Question

Find all values of \(p\) for which the integral converges. (a) \(\int_{0}^{\pi / 2} \frac{\sin x}{x^{p}} d x\) (b) \(\int_{0}^{\pi / 2} \frac{\cos x}{x^{p}} d x\) (c) \(\int_{0}^{\infty} x^{p} e^{-x} d x\) (d) \(\int_{0}^{\pi / 2} \frac{\sin x}{(\tan x)^{p}} d x\) (e) \(\int_{1}^{\infty} \frac{d x}{x(\log x)^{p}}\) (f) \(\int_{0}^{1} \frac{d x}{x(|\log x|)^{p}}\) (g) \(\int_{0}^{\pi} \frac{x d x}{(\sin x)^{p}}\)

Short answer

Based on the step-by-step solution, provide the values of \(p\) for which the following integrals converge. (a) For the integral \(\int_{0}^{\pi / 2} \frac{\sin x}{x^{p}} d x\), it converges when \(p<2\). (b) For the integral \(\int_{0}^{\pi / 2} \frac{\cos x}{x^{p}} d x\), it converges when \(p>1\). (c) For the integral \(\int_{0}^{\infty} x^{p} e^{-x} d x\), it converges when \(p>-1\). (d) For the integral \(\int_{0}^{\pi / 2} \frac{\sin x}{(\tan x)^{p}} d x\), it converges when \(p<2\). (e) For the integral \(\int_{1}^{\infty} \frac{d x}{x(\log x)^{p}}\), it converges when \(p>1\). (f) For the integral \(\int_{0}^{1} \frac{d x}{x(|\log x|)^{p}}\), it converges when \(p<1\). (g) For the integral \(\int_{0}^{\pi} \frac{x d x}{(\sin x)^{p}}\), it converges when \(p>1\).

Step-by-step solution

Apply the comparison test for integral convergence

Using the comparison test, we want to compare our given integral with another integral whose convergence is known. Near \(x=0\), we know that \(\sin x \sim x\). So we can compare the given integral with \(\int_{0}^{\pi / 2} \frac{x}{x^{p}} d x\).

Simplify the reference integral and apply conditions for convergence

Simplifying the reference integral, we get \(\int_{0}^{\pi / 2} x^{1-p} d x\). This integral converges if \(1-p > -1\), i.e., \(p<2\). Thus, for the given integral to converge, we must have \(p<2\). For part (b): \(\int_{0}^{\pi / 2} \frac{\cos x}{x^{p}} d x\)

Apply the comparison test for integral convergence

Using the comparison test, we want to compare our given integral with another integral whose convergence is known. Near \(x=0\), we know that \(\cos x \sim 1\). So, we can compare the given integral with \(\int_{0}^{\pi / 2} \frac{1}{x^{p}} d x\).

Apply conditions for convergence

The integral \(\int_{0}^{\pi / 2} \frac{1}{x^{p}} d x\) converges if \(p > 1\). Thus, for the given integral to converge, we must have \(p > 1\). For part (c): \(\int_{0}^{\infty} x^{p} e^{-x} d x\) This integral belongs to gamma function form. We can directly compute the condition.

Convergence condition for gamma function integral

The gamma function integral converges if \(p > -1\). So, for the given integral to converge, we must have \(p>-1\). For the rest of the integrals, we will follow a similar approach. For part (d): \(\int_{0}^{\pi / 2} \frac{\sin x}{(\tan x)^{p}} d x\) Upon comparing with \(\int_{0}^{\pi / 2} \frac{\sin x}{x^{p}} d x\), we find that the integral converges for \(p < 2\). For part (e): \(\int_{1}^{\infty} \frac{d x}{x(\log x)^{p}}\) Upon comparing with \(\int_{1}^{\infty} \frac{d x}{x x^{p}} d x\), we find that the integral converges for \(p > 1\). For part (f): \(\int_{0}^{1} \frac{d x}{x(|\log x|)^{p}}\) Considering the integral converges if \(p < 1\). For part (g): \(\int_{0}^{\pi} \frac{x d x}{(\sin x)^{p}}\) Upon comparing with \(\int_{0}^{\pi} \frac{x d x}{x^{p}} d x\), we find that the integral converges for \(p > 1\). In summary, we have the following values of \(p\) for which the integrals converge: (a) \(p<2\) (b) \(p>1\) (c) \(p>-1\) (d) \(p<2\) (e) \(p>1\) (f) \(p<1\) (g) \(p>1\)

Key concepts

Comparison Test

Integral convergence is a widespread concern in calculus, especially when dealing with improper integrals. To ascertain whether an integral converges, one powerful tool at our disposal is the Comparison Test.

This mathematical litmus test allows us to determine the convergence of an integral by comparing it to another integral with known behavior. Essentially, if we can show that our integral of interest is less than or equal to a converging integral, then our integral also converges. Conversely, if our integral is greater than or equal to a diverging integral, it too will diverge.

For instance, consider the integral \(\int_{0}^{\pi / 2} \frac{\sin x}{x^{p}} d x\). Near \(x=0\), where most concerns of convergence arise, we can approximate \(\sin x \approx x\). This approximation lets us compare our integral to \(\int_{0}^{\pi / 2} x^{1-p} d x\), simplifying our convergence investigation.

However, knowing when to employ such comparison often depends on recognizing the behavior of functions within the integral near potential points of trouble, such as \(x=0\) or as \(x\) approaches infinity. Mastery of the Comparison Test is therefore pivotal in the toolbox of anyone wrestling with the convergence of integrals.

Gamma Function

The Gamma Function holds significant importance in various branches of mathematics and science because of its relationship with factorials and its flexibility in handling complex integration problems.

Expressed as \(\Gamma (z)\) for a complex number \(z\), it is defined as: \(\Gamma (z) = \int_{0}^{\infty} x^{z-1} e^{-x} d x\), where the real part of \(z\) has to be greater than zero for the integral to converge.

Taking the example from our exercise, \(\int_{0}^{\text{\infty}} x^{p} e^{-x} d x\) fits the form of the Gamma Function, where the integral will converge provided that \(p > -1\). This condition stems from the definition of the Gamma Function and ensures the exponential factor \(e^{-x}\) adequately dampens the integrable function as \(x\) goes to infinity, thus securing convergence.

The Gamma Function is a wonderful example of how seemingly complicated integrals can be tamed using established mathematical functions. Recognizing integrals that fit into the Gamma Function framework can save valuable time and effort when determining their convergence.

Convergence Conditions

When faced with the task of determining the convergence of integrals, one must pay close attention to the Convergence Conditions that apply. These conditions are guidelines to ascertain whether an integral will yield a finite value (converge) or not (diverge).

In the context of improper integrals, convergence conditions often boil down to the behavior of the integrand at the endpoints of integration. For example, when integrating over an unbounded domain, the decay rate of the integrand as it approaches infinity is a crucial factor. If the rate of decay is not swift enough (for instance, if the exponent of \(x\) in \(x^p\) does not sufficiently counterbalance the 'growth' of \(x\)), the integral will diverge.

Similarly, considering integrals with singularities, it is essential to scrutinize the behavior near the point of singularity. As we discovered with the integrals in the original exercise, conditions such as \(p<2\) emerged from evaluating the growth or decay rates of the integrand near the troublesome points.

Ultimately, understanding and applying convergence conditions is pivotal for anyone seeking to navigate the complex terrain of integrals. It is this comprehension that enables us to distinguish between the finite and the infinite, thereby deepening our understanding of calculus and its applications.