Question

Evaluate the integrals. If the integral diverges, answer "diverges."\(\int_{-\infty}^{0} \frac{d x}{x^{2}+1}\)

Short answer

The integral converges to \( \frac{\pi}{2} \).

Step-by-step solution

Identify the Type of Integral

We are given the integral \( \int_{-\infty}^{0} \frac{d x}{x^{2}+1} \). This is an improper integral because the lower limit of integration is \(-\infty\). We need to determine if it converges or diverges.

Rewrite the Integral with a Limit

To handle the improper integral, replace \(-\infty\) with a limit \(a\) and take the limit as \(a\) approaches \(-\infty\):\[ \lim_{a \to -\infty} \int_{a}^{0} \frac{d x}{x^2 + 1}. \]

Compute the Integral

The integral \( \int \frac{d x}{x^2 + 1} \) is a standard form that results in \( \tan^{-1}(x) + C \). Therefore, evaluating our limit gives us:\[ \lim_{a \to -\infty} \left[ \tan^{-1}(x) \right]_{a}^{0}. \]

Evaluate the Definite Integral

Substitute the limits of integration into the antiderivative:\[ \lim_{a \to -\infty} \left( \tan^{-1}(0) - \tan^{-1}(a) \right). \] Since \( \tan^{-1}(0) = 0 \), it simplifies to \( - \tan^{-1}(a) \).

Determine the Convergence of the Limit

Analyze the limit \( \lim_{a \to -\infty} - \tan^{-1}(a) \). As \( a \) approaches \(-\infty\), \( \tan^{-1}(a) \) approaches \(-\frac{\pi}{2} \), so \(- \tan^{-1}(a)\) approaches \(\frac{\pi}{2}\).

Conclusion

Since the limit results in a finite value \( \frac{\pi}{2} \), we conclude that the integral converges to \( \frac{\pi}{2} \).

Key concepts

Convergence

Improper integrals often have limits of integration that extend to infinity, leading to potential infinite values. The primary goal with these integrals is to determine if they "converge" to a finite number, or "diverge" to infinity. In our given exercise, we started with an integral from \(-\infty\) to 0, which means it has an infinite lower bound. An integral that converges means its value settles down to a specific finite number as we approach these bounds.

• To determine convergence, we transform the original integral into a limit problem.
• We replace the infinite limit with a variable, say \(a\), and then see what happens as \(a\) approaches the infinite boundary.

In our example, the improper integral converged to \( \frac{\pi}{2} \). When solving improper integrals, remember:- If the limit results in a finite number, the integral converges.- If the limit does not exist or goes to infinity, the integral diverges.

Limit Process

The limit process is a powerful tool in calculus used to evaluate improper integrals. When you encounter an integral with infinite limits, the limit process can turn it into a solvable form. Here's how it applies to the integral in our exercise:1. **Set up the Limit:** The integral \( \int_{-\infty}^{0} \frac{d x}{x^2 + 1} \) is not immediately solvable because of its infinite lower bound. By setting it as a limit \( \lim_{a \to -\infty} \int_{a}^{0} \frac{d x}{x^2 + 1} \), we manage this.

2. **Solve the Integral:** Evaluate the integral \( \int \frac{d x}{x^2 + 1} \) to get the antiderivative \( \tan^{-1}(x) \).

3. **Apply the Limits:** Substitute the bounds within the limit: \( \lim_{a \to -\infty} ( \tan^{-1}(0) - \tan^{-1}(a) ) \). The known values help finalize the solution.By applying this process, you're essentially converting an unsolvable infinite problem into a finite one through limits.

Definite Integrals

Definite integrals have both an upper and lower limit, and their result represents the area under the curve of a function over a specific interval. They are often easier to compute than improper ones due to their finite intervals. In the context of the exercise, we first needed to transform the improper integral into a form that resembles a definite integral. This was done using the limit process, which allowed us to evaluate the specific bounds.

• The integral \( \int_{a}^{0} \frac{d x}{x^2 + 1} \) resembled a definite integral between finite points.
• The expression for a definite integral is usually \( F(b) - F(a) \) where \( F \) is an antiderivative of the integrand.

See here, the direct computation of \( \tan^{-1}(x) \) from \(a\) to \(0\) required plugging in limits, showcasing a definitive evaluation process in action.Ultimately, by utilizing the steps of computing the antiderivative and evaluating boundaries, we demonstrated how improper integrals can be treated similarly to definite ones, within the context of well-applied limits.