Question

Evaluating an Improper Integral over an Infinite Interval Evaluate \(\int_{-\infty}^{0} \frac{1}{x^{2}+4} d x\). State whether the improper integral converges or diverges.

Short answer

The integral converges to \(\frac{\pi}{4}\).

Step-by-step solution

Recognize the Improper Integral

The given integral is improper because it has an infinite limit of integration. Specifically, it extends from \(-\infty\) to 0, which means the lower limit is not finite.

Set Up the Limit Definition

To evaluate the improper integral, we express it as a limit:\[\lim_{a \to -\infty} \int_{a}^{0} \frac{1}{x^{2}+4} \, dx.\]This redefines the integral over a finite interval \([a,0]\), where \(a\) approaches \(-\infty\).

Evaluate the Indefinite Integral

Calculate the indefinite integral \(\int \frac{1}{x^2 + 4} \, dx\). Use the formula for the integral of the form \(\frac{1}{x^2 + a^2}\) which is \(\frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\). Here, \(a\) is 2.Thus, \[\int \frac{1}{x^2 + 4} \, dx = \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C.\]

Apply the Limits of Integration

Substitute the limits of integration into the antiderivative:\[\lim_{a \to -\infty} \left[ \frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) \right]_{a}^{0}.\]This evaluates to:\[\lim_{a \to -\infty} \left( \frac{1}{2} \tan^{-1}\left(0\right) - \frac{1}{2} \tan^{-1}\left(\frac{a}{2}\right) \right).\]

Calculate the Definite Integral

Compute the value using the properties of \(\tan^{-1}\):\(\tan^{-1}(0) = 0\).As \(a\) approaches \(-\infty\), \(\tan^{-1}\left(\frac{a}{2}\right)\) approaches \(-\frac{\pi}{2}\).Thus we have:\[\lim_{a \to -\infty} \left( 0 - \frac{1}{2} \times (-\frac{\pi}{2}) \right) = \frac{\pi}{4}.\]

Conclusion: Determine Convergence

Since the limit equals \(\frac{\pi}{4}\), the improper integral converges to this finite value.

Key concepts

Understanding Calculus in Improper Integrals

Calculus is a branch of mathematics that helps us understand changes. It consists of differentiation and integration. Specifically, integration is vital when dealing with areas under a curve or cumulative sums.
In improper integrals, calculus allows us to evaluate integrals with infinite limits or unbounded functions. This means we deal with limits that extend to infinity or involve discontinuities.
A familiar tool in these evaluations is the limit, which lets us redefine improper integrals as a limit of definite integrals. This approach turns the problem into something more manageable.
In essence, calculus provides the toolkit for handling integrals that go beyond the conventional boundaries, enabling us to find actual values or ascertain whether they extend infinitely.

Dealing with Infinite Intervals

The concept of infinite intervals appears frequently in improper integrals. These occur when the range of integration extends to infinity, either positively or negatively.
An example is when we need to integrate a function from \(-\infty\) to a certain point. Since it's impossible to reach infinity, we use limits to approach it. This transforms the integral into a limit problem.
This approach allows us to evaluate these types of integrals by progressively considering larger and larger finite intervals. Only then can we determine whether they have a finite sum or not, reflecting the behavior of the function over that range.
Infinite intervals challenge us to think beyond standard bounds, literally stretching our mathematical thinking to beyond finite limits.

The Role of Convergence

Convergence is crucial in understanding improper integrals. It tells us whether an integral reaches a specific, finite value or not.
An integral converges if its value approaches a fixed number as the limits extend to infinity. If not, it diverges, indicating that the integral does not settle on a finite value.
This determination relies heavily on the behavior of the function and the interval being considered. Calculus provides methods, such as comparison tests, to help decide this.

• If the integral's limit exists and is finite, it's said to converge.
• If the limit doesn't exist or is not finite, it diverges.

Understanding convergence is essential, as it ensures that the calculations have meaningful and usable results. Without convergence, the integral could imply infinite or undefined values, which are important considerations in calculus.