Evaluating The Limit Of Logarithmic Integral Expression

by ADMIN 56 views
Iklan Headers

In this article, we will delve into the evaluation of a challenging limit problem that involves a logarithmic function and a definite integral. Specifically, we aim to find the value of the following limit:

lim⁑nβ†’βˆžlog⁑n(∫01(1βˆ’x3)ndx)\lim_{n\to \infty}\log_{n}\left(\int_{0}^{1}(1-x^{3})^{n}dx\right)

This intriguing limit has appeared in the MIT BEE (presumably a mathematics competition or exam), and it presents a fascinating exercise in calculus and limit evaluation techniques. The problem requires a blend of integral calculus, limit manipulation, and potentially some clever substitutions or approximations. The goal is to unravel the behavior of the integral as nn approaches infinity and then to apply the logarithmic function to determine the ultimate limit. Let's embark on a journey to dissect this problem and arrive at a solution.

One might initially attempt to tackle this problem using standard integral calculus techniques, such as direct integration or trigonometric substitutions. However, the presence of the exponent 'nn' within the integrand (1βˆ’x3)n(1-x^3)^n makes direct integration quite challenging. As nn grows large, the function (1βˆ’x3)n(1-x^3)^n rapidly approaches zero for most values of xx in the interval [0,1][0, 1], except for values close to x=0x=0. This behavior suggests that we might need to employ techniques that focus on the behavior of the integrand near x=0x=0. Trigonometric substitutions, while sometimes useful for simplifying integrals, don't seem to offer a straightforward path to simplification in this particular case. Therefore, alternative approaches, such as approximation methods or the use of special functions, may be necessary to effectively evaluate this limit. Let's explore some of these avenues to unravel the intricacies of this problem.

To evaluate the limit, we can begin by focusing on the integral part of the expression. Let's define the integral as follows:

In=∫01(1βˆ’x3)ndxI_n = \int_{0}^{1}(1-x^{3})^{n}dx

The key to solving this problem lies in understanding how this integral behaves as nn approaches infinity. As nn becomes very large, the term (1βˆ’x3)n(1-x^3)^n approaches zero rapidly for x>0x > 0. This suggests that the primary contribution to the integral comes from the region very close to x=0x = 0. To exploit this behavior, we can make a suitable substitution that magnifies the region near x=0x = 0.

Let's make the substitution:

u=x3u = x^3

Then, $x = u^{1/3}$ and $dx = \frac{1}{3}u^{-2/3}du$

When x=0x = 0, u=0u = 0, and when x=1x = 1, u=1u = 1. Thus, the integral InI_n can be rewritten in terms of uu as follows:

In=∫01(1βˆ’u)n13uβˆ’2/3du=13∫01(1βˆ’u)nuβˆ’2/3duI_n = \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

Now, this integral resembles the Beta function. Recall that the Beta function is defined as:

B(m,n)=∫01tmβˆ’1(1βˆ’t)nβˆ’1dtB(m, n) = \int_{0}^{1}t^{m-1}(1-t)^{n-1}dt

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

mβˆ’1=βˆ’23β€…β€ŠβŸΉβ€…β€Šm=13m - 1 = -\frac{2}{3} \implies m = \frac{1}{3}

nβˆ’1=nβ€…β€ŠβŸΉβ€…β€ŠTheΒ exponentΒ ofΒ (1βˆ’u)Β shouldΒ beΒ n+1,Β soΒ weΒ haveΒ aΒ slightΒ mismatch.n - 1 = n \implies \text{The exponent of }(1-u)\text{ should be } n+1, \text{ so we have a slight mismatch.}

To align our integral with the Beta function, we can rewrite InI_n as:

In=13∫01u13βˆ’1(1βˆ’u)ndu=13B(13,n+1)I_n = \frac{1}{3}\int_{0}^{1}u^{\frac{1}{3}-1}(1-u)^{n}du = \frac{1}{3}B\left(\frac{1}{3}, n+1\right)

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

B(m,n)=Ξ“(m)Ξ“(n)Ξ“(m+n)B(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}

Therefore,

In=13Ξ“(13)Ξ“(n+1)Ξ“(n+43)I_n = \frac{1}{3}\frac{\Gamma(\frac{1}{3})\Gamma(n+1)}{\Gamma(n+\frac{4}{3})}

Now we need to analyze the behavior of InI_n as nn approaches infinity. We can use the property of the Gamma function:

Ξ“(z+1)=zΞ“(z)\Gamma(z+1) = z\Gamma(z)

Applying this property repeatedly, we can write:

Ξ“(n+43)=(n+13)(nβˆ’23)β‹―(43)Ξ“(43)\Gamma(n+\frac{4}{3}) = \left(n+\frac{1}{3}\right)\left(n-\frac{2}{3}\right)\cdots\left(\frac{4}{3}\right)\Gamma(\frac{4}{3})

For large nn, we can approximate the Gamma function ratio using Stirling's approximation or by considering the leading terms:

Ξ“(n+1)Ξ“(n+43)β‰ˆn!(n+13)(nβˆ’23)β‹―(43)Ξ“(43)β‰ˆnβˆ’13\frac{\Gamma(n+1)}{\Gamma(n+\frac{4}{3})} \approx \frac{n!}{\left(n+\frac{1}{3}\right)\left(n-\frac{2}{3}\right)\cdots\left(\frac{4}{3}\right)\Gamma(\frac{4}{3})} \approx n^{-\frac{1}{3}}

Thus, for large nn, we have:

Inβ‰ˆ13Ξ“(13)nβˆ’13I_n \approx \frac{1}{3}\Gamma\left(\frac{1}{3}\right)n^{-\frac{1}{3}}

Now we can substitute this approximation back into the original limit expression:

lim⁑nβ†’βˆžlog⁑n(∫01(1βˆ’x3)ndx)=lim⁑nβ†’βˆžlog⁑n(13Ξ“(13)nβˆ’13)\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}\Gamma\left(\frac{1}{3}\right)n^{-\frac{1}{3}}\right)

Using the properties of logarithms, we can rewrite the expression as:

lim⁑nβ†’βˆžlog⁑n(13Ξ“(13)nβˆ’13)=lim⁑nβ†’βˆž[log⁑n(13Ξ“(13))+log⁑n(nβˆ’13)]\lim_{n\to \infty}\log_{n}\left(\frac{1}{3}\Gamma\left(\frac{1}{3}\right)n^{-\frac{1}{3}}\right) = \lim_{n\to \infty}\left[\log_{n}\left(\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\right) + \log_{n}\left(n^{-\frac{1}{3}}\right)\right]

lim⁑nβ†’βˆž[log⁑n(13Ξ“(13))βˆ’13log⁑n(n)]\lim_{n\to \infty}\left[\log_{n}\left(\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\right) - \frac{1}{3}\log_{n}(n)\right]

Since log⁑n(n)=1\log_{n}(n) = 1, we have:

lim⁑nβ†’βˆž[log⁑n(13Ξ“(13))βˆ’13]\lim_{n\to \infty}\left[\log_{n}\left(\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\right) - \frac{1}{3}\right]

As nn approaches infinity, the term log⁑n(13Ξ“(13))\log_{n}\left(\frac{1}{3}\Gamma\left(\frac{1}{3}\right)\right) approaches 0 because the argument of the logarithm is a constant. Therefore, the limit is:

lim⁑nβ†’βˆžlog⁑n(∫01(1βˆ’x3)ndx)=βˆ’13\lim_{n\to \infty}\log_{n}\left(\int_{0}^{1}(1-x^{3})^{n}dx\right) = -\frac{1}{3}

In conclusion, we have successfully evaluated the given limit by employing a combination of integral calculus techniques, substitution methods, and approximations involving the Beta and Gamma functions. The key steps included recognizing the dominant contribution to the integral near x=0x = 0, making a suitable substitution to transform the integral into a form involving the Beta function, and then using properties of the Gamma function to approximate the behavior of the integral as nn approaches infinity. Finally, applying the properties of logarithms, we arrived at the result:

lim⁑nβ†’βˆžlog⁑n(∫01(1βˆ’x3)ndx)=βˆ’13\lim_{n\to \infty}\log_{n}\left(\int_{0}^{1}(1-x^{3})^{n}dx\right) = -\frac{1}{3}

This problem exemplifies the power of combining different mathematical tools and techniques to tackle challenging problems in calculus and analysis. It showcases the importance of understanding the behavior of integrals and special functions in the limit as variables tend to infinity. The journey to the solution has been a testament to the beauty and intricacy of mathematical problem-solving.