplify the integral, related the integral to the Gamma function, applied Stirling's approximation to handle the limit as n approaches infinity, and finally, used logarithmic properties and limit rules to evaluate the final limit. The result is 0.

This problem serves as an excellent example of how various mathematical concepts can come together to solve a challenging problem. It also emphasizes the importance of strategic problem-solving, careful application of techniques, and a deep understanding of fundamental principles. The journey through this problem has not only provided us with a solution but also enhanced our mathematical intuition and problem-solving skills.