Question

Compute (a) \(\lim _{t \rightarrow 0+} \int_{t}^{1} \frac{1}{x} \mathrm{dx}\) (b) \(\lim _{t \rightarrow 1-} \int_{0}^{t} \tan \frac{\pi}{2} x d x\). How does the result give insight into the fact that neither integrand is integrable over the interval \([0,1] ?\)

Short answer

As t approaches 0+ in part (a), the integral of (1/x) approaches infinity, and as t approaches 1- in part (b), the integral of (tan(π/2)x) also approaches infinity. This indicates that the functions (1/x) and (tan(π/2)x) have infinite areas near the endpoints 0 and 1, respectively, and are therefore not integrable over the interval [0, 1] as they do not have finite areas over this interval.

Step-by-step solution

Find the integral

Evaluate the integral \(\int_{t}^{1} \frac{1}{x} \mathrm{dx}\). To do this, recognize that the antiderivative of \(\frac{1}{x}\) is \(\ln|x|\). So, we get: \(\int_{t}^{1} \frac{1}{x} \mathrm{dx} = \ln|x| \Big|_{t}^{1} = \ln(1) - \ln(t) = -\ln(t)\).

Compute the limit

Now, we'll find the limit as \(t \rightarrow 0+\): \(\lim _{t \rightarrow 0+} -\ln(t)\). Since the natural logarithm function (\(\ln\)) approaches \(-\infty\) as its argument approaches 0 from the right, we have: \(\lim _{t \rightarrow 0+} -\ln(t) = +\infty\). #Part (b)#

Find the integral

Evaluate the integral \(\int_{0}^{t} \tan \frac{\pi}{2} x d x\). To do this, recognize that the antiderivative of \(\tan \frac{\pi}{2} x\) is \(-\frac{2}{\pi} \ln | \cos (\frac{\pi}{2}x)|\). So, we get: \(\int_{0}^{t} \tan \frac{\pi}{2} x d x = -\frac{2}{\pi}\ln| \cos (\frac{\pi}{2}x)| \Big|_{0}^{t} = -\frac{2}{\pi}\ln| \cos (\frac{\pi}{2}t)| + \frac{2}{\pi}\ln(\cos(0)) = -\frac{2}{\pi}\ln(\cos(\frac{\pi}{2}t))\).

Compute the limit

Now, we'll find the limit as \(t \rightarrow 1-\): \(\lim_{t \rightarrow 1-} -\frac{2}{\pi}\ln(\cos(\frac{\pi}{2}t))\). Since the cosine function approaches 0 as its argument approaches \(\frac{\pi}{2}\) from the left (since \(\cos(\frac{\pi}{2}) = 0\)), the natural logarithm term approaches \(-\infty\). Therefore, we have: \(\lim_{t \rightarrow 1-} -\frac{2}{\pi}\ln(\cos(\frac{\pi}{2}t)) = +\infty\).

Discuss the insights

The results from both parts show that the integrals approach infinity as t approaches 0 from the right (in part (a)) and 1 from the left (in part (b)). This means that the functions \(\frac{1}{x}\) and \(\tan \frac{\pi}{2} x\) have infinite areas (or, rather, they "blow up") near the endpoints 0 and 1, respectively. This insight tells us that neither of these functions is integrable over the interval \([0, 1]\) since they do not have finite areas over this interval.

Key concepts

Limits of Integration

In integral calculus, limits of integration define the interval over which we are integrating a function. They pinpoint where the integration starts and ends, usually denoted by the lower and upper limits on the integral sign. For example, if we want to integrate a function f(x) from a to b, we write \[ \int_{a}^{b} f(x) \, dx. \] These limits are crucial because they can radically affect the result of the integral. In some cases, as a approaches a certain value, it's possible that the integral tends toward infinity, which suggests that the function is not integrable over that range. This is often encountered in improper integrals, where one or both limits of integration approach a finite number or infinity.

Antiderivative

An antiderivative of a function f(x) is another function whose derivative is f(x). It represents the most general form of integrals. There is a family of functions that are antiderivatives of f(x), all of which differ by a constant. When we compute the definite integral of a function, we are essentially finding the difference between the antiderivatives evaluated at the upper and lower limits of integration. For continuous functions, these are guaranteed to exist. However, when the function becomes unbounded near a point, as with \( \frac{1}{x} \) as it nears zero, the antiderivative can lead to an infinite result when evaluated at that point.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a fundamental mathematical function that is the inverse of the exponential function e^x. It has unique properties, with one of the most prominent being its behavior near zero; \( \ln(x) \) tends towards negative infinity as x approaches zero from the right. This is depicted in the integrals evaluated in the original exercise with \( \frac{1}{x} \) and emphasizes that the natural logarithm is an essential tool when dealing with growth processes or decay in various scientific fields. It plays a pivotal role in solving integrals involving 1/x, like those in improper integrals where a limit needs to be taken due to an undefined or infinite boundary.

Improper Integrals

An improper integral occurs when one or both limits of integration are infinite, or the integrand (function being integrated) becomes infinite within the limits of integration. To evaluate an improper integral, we usually take the limit of a related integral with finite bounds. As displayed in the original exercise, limits are calculated as the bounds approach the point where the integrand becomes unbounded or infinity. If the limit exists and is finite, we consider the improper integral to be convergent; otherwise, it is divergent. This illustrates the concept that the area under the curve of certain functions can be infinite, challenging the conventional concept of integrals always representing an area or volume.

Integrability

Integrability refers to the possibility of computing the integral of a function in a given interval. A critical criterion for integrability is that the function must be bounded within the interval. Functions that have certain types of discontinuities or that go to infinity within the interval may not have a finite integral, hence they are not integrable over that range. The original exercise brings to light such a scenario where the functions \( \frac{1}{x} \) and \( \tan \frac{\pi}{2} x \) are not integrable over the interval [0,1] due to their behavior at the endpoints. Their respective areas 'blow up,' indicating an infinite accumulation of value in a finite space, which in practical terms, means these functions are not integrable over the specified interval.