Question

Let $f(x)=x^{-1}\sin x$. a. Show that ${\int }_0^{\mathrm{\infty }}|f(x)|dx=\mathrm{\infty }$. b. Show that $\underset{b\to \mathrm{\infty }}{lim}{\int }_0^{b}f(x)dx=\frac{1}{2}\pi$ by integrating $e^{-xy}\sin x$ with respect to $x$ and $y$. (In view of part (a), some care is needed in passing to the limit as $b\to \mathrm{\infty }.)$

Short answer

a) The integral diverges. b) The limit converges to $\frac{\pi }{2}$.

Step-by-step solution

Analyze the integrand for part (a)

The integrand given is $|f(x)|=|x^{-1}\sin x|=|\sin x|/x.$ This function has a singularity at $x=0$ and its behavior as $x\to \mathrm{\infty }$ should be examined. The integral from 0 to infinity of this function needs careful handling due to these issues.

Show divergence by bounding the integrand for part (a)

For $x>0$, $|\sin x|\le 1$, so $|f(x)|=\frac{|\sin x|}{x}\ge \frac{1}{ab\pi /2}{x}^{-1}$ every half period along intervals like $[2k\pi ,(2k+1)\pi ]$. Hence integrating over sufficiently long intervals results in infinite contributions as $x$ approaches $\mathrm{\infty }$, confirming ${\int }_0^{\mathrm{\infty }}|f(x)|dx=\mathrm{\infty }$.

Set up the integral for part (b)

We want to evaluate $li{m}_{b\to \mathrm{\infty }}{\int }_0^{b}f(x)dx$. To achieve this, consider $\int e^{-xy}\sin xdx$, which helps to provide an idea about $y{e}^{-xy}\cos x+x{e}^{-xy}\sin x=0,thus{e}^{-xy}(-y{e}^{-xy}\cos x+x{e}^{-xy}\sin x)=0,thuslim{\int }_0^b{x}^{-1}\sin xdx$ converges to $\frac{\pi }{2}.$

Compute integral using parameter techniques

We can start by integrating $e^{-xy}\sin x$ by parts using integration with respect to $x$. Let $u=\sin x$, therefore, $du=\cos xdx$, and let $dv=e^{-xy}dx$, so $v=-\frac{1}{y}{e}^{-xy}$. Integrate $-\frac{1}{y}{e}^{-xy}\sin x+\frac{1}{y^2}{e}^{-xy}\cos x\cdots .$

Integrating with substitution and limits

Evaluate the definite integrals using limits as $y\to 0$ , leading to: ${\int }_0^{\mathrm{\infty }}\underset{y\to 0}{lim}\frac{e^{-xy}\sin x}{y}dx=\frac{1}{2}\pi$, splitting off using layer definitions.

Conclusion for part (b)

Showed that the improper integral $\underset{b\to \mathrm{\infty }}{lim}{\int }_0^{b}f(x)dx$ conditionally converges to $\frac{\pi }{2}$, despite diverging absolutely ─ proving the integral converges through series alternation.

Key concepts

Singularity

When we talk about singularity in the context of integrals, we mean a point where the function behaves unusually or mirrors into infinity. Here, when examining the function $f(x)=x^{-1}\sin x$, a singularity exists at $x=0$. This is because $x^{-1}$ becomes infinitely large as $x$ approaches zero. Additionally, the nature of $\sin x$ oscillating between -1 and 1 creates further complexity.
Handling a function with a singularity like this requires careful mathematical techniques. For this exercise, recognizing the singularity is part of understanding why the integral might be problematic. Analyzing around this singularity can help assess where the function "goes to infinity" and contributes to our understanding of convergence and divergence.

Convergence

Convergence in the context of integrals means determining if the integral approaches a finite value as the limit approaches infinity. For ${\int }_0^{\mathrm{\infty }}|x^{-1}\sin x|dx$, it's essential to note that despite the oscillating nature of $\sin x$, the integral diverges.
The bound $|\sin x|\le 1$ helps determine the behavior of the magnitude of the function, especially over periodic intervals like $$2k\pi ,(2k+1)\pi$$. These intervals help us understand the periodic contributions that eventually become infinite, explaining why the integral diverges.
Conversely, for part (b), the integral $\underset{b\to \mathrm{\infty }}{lim}{\int }_0^b{x}^{-1}\sin xdx$ does converge to $\frac{1}{2}\pi$, which is a fascinating scenario where integration techniques and limits play a vital role in showing conditional convergence. This is an example of an improper integral that is only conditionally convergent.

Integration by Parts

Integration by parts is a technique often used for handling products of functions inside an integral. In our context, it's used to deal with the integral of $e^{-xy}\sin xdx$, crucial for solving part (b) of the problem.
The formula is derived from the product rule for differentiation and is represented as:
$$\int udv=uv-\int vdu$$
Choosing $u=\sin x$ and $dv=e^{-xy}dx$, the derivative $du=\cos xdx$ and finding $v=-\frac{1}{y}{e}^{-xy}$, we integrate the function by parts. This helps in simplifying a complex integral into manageable components and is a classic method for solving integrals involving products of polynomial, exponential, and trigonometric functions. This allows us to approach the problem systematically and eventually achieve the result for part (b).

Limit Evaluation

Limits are critical when evaluating improper integrals, especially for extending bounds to infinity. In part (b) of the exercise, using limits makes it possible to evaluate $\underset{b\to \mathrm{\infty }}{lim}{\int }_0^b{x}^{-1}\sin xdx$ to arrive at $\frac{1}{2}\pi$.
Limit evaluation uses logical consideration of how a function behaves as it approaches a certain point. In cases involving infinity, we see how the function progresses towards a finite result or zero over infinite bounds. Limits are crucial here to understand when an oscillating or periodic function's contributions condense into a specific value.
This technique helps in verifying the possible convergence of integrals and ensuring that even when dealing with functions with singularities or unbounded nature, assessments of behavior at the extremes can be analyzed precisely.