Question

Find all values of \(p\) for which the integral converges (i) absolutely; (ii) conditionally. (a) \(\int_{0}^{1} x^{p} \sin 1 / x d x\) (b) \(\int_{0}^{1}|\log x|^{p} d x\) (c) \(\int_{1}^{\infty} x^{p} \cos (\log x) d x\) (d) \(\int_{1}^{\infty}(\log x)^{p} d x\) (e) \(\int_{0}^{\infty} \sin x^{p} d x\)

Short answer

a) The integral converges absolutely and conditionally for \(p > -1\). b) The integral converges absolutely and conditionally for \(p > 0\). c) The integral converges absolutely and conditionally for \(p < -1\). d) The integral converges absolutely and conditionally for \(p > -1\). e) The integral converges absolutely for all \(p\) and converges conditionally for \(p\ge 1\).

Step-by-step solution

a) \(\int_{0}^{1} x^{p} \sin \frac{1}{x} d x\)

First, we need to find whether the integral converges absolutely. To do that, check the absolute value of the function: $$\int_{0}^{1} \left| x^{p} \sin \frac{1}{x} \right| d x$$ Since \(x^{p}>0\) on the interval \((0,1)\), the function can also be written as \(x^{p} |\sin \frac{1}{x}|\). Now we apply the Comparison test. We know that: $$0 \le |\sin \frac{1}{x}| \le 1$$ Multiply all sides by \(x^{p}\): $$0 \le x^{p} |\sin \frac{1}{x}| \le x^{p}$$ Now integrate from \(0\) to \(1\): $$0 \le \int_{0}^{1} x^{p} |\sin \frac{1}{x}| d x \le \int_{0}^{1} x^{p} d x$$ The right integral converges when \(p > -1\), so the absolute convergence happens when \(p > -1\). To find the conditional convergence, we apply the Dirichlet's criterion for convergence. Assume \(f(x) = x^{p}\) and \(g(x) = \sin \frac{1}{x}\): * The integral of \(f(x)\) from \(0\) to \(1\) converges when \(p > -1\). * The integral of \(g'(x)\) has bounded variation in the interval \((0, 1)\). The derivative of \(g(x)\) is \(g'(x) = -\frac{1}{x^2} \cos \frac{1}{x}\). The variation of \(g'\) from \(0\) to \(1\) is bounded when \(p > -1\). Therefore, the integral converges conditionally when \(p > -1\). So both absolute convergence and conditional convergence occur when \(p > -1\).

b) \(\int_{0}^{1}|\log x|^{p} d x\)

To check if the integral converges absolutely, evaluate the integral of \(|\log x|^p\) from \(0\) to \(1\) to see if it is finite: $$\int_{0}^{1} |\log x|^p d x$$ Make the substitution \(t = -\ln{x}\), \({d}t = -\frac{1}{x}{d}x\): $$\int_{1}^{\infty} p\log(t)^{(p-1)} dt$$ This integral converges when \(p-1 > -1\); that is, when \(p>0\). So the absolute convergence occurs when \(p > 0\). Since any positive \(p\) makes the integrand non-negative in \((0, 1)\), the conditional convergence is the same as the absolute convergence. Thus, it converges conditionally when \(p > 0\).

c) \(\int_{1}^{\infty} x^{p} \cos (\log x) d x\)

Let's first check for the absolute convergence: $$\int_{1}^{\infty} |x^{p} \cos{(\log x)}| d x \le \int_{1}^{\infty} |x^{p}| d x$$ The integral on the right converges only when \(p<-1\). Therefore, the absolute convergence occurs when \(p<-1\). To check for the conditional convergence, we will use the Dirichlet's test. Let \(f(x) = x^{p}\) and \(g(x) = \cos (\log x)\). Then: * The integral of \(f(x)\) from \(1\) to \(\infty\) converges when \(p < -1\). * The integral of \(g'(x)\) has a bounded variation in the interval \((1, \infty)\). The derivative of \(g(x)\) is \(g'(x) = -\frac{1}{x} \sin(\log x)\). The integral converges conditionally when \(|\int g'(x) d_x|\) has a bounded variation. So the conditional convergence occurs when \(p < -1\).

d) \(\int_{1}^{\infty}(\log x)^{p} d x\)

For absolute convergence, we consider the absolute value of the integrand function: $$\int_{1}^{\infty} |(\log x)^{p}| dx = \int_{1}^{\infty} (\log x)^{p} dx$$ Since \(\log(x) > 0\) for \(x>1\), we find that the integral converges when \(p>-1\). Therefore, the absolute convergence occurs when \(p > -1\). Similarly, since the integrand is non-negative, the conditional convergence is the same as the absolute convergence. So it converges conditionally when \(p > -1\).

e) \(\int_{0}^{\infty} \sin x^{p} d x\)

To find the absolute convergence values, consider the integral of \(|\sin(x^{p})|\) from \(0\) to \(\infty\): $$\int_{0}^{\infty} |\sin x^{p}| dx$$ Since \(|\sin x^{p}| \le 1\), the integral is finite for all \(p\in \mathbb{R}\). Therefore, the absolute convergence occurs for all \(p\). For the conditional convergence, we evaluate the original integral: $$\int_{0}^{\infty} \sin x^{p} dx$$ The integral converges conditionally when \(p\ge 1\). Therefore, the conditional convergence occurs when \(p\ge 1\). So, in summary: (a) The integral converges absolutely and conditionally for \(p > -1\). (b) The integral converges absolutely and conditionally for \(p > 0\). (c) The integral converges absolutely and conditionally for \(p < -1\). (d) The integral converges absolutely and conditionally for \(p > -1\). (e) The integral converges absolutely for all \(p\) and converges conditionally for \(p\ge 1\).

Key concepts

Absolute Convergence

Absolute convergence ensures that an integral of the absolute value of a function is finite. Consider an integral of a function \( f(x) \). To determine absolute convergence, we evaluate \( \int |f(x)| dx \). If this integral is finite, then \( f(x) \) is absolutely convergent. Absolute convergence is quite strong because when a function converges absolutely, it also converges in the regular sense. For instance, if \( \int_{0}^{1} x^{p} \sin \frac{1}{x} dx \) needs to be checked for absolute convergence, consider \( \int_{0}^{1} x^{p} |\sin \frac{1}{x}| dx \). By employing comparison tests, and knowing \(|\sin \frac{1}{x}| \leq 1\), we can determine the parameters \( p \) for which the integral converges to a finite number. This approach applies broadly to any definite integral, and absolute convergence often requires careful consideration of how the absolute value impacts the function's behavior over its integration range.

Conditional Convergence

Conditional convergence happens when an integral converges but does not do so absolutely. It implies that while the integral \( \int f(x) dx \) is finite, \( \int |f(x)| dx \) is not. This situation might initially seem contradictory because the integral still yields a finite value without the absolute value requirement.The significance of conditionally convergent integrals lies in their sensitivity to changes in limits or segments. A conditionally convergent function relies on a delicate balance between positive and negative areas. For example, the integral \( \int_{1}^{\infty} x^{p} \cos (\log x) dx \) involves the oscillating function \( \cos (\log x) \), contributing to the conditional nature of convergence. Analyzing the convergence involves checking if \( \int f(x) dx \) alone (without absolute values) converges, often using tests like Dirichlet's test to assess the convergence under specific conditions.

Convergence Tests

Convergence tests are pivotal tools that help determine if an integral (or series) converges or diverges. Many techniques are available depending on the nature of the function involved, and each can provide insights into the behavior of the integral.

• Comparison Test: This involves comparing the given function to another function whose convergence behavior is already known. If \( |f(x)| \leq g(x) \) and \( \int g(x) \) converges, so does \( \int f(x) \).
• Dirichlet's Test: Often used when dealing with oscillating functions like sine or cosine. It utilizes the properties of integration where \( f'(x) \) has bounded variation, guaranteeing the convergence if the integral of \( f(x) \) itself decreases to zero.
• Integral Test: This method is closely aligned with the behavior of infinite series. It uses the integral of the function over an interval to determine convergence similar to the convergence of a series.

Each test serves distinct scenarios in analyzing integrals, helping reveal whether the integral remains finite over specific values or ranges and guiding us in understanding the behavior under varying conditions.