Math Problem Statement

\lim _{x\to \infty }\left(\int _0^{\pi :}:\left(t/\left(3\cdot ::cos^2\left(x\cdot ::t\right)+1\right)\right)dt\right)

Solution

We are tasked with evaluating the following limit:

limx(0πt3cos2(xt)+1dt)\lim _{x \to \infty} \left(\int_0^\pi \frac{t}{3 \cos^2(x \cdot t) + 1} \, dt\right)

Step-by-step breakdown:

  1. Expression Analysis: The integral involves a term cos2(xt)\cos^2(x \cdot t), which oscillates as xx \to \infty. For large xx, the cosine function rapidly oscillates for any non-zero tt, and thus we expect the integral to become an average behavior over one period of oscillation.

  2. Rough Intuition for the Integral:

    • The denominator 3cos2(xt)+13 \cos^2(x \cdot t) + 1 oscillates between values:
      • 3(0)2+1=13(0)^2 + 1 = 1 when cos(xt)=0\cos(x \cdot t) = 0,
      • 3(1)2+1=43(1)^2 + 1 = 4 when cos(xt)=±1\cos(x \cdot t) = \pm 1.
    • As xx \to \infty, the integral will smooth out because of the oscillations of cos(xt)\cos(x \cdot t). The denominator's value will, on average, be dominated by the mean behavior of cos2(xt)\cos^2(x \cdot t).
  3. Averaging the Function: The average value of cos2(xt)\cos^2(x \cdot t) over any interval of length π\pi is 12\frac{1}{2}, because: cos2(xt)=1π0πcos2(xt)dt=12.\langle \cos^2(x \cdot t) \rangle = \frac{1}{\pi} \int_0^\pi \cos^2(x \cdot t) \, dt = \frac{1}{2}. Hence, for large xx, the denominator can be approximated by: 3cos2(xt)+1=312+1=32+1=52.3 \langle \cos^2(x \cdot t) \rangle + 1 = 3 \cdot \frac{1}{2} + 1 = \frac{3}{2} + 1 = \frac{5}{2}.

  4. Final Approximation: The integral for large xx can be approximated by: 0πt52dt=250πtdt.\int_0^\pi \frac{t}{\frac{5}{2}} \, dt = \frac{2}{5} \int_0^\pi t \, dt. The integral of tt from 0 to π\pi is: 0πtdt=π22.\int_0^\pi t \, dt = \frac{\pi^2}{2}. Therefore, the integral becomes: 25π22=π25.\frac{2}{5} \cdot \frac{\pi^2}{2} = \frac{\pi^2}{5}.

Conclusion:

Thus, the value of the limit is: limx(0πt3cos2(xt)+1dt)=π25.\lim _{x \to \infty} \left( \int_0^\pi \frac{t}{3 \cos^2(x \cdot t) + 1} \, dt \right) = \frac{\pi^2}{5}.


Would you like further details on any specific part of the solution? Here are some related questions for further exploration:

  1. How do rapid oscillations affect integrals like this one?
  2. What is the average value of trigonometric functions like cos2(xt)\cos^2(x \cdot t) over different intervals?
  3. Can we apply a similar approach to other integrals with oscillatory components?
  4. How do limits of integrals involving oscillatory functions behave when the frequency tends to infinity?
  5. What techniques are commonly used for handling integrals involving rapidly oscillating terms?

Tip: When dealing with oscillatory integrals, averaging techniques can be particularly helpful for finding the long-term behavior as the frequency tends to infinity.

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Math Problem Analysis

Mathematical Concepts

Limit
Integral Calculus
Oscillatory Functions
Averaging Techniques

Formulas

\int_0^\pi t \, dt = \frac{\pi^2}{2}
\langle \cos^2(x \cdot t) \rangle = \frac{1}{2}

Theorems

Limit of oscillatory integrals
Averaging of periodic functions

Suitable Grade Level

College/Advanced High School