Problem 406 - Calculus Volume 1

Question

Explain what is wrong with the following integral: $$\int_{-1}^{1} \frac{d t}{|t| \sqrt{t^{2}-1}}$$

Step-by-Step Solution

Answer

The integrand is undefined within \([-1, 1]\), making the integral impossible to evaluate.

Step 1: Identify the Domain of the Integrand

The integrand is \( \frac{1}{|t| \sqrt{t^2 - 1}} \). We first need to determine where this expression is defined. The term \( \sqrt{t^2 - 1} \) implies that \( t^2 - 1 \geq 0 \). Solving this inequality, we get \( t^2 \geq 1 \). This means \( t \leq -1 \) or \( t \geq 1 \). Since the integration limits are from \(-1\) to \(1\), it excludes these conditions, resulting in the function not defined within the interval.

Step 2: Analyze |t| in the Denominator

The presence of \(|t|\) in the denominator indicates that the integrand is undefined when \( t = 0 \). This means that the point \( t = 0 \) is a point of discontinuity within the integration limits, further complicating the possibility of evaluating this integral.

Step 3: Conclusion

Because the integrand is not defined within the interval \([-1, 1]\), and the presence of both a discontinuity at \( t = 0 \) and the undefined term \( \sqrt{t^2 - 1} \) in the interval, the integral cannot be evaluated as a standard definite integral.

Key Concepts

Understanding the Domain of the Integrand

The domain of an integrand refers to the set of all possible values of the variable for which the integrand (the function being integrated) is defined.
For the given integral, the integrand is \( \frac{1}{|t| \sqrt{t^2 - 1}} \). We need to check where this expression holds valid values without discrepancies.

• The term \( \sqrt{t^2 - 1} \) implies that we need \( t^2 - 1 \geq 0 \), leading to \( t^2 \geq 1 \). Therefore, \( t \leq -1 \) or \( t \geq 1 \). This means the function is not defined between -1 and 1, which is problematic because those are the limits of our integral.
• Additionally, \( |t| \) in the denominator requires \( t eq 0 \), because anything divided by zero is undefined.

When calculating definite integrals, it is crucial for the integrand to be defined over the entire interval of integration. Here, that condition is not satisfied, causing issues in computing the integral.

Discontinuity in the Integrand

Discontinuity refers to points where a function is not continuous, often causing breaks or jumps in the graph of the function. In integrals, discontinuities within the range of integration can lead to undefined behavior.
In this exercise, the integrand \( \frac{1}{|t| \sqrt{t^2 - 1}} \) has a notable point of discontinuity at \( t = 0 \). This is because the absolute value term \( |t| \) becomes zero, rendering the fraction undefined.

• Since the interval runs from \(-1\) to \(1\), \( t \) will pass through 0, creating a discontinuity right within the integration limits.
• This discontinuity means that the integral cannot be evaluated in the standard way because the function isn't smoothly behaving across the interval.

In summary, discontinuities introduce complexities in definite integrals and often require methods like limit approximation or redefined intervals to be handled properly.

The Nature of Definite Integrals

A definite integral calculates the net area under a curve within a specified range. It's tied to the idea of accumulation and typically requires the integrand to be smooth and continuous over the interval.
In the context of the exercise, the definite integral \( \int_{-1}^{1} \frac{d t}{|t| \sqrt{t^{2}-1}} \), smoothness and continuity are compromised.

• The integrand is not defined over the given interval \([-1, 1]\) due to its domain and discontinuity issues.
• Definite integrals, therefore, rely on the precise adherence to these conditions for correct evaluation.
• In improper cases like this, either the interval of integration must be adjusted, or the integral considered within a more complex framework to account for the undefined nature.

Therefore, in problems like these, understanding when standard definite integral techniques fail is essential for applying the correct mathematical approaches.