91Ó°ÊÓ

Improper integrals occur when either the interval of integration is infinite or the function being integrated has a vertical asymptote within the interval. In the problem given, the integral \(\int_{0}^{\pi/2} \tan \theta \, d\theta\) is improper because \(\tan \theta\) becomes undefined at \(\theta = \pi/2\). This is due to \(\cos \theta\) approaching zero, which makes \(\tan \theta\) approach infinity. To handle such cases, we substitute the upper limit with a variable that approaches the problematic point, evaluating it as a limit. This turns the problem into \(\lim_{b \to \pi/2^-} \int_{0}^{b} \tan \theta \, d\theta\). This approach helps in assessing whether the integral converges to a specific value (converges) or grows without bound (diverges). In this context, convergence implies that as we approach the boundary point, the integral settles at a finite number. Divergence indicates otherwise, that the integral grows infinitely large as it nears the problematic point.

The Direct Comparison Test is a method used to determine the convergence or divergence of an improper integral by comparing it with a second, more familiar integral. The idea is simple: if a function \(f(x)\) is known and another function \(g(x)\) is related such that \(0 \leq f(x) \leq g(x)\) within the interval, then:

• If \(\int g(x) \, dx\) converges, then \(\int f(x) \, dx\) also converges.
• If \(\int f(x) \, dx\) diverges, then \(\int g(x) \, dx\) also diverges.

In our scenario, we set \(f(\theta) = \tan \theta\) and selected \(g(\theta) = \frac{1}{\cos \theta}\). Since \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and as \(\theta\) increases towards \(\pi/2\), \(\sin \theta\) approaches 1, making these functions comparable. On evaluating \(\int_{0}^{\pi/2} \frac{1}{\cos \theta} \, d\theta\), we find that it diverges. Since \(\tan \theta\) and \(\frac{1}{\cos \theta}\) are comparable and the latter diverges, we conclude based on the Direct Comparison Test that \(\int_{0}^{\pi/2} \tan \theta \, d\theta\) also diverges.

The Limit Comparison Test is a handy tool for determining the convergence of an integral by comparing it with another integral that is easier to evaluate. The test involves calculating the limit of the ratio of two functions. Let's say we want to compare \(f(x)\) and \(g(x)\). We compute the following:\[\lim_{x \to c} \frac{f(x)}{g(x)} = L,\]where \(L\) is a finite number and \(c\) is the point of interest, such as infinity or where the function becomes undefined. If \(L\) exists and is positive, then both \(\int f(x) \, dx\) and \(\int g(x) \, dx\) either both converge or both diverge.Although the Limit Comparison Test was not directly used in our specific exercise, it's possible to employ this as an alternative method if the Direct Comparison Test encounters issues or doesn't provide a clear conclusion. The integral comparison in this test offers flexibility, especially when selecting comparison functions that simplify evaluations while keeping the focus on the relationship between the behaviors of \(f(x)\) and \(g(x)\). For our given integral, if an alternative comparison such as \(c \cdot \frac{1}{\cos \theta}\) results in a manageable limit, we can infer results on the convergence of \(\int_{0}^{\pi/2} \tan \theta \, d\theta\).