Evaluate Limit Log_n(integral (1-x^3)^n Dx) As N Approaches Infinity

This article delves into the intricate process of evaluating the limit of a logarithmic integral expression, a problem encountered in the MIT BEE competition. We will explore a step-by-step solution, providing a comprehensive understanding of the techniques involved. The specific limit we aim to evaluate is:

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

Initial Approaches and Challenges

When first encountering such a limit, several approaches might come to mind. One common technique is to attempt simplification using trigonometric substitutions. However, this method often leads to complex expressions that are difficult to manage, especially within the context of a limit as n approaches infinity. Direct integration also poses a challenge, as the integrand

$(1-x^3)^n$

becomes increasingly complex with larger values of n. Therefore, a more strategic approach is required to tackle this problem effectively.

A Strategic Approach: Substitution and the Gamma Function

To navigate this challenging limit, we employ a strategic combination of substitution and the Gamma function. The initial step involves a clever substitution that simplifies the integral and allows us to express it in a more manageable form. Specifically, we substitute:

$u = x^3$

This substitution transforms the integral, making it more amenable to evaluation. Let's delve into the step-by-step process of this transformation.

Step 1: Performing the Substitution

With the substitution $u = x^3$, we also have $x = u^{1/3}$. Differentiating both sides with respect to x, we get:

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

Which can be rewritten as:

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

Now, we need to change the limits of integration. When $x = 0$, $u = 0^3 = 0$, and when $x = 1$, $u = 1^3 = 1$. Thus, the integral transforms as follows:

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

This substitution has successfully transformed our integral into a form that involves the product of a power of $(1-u)$ and a power of $u$, which is highly suggestive of the Beta function.

Step 2: Relating to the Beta Function

The Beta function, denoted by $B(x, y)$, is defined as:

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

Comparing our transformed integral with the definition of the Beta function, we can see a clear connection. To express our integral in terms of the Beta function, we need to match the exponents. In our integral, we have $(1-u)^n$ and $u^{-2/3}$. Thus, we can identify the following correspondences:

$y - 1 = n$ and $x - 1 = -\frac{2}{3}$

Solving these equations, we find:

$y = n + 1$ and $x = \frac{1}{3}$

Therefore, our integral can be expressed in terms of the Beta function as:

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

Step 3: Expressing the Beta Function in terms of the Gamma Function

The Beta function is closely related to the Gamma function, denoted by $\Gamma(z)$. The relationship is given by:

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

Using this relationship, we can express our integral in terms of the Gamma function:

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

Recall that $\Gamma(n+1) = n!$ for positive integers n. Thus, we have:

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

Now, we have successfully expressed the integral in terms of the Gamma function and factorials, which is a crucial step in evaluating the limit.

Evaluating the Limit

Now that we have simplified the integral, we can focus on evaluating the original limit:

$$\lim_{n\to \infty}\log_{n}\left(\int_{0}^{1}(1-x^{3})^{n}dx\right) = \lim_{n\to \infty}\log_{n}\left(\frac{1}{3}\frac{\Gamma(\frac{1}{3})n!}{\Gamma(n+\frac{4}{3})}\right)$$

To handle the Gamma functions in the limit, we can use Stirling's approximation, which provides an asymptotic formula for the Gamma function for large arguments.

Step 4: Applying Stirling's Approximation

Stirling's approximation states that for large z:

$\Gamma(z) \approx \sqrt{\frac{2\pi}{z}}\left(\frac{z}{e}\right)^{z}$

Applying Stirling's approximation to $\Gamma(n+\frac{4}{3})$, we get:

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

Also, we can use the approximation $n! \approx \sqrt{2\pi n}(\frac{n}{e})^n$ which is also derived from Stirling's approximation.

Substituting these approximations into our expression, we have:

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

Step 5: Simplifying the Expression

Let's simplify the expression inside the logarithm. We can rewrite the expression as:

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

Now, consider the term:

$$\frac{n^n}{(n+\frac{4}{3})^{n+\frac{4}{3}}} = \frac{n^n}{(n+\frac{4}{3})^n(n+\frac{4}{3})^{\frac{4}{3}}} = \frac{1}{(1+\frac{4}{3n})^n(n+\frac{4}{3})^{\frac{4}{3}}}$$

As $n \to \infty$, $(1+\frac{4}{3n})^n$ approaches $e^{\frac{4}{3}}$. Thus, our expression becomes:

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

Taking the logarithm base n, we have:

$$\log_{n}\left(\frac{1}{3}\Gamma(\frac{1}{3})\frac{n}{\frac{4}{3}}\right) = \log_{n}\left(\frac{1}{3}\Gamma(\frac{1}{3})n^{1-\frac{4}{3}}\right) = \log_{n}\left(\frac{\Gamma(\frac{1}{3})}{3}\right)+\log_{n}(n) - \log_{n}(n^{\frac{4}{3}})$$

Step 6: Evaluating the Final Limit

Now, we can rewrite our limit as:

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

As $n \to \infty$, $\frac{\log(\frac{\Gamma(\frac{1}{3})}{3})}{\log(n)}$ goes to 0.

Therefore, the limit simplifies to:

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

Conclusion

Through a combination of strategic substitution, the application of the Beta and Gamma functions, and the use of Stirling's approximation, we have successfully evaluated the limit:

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

This problem highlights the power of these mathematical tools in tackling complex limit problems, especially those involving integrals and special functions. The step-by-step approach outlined in this article provides a clear and comprehensive understanding of the techniques involved, making it a valuable resource for anyone studying advanced calculus and limit evaluation.