Evaluating Limits Of Logarithmic Integrals A Step By Step Guide

Hey guys! Today, we're diving deep into a fascinating problem from the realm of calculus, specifically one that involves evaluating the limit of a logarithmic integral. This problem, which popped up in a MIT BEE exam, is a real head-scratcher, but fear not, we'll break it down step by step. We will explore the intricacies of limits, integration, and logarithmic functions, providing a comprehensive guide on how to approach and solve such challenges. Let's get started!

The Challenge: Unpacking the Limit

The problem at hand is to evaluate the following limit:

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

At first glance, this looks intimidating. We've got a limit as n approaches infinity, a logarithmic function with a base of n, and an integral nestled inside. It's like a mathematical Matryoshka doll! The key to cracking this is to systematically unpack each layer and simplify the expression. Understanding the behavior of the integral as n gets larger is crucial, as is knowing how logarithms interact with exponential growth. We'll start by dissecting the integral and then work our way outwards, using a combination of substitution, estimation, and limit properties. So, buckle up, it's going to be a fun ride!

My Initial Approach: A Trigonometric Tango (and Why It Didn't Quite Work)

When faced with integrals involving expressions like $(1 - x^3)$, a natural instinct might be to try a trigonometric substitution. The user initially attempted this approach, likely aiming to simplify the integrand and make it more manageable. The idea probably revolved around substituting $x = \sin^{2/3}(\theta)$ or a similar variant, hoping to leverage trigonometric identities. This substitution can be a good starting point because it transforms the $(1 - x^3)$ term into a trigonometric expression that might be easier to handle. However, while trigonometric substitutions are powerful tools, they don't always lead to a straightforward solution, especially when dealing with limits and integrals combined. Sometimes, the resulting trigonometric integrals can be just as complex, or even more so, than the original problem.

In this particular case, while the trigonometric substitution might simplify the integrand locally, it doesn't address the core issue of how the integral behaves as n approaches infinity. The complexity introduced by the substitution might obscure the dominant behavior of the function, making it difficult to evaluate the limit. Therefore, while the initial approach demonstrates a good problem-solving instinct (trying a common technique), it highlights the importance of recognizing when a particular method might not be the most efficient or effective. We need a strategy that directly tackles the limit and the integral's asymptotic behavior.

A More Strategic Approach: Focusing on the Integral's Behavior

Instead of diving into trigonometric substitutions, let's take a step back and think about what happens to the integral $\int_{0}^{1}(1-x{3})^{n}dx$ as n becomes very large. This is where the magic happens! The function $(1 - x^3)$ is always between 0 and 1 on the interval [0, 1]. As we raise it to higher and higher powers (n), it gets squeezed closer and closer to zero, except when x is very close to zero. This is a crucial observation.

Think about it: if x is, say, 0.5, then $(1 - x^3)$ is 0.875. Raise that to a large power, and it becomes tiny. But if x is 0.01, then $(1 - x^3)$ is very close to 1, and raising it to a large power still leaves it close to 1. This suggests that the integral's main contribution comes from the region near x = 0. This insight is the key to unlocking the problem. We can use this behavior to our advantage by making a clever substitution that zooms in on this region. The goal here is to find a way to approximate the integral as n goes to infinity, making the limit evaluation much more tractable.

The Substitution That Saves the Day: Unveiling the Asymptotic Behavior

The strategic move here is to make a substitution that focuses our attention on the region near x = 0. Let's use the substitution:

$$u = nx^{3}$$

Why this substitution? Notice that it scales the variable x by a factor of n, effectively magnifying the region near x = 0. As n gets large, this substitution stretches the interval of integration, allowing us to better analyze the integral's behavior in this crucial region. Let's see how this substitution transforms our integral. First, we need to find dx in terms of du:

$$du = 3nx^{2}dx$$

$$dx = \frac{du}{3nx^{2}}$$

Now, we need to express x in terms of u. From our substitution, we have:

$$x = \left(\frac{u}{n}\right)^{1/3}$$

Substituting this back into our expression for dx, we get:

$$dx = \frac{du}{3n\left(\frac{u}{n}\right)^{2/3}} = \frac{du}{3n^{1/3}u^{2/3}}$$

We also need to change the limits of integration. When x = 0, u = 0. When x = 1, u = n. So, our integral becomes:

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

This looks a bit messy, but we've made significant progress. The key now is to recognize a familiar limit within the integral.

Spotting a Familiar Limit: The Exponential Connection

Within our transformed integral, we have the expression $(1 - \frac{u}{n})^n$. This should ring a bell! This is a classic expression that converges to $e^{-u}$ as n approaches infinity. Remember the fundamental limit:

$$\lim_{n\to \infty} \left(1 + \frac{x}{n}\right)^n = e^x$$

In our case, x is -u. This means that as n gets very large, the term $(1 - \frac{u}{n})^n$ inside the integral becomes very close to $e^{-u}$. This is a crucial simplification because it replaces a complicated expression with a much more manageable exponential function. We're now in a position to approximate our integral and evaluate the original limit.

Approximating the Integral: Taming the Infinite

Using the limit we just identified, we can approximate our integral as n approaches infinity:

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

Now, a subtle but important point: as n goes to infinity, the upper limit of integration, n, also goes to infinity. So, we can further approximate the integral by extending the upper limit to infinity:

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

This step might seem a bit hand-wavy, but it's justified because the exponential term $e^{-u}$ decays very rapidly as u increases. This means that the contribution to the integral from large values of u is negligible. Extending the upper limit to infinity simplifies the calculation without significantly affecting the result. Now, we have a much simpler integral to deal with, and it looks suspiciously like something we might recognize...

Recognizing the Gamma Function: A Special Guest Star

The integral $\int_{0}^{\infty}e{-u}u^{-2/3}du$ is a special one. It's a close relative of the Gamma function, a generalization of the factorial function to complex numbers. The Gamma function is defined as:

$$\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt$$

Comparing our integral to the Gamma function definition, we see that it's equal to $\Gamma(1/3)$. Specifically, if we let z = 1/3, then z - 1 = -2/3, and we have exactly our integral. The Gamma function is a well-studied function with known values and properties. While we don't need to know the exact value of $\Gamma(1/3)$ for this problem, recognizing the Gamma function is a crucial step. It allows us to express our integral in a compact and meaningful way.

Putting It All Together: The Final Calculation

Now we're ready to assemble all the pieces and evaluate the original limit. We have:

$$\int_{0}^{1}(1-x^{3})^{n}dx \approx \frac{1}{3n^{1/3}}\Gamma(\frac{1}{3})$$

Plugging this back into our original limit expression, we get:

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

Now we can use the properties of logarithms to simplify this expression. Recall that $\log_{b}(xy) = \log_{b}(x) + \log_{b}(y)$ and $\log_{b}(x^a) = a\log_{b}(x)$. Applying these properties, we get:

$$\lim_{n\to \infty}\left[\log_{n}(\frac{\Gamma(\frac{1}{3})}{3}) + \log_{n}(n^{-1/3})\right]$$

The first term, $\log_{n}(\frac{\Gamma(\frac{1}{3})}{3})$, goes to zero as n approaches infinity because the logarithm of a constant (with respect to n) divided by $\log(n)$ goes to zero. The second term simplifies to:

$$\log_{n}(n^{-1/3}) = -\frac{1}{3}\log_{n}(n) = -\frac{1}{3}$$

Therefore, our final answer is:

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

Conclusion: Triumph Through Strategic Thinking

Whew! That was quite a journey, guys. We started with a daunting limit involving an integral and a logarithm, and we conquered it by strategically analyzing the integral's behavior, making a clever substitution, recognizing a familiar limit, and connecting it to the Gamma function. The key takeaway here isn't just the answer, but the process. We saw how crucial it is to:

• Understand the asymptotic behavior of functions.
• Make strategic substitutions to simplify integrals.
• Recognize familiar limits and special functions.
• Apply the properties of logarithms to manipulate expressions.

This problem is a testament to the power of mathematical thinking and problem-solving. So, the next time you encounter a challenging limit, remember these strategies, and you'll be well on your way to cracking it! Keep exploring, keep learning, and keep those mathematical gears turning!