Question

The family \(f(x)=\frac{1}{x^{p}}\) revisited Consider the family of functions \(f(x)=\frac{1}{x^{p}},\) where \(p\) is a real number. For what values of \(p\) does the integral \(\int_{0}^{1} f(x) d x\) exist? What is its value?

Short answer

Answer: The integral exists for \(p > 0\), and its value is infinity.

Step-by-step solution

Integrate by substitution

Let \(u = x^p\). Then, the integral becomes \(\int \frac{1}{u} du\). To find the limits for this integral, when \(x = 0\), we have \(u = 0^p = 0\) and when \(x = 1\), we have \(u = 1^p = 1\). Hence, the integral is: $$\int_{0}^{1} \frac{1}{x^p} dx = \int_{0}^{1} \frac{1}{u} du.$$

Integrate and determine the conditions for existence

Now, let's find the antiderivative of \(\frac{1}{u}\) and apply the Fundamental Theorem of Calculus: $$\int_{0}^{1} \frac{1}{u} du = \left[ \ln |u| \right]_0^1.$$ The integral will exist only if the limits exist. When \(u = 0\), \(\ln |u|\) will be undefined. So, we need to find a value for \(p\) that makes this log function well-defined at the endpoint \(u=0\). Recall that \(u = x^p\). As \(x \to 0^+\), \(u\) also goes to \(0^+\) if \(p > 0\). Therefore, we need \(p\) to be greater than zero for the integral to exist.

Evaluate the integral for valid values of \(p\)

Now that we know the integral exists for \(p > 0\), let's evaluate the integral for those values: $$\int_{0}^{1} \frac{1}{u} du = \left[ \ln |u| \right]_0^1 = \ln |1| - \ln ( \lim_{x \to 0^+} x^p ) = \ln 1 - \lim_{x \to 0+} \ln x^p.$$ Since \(\lim_{x \to 0+} \ln x^p = -\infty\), for \(p > 0\), we get: $$\int_{0}^{1} \frac{1}{x^p} dx = \ln 1 - (-\infty) = \boxed{\infty}.$$ So, the integral \(\int_{0}^{1} \frac{1}{x^p} dx\) exists for \(p > 0\), but its value is infinity.

Key concepts

Integration Techniques

When tackling any integral, particularly those categorized as improper integrals, it's crucial to select the correct integration techniques. These techniques allow us to simplify complex integrals or rewrite them in a form that can be easily managed. In the case of the integral presented, \( \int_{0}^{1} \frac{1}{x^p} dx \), we handle the potential issue of evaluating the integral at a point of discontinuity by performing a substitution. The substitution \( u = x^p \) transforms the original integral into one in terms of \( u \) which is easier to evaluate. This step showcases how substitution can help us deal with points where the function might not be defined, helping us to proceed further with the evaluation process.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is pivotal in evaluating definite integrals. It connects differentiation and integration, providing a way to evaluate the area under a curve. In simple terms, the theorem states that if you have a continuous function on an interval, the definite integral over that interval is equal to the antiderivative evaluated at the upper limit minus the antiderivative evaluated at the lower limit. For our function \( f(x) = \frac{1}{x^{p}} \), we search for an antiderivative which, in this case, leads to the natural logarithm function. However, it's crucial to ensure that the function is continuous in the interval, or at the very least, we can define a limit that gives us a finite area. This concept is the backbone behind step 2 of the provided solution, which probes into the existence of the integral by evaluating the antiderivative at the bounds of the integral.

Convergence of Integrals

Dealing with improper integrals often raises the question of convergence. An integral is said to converge if it results in a finite value as the limits approach their bounds; otherwise, it diverges. In the context of the exercise, the integral \( \int_{0}^{1} \frac{1}{x^p} dx \) converges only if the limits of integration lead to a definite, finite result. The solution steps showed a condition for \( p \) that makes the integral convergent, at least theoretically. However, upon closer inspection, it's discovered that even though the integral exists for \( p > 0 \), it actually diverges to infinity. This examination underscores the importance of thoroughly assessing the limits and the behavior of the function at these limits to determine the convergence of an integral.

Limits and Continuity

In calculus, the concepts of limits and continuity are tightly linked, especially when dealing with improper integrals. A limit describes the behavior of a function as the input approaches a certain value, while continuity requires that the function is defined and follows an unbroken path at that point. When we're confronted with values such as \( x = 0 \) in our function \( f(x) = \frac{1}{x^{p}} \), we can't plug this value directly due to a discontinuity. Hence, we explore the limit of the function as \( x \) approaches zero from the right (\( x \to 0^+ \)). The condition \( p > 0 \) ensures that the logarithm doesn't deal with a log of zero (which would be undefined), maintaining the requirement for this method of integration. Evaluating the limit carefully is key in determining both the existence and value of improper integrals, as demonstrated in the solution steps known as l'Hôpital's rule or various limit properties could have also been applied here to navigate around the discontinuity at \( x=0 \) for the original function.