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Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{\ln 2} x^{-2} e^{-1 / x} d x$$

Short Answer

Expert verified
The integral diverges.

Step by step solution

01

Analyze the Given Integral

We are given the integral \( \int_{0}^{\ln 2} x^{-2} e^{-1/x} \, dx \). The function \( x^{-2} e^{-1/x} \) is defined and finite for all \( x \in (0, \ln 2] \), however, it poses a singularity at \( x = 0 \) because \( x^{-2} \) becomes undefined.
02

Set up the Limit Comparison Test

To use the Limit Comparison Test, we need a simpler comparison function. A suitable option is \( f(x) = x^{-2} \) because both functions \( x^{-2} e^{-1/x} \) and \( x^{-2} \) have similar behavior near the point \( x = 0 \). We are interested in the convergence properties:\[ \int_{0}^{\ln 2} x^{-2} \, dx \]
03

Evaluate the Comparison Function

The integral \( \int_{0}^{\ln 2} x^{-2} \, dx \) is known to diverge as \( \int_{0}^{a} x^{-2} \, dx = \lim_{b \to 0^{+}} \left[ -x^{-1} \right]_{b}^{a} \), which leads to \(+\infty\). This divergence near \( x = 0 \) suggests \( x^{-2} \) is a proper comparison.
04

Compute the Limit for Comparison

For the Limit Comparison Test, we compute:\[ \lim_{x \to 0^{+}} \frac{x^{-2} e^{-1/x}}{x^{-2}} = \lim_{x \to 0^{+}} e^{-1/x} = 0 \]Since \( e^{-1/x} \rightarrow 0 \) and \( 0 < \infty \), the original integral shares the same convergence properties with \( \int_{0}^{\ln 2} x^{-2} \, dx \).
05

Conclude the Behavior of the Integral

The limit comparison establishes that both integrals behave similarly near \( x = 0 \). Since \( \int_{0}^{\ln 2} x^{-2} \, dx \) diverges, the given integral \( \int_{0}^{\ln 2} x^{-2} e^{-1/x} \, dx \) also diverges as \( x \to 0^{+} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Comparison Test
The Limit Comparison Test is a handy tool for determining the convergence of an integral by comparing it with a simpler function. If you have an integral that's complicated, you can choose a simpler function that behaves similarly near the point of difficulty.
Simply divide the given function by the chosen simpler function and compute their limit as the variable approaches the point of singularity. For example, if both functions diverge or converge, the original function behaves like the comparison function.
This test is especially useful when the integral involves complex expressions where direct integration might be challenging.

However, it's critical to choose an appropriate comparison function. A good choice shares the key features of the original function near the point of interest. Once you've selected it, the rest involves simple algebraic manipulation and finding limits.
Direct Comparison Test
The Direct Comparison Test involves directly comparing the integral of interest with another integral whose convergence is already known. If you find a function that bounds your given function, you can use this clear relationship to conclude convergence.
Specifically, if your function is always smaller than a convergent function, it must also converge. Conversely, if it's larger than a divergent function, then it diverges.
However, you need to be cautious when using this test because choosing functions that are too different might lead to incorrect conclusions.

This test provides a straightforward way to prove convergence but depends heavily on finding the right comparisons.
Thus, it's crucial to have a solid understanding of the behavior of common functions and their integrals to apply this test efficiently.
Improper Integrals
Improper integrals occur when either the interval of integration extends to infinity or the function blows up at one or more points within the interval. These challenging scenarios require careful analysis before drawing conclusions about convergence.
In the case of having singularities, such as an integral over (0, a] with behavior like 1/x, it often involves examining limits as the variable approaches the point of singularity.
For instance, dealing with these integrals might require integration and taking limits to suite the particular peculiar behavior of the function.

While they can provide valuable insights into particular solutions, improper integrals need more effort and technique compared to standard, or 'proper', integrals.
Singularities in Integrals
Singularities in integrals refer to points where the function being integrated tends to infinity or becomes undefined, making the evaluation of the integral tricky. For example, if you have a function like 1/x when x is near zero, this creates a signularity that must be examined meticulously.
When singularities are present, special techniques, such as comparison tests, are often necessary to assess whether an integral converges or diverges.

The presence of singularities usually necessitates the use of limits to properly evaluate the behavior of the function near these challenging points.
  • Using tools like the Limit Comparison Test can simplify the complexity involved in dealing with singularities by allowing comparison to functions of known behavior.
  • It's essential to approach these points carefully, ensuring accurate conclusions regarding convergence.
Addressing singularities is crucial in both theoretical and practical integration problems.

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Most popular questions from this chapter

In Exercises \(21-32,\) express the integrand as a sum of partial fractions and evaluate the integrals. $$\int \frac{2 s+2}{\left(s^{2}+1\right)(s-1)^{3}} d s$$

As Example 3 shows, the integral \(\int_{1}^{\infty}(d x / x)\) diverges. This means that the integral $$\int_{1}^{\infty} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x$$ which measures the surface area of the solid of revolution traced out by revolving the curve \(y=1 / x, 1 \leq x,\) about the \(x\)-axis, diverges also. By comparing the two integrals, we see that, for every finite value \(b>1,\) $$\int_{1}^{b} 2 \pi \frac{1}{x} \sqrt{1+\frac{1}{x^{4}}} d x>2 \pi \int_{1}^{b} \frac{1}{x} d x$$ (Check your book to see image) However, the integral $$\int_{1}^{\infty} \pi\left(\frac{1}{x}\right)^{2} d x$$ for the volume of the solid converges. a. Calculate it. b. This solid of revolution is sometimes described as a can that does not hold enough paint to cover its own interior. Think about that for a moment. It is common sense that a finite amount of paint cannot cover an infinite surface. But if we fill the horn with paint (a finite amount), then we will have covered an infinite surface. Explain the apparent contradiction.

The function $$f(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}$$ is called the normal probability density function with mean \(\mu\) and standard deviation \(\sigma .\) The number \(\mu\) tells where the distribution is centered, and \(\sigma\) measures the "scatter" around the mean. From the theory of probability, it is known that $$\int_{-\infty}^{\infty} f(x) d x=1$$ In what follows, let \(\mu=0\) and \(\sigma=1\) a. Draw the graph of \(f .\) Find the intervals on which \(f\) is increasing, the intervals on which \(f\) is decreasing, and any local extreme values and where they occur. b. Evaluate $$\int_{-n}^{n} f(x) d x$$ for \(n=1,2,\) and 3. c. Give a convincing argument that $$\int_{-\infty}^{\infty} f(x) d x=1$$ (Hint: Show that \(0< f(x)< e^{-x / 2}\) for \(x>1\), and for \(b>1\), $$\int_{b}^{\infty} e^{-x / 2} d x \rightarrow 0 \quad \text { as } \quad b \rightarrow \infty.)$$

Integration by parts leads to a rule for integrating es that usually gives good results:$$\begin{aligned}\int f^{-1}(x) d x &=\int y f^{\prime}(y) d y \\\&=y f(y)-\int f(y) d y \\ &=x f^{-1}(x)-\int f(y) d y\end{aligned}$$. The idea is to take the most complicated part of the integral, in this case \(f^{-1}(x)\), and simplify it first. For the integral of \(\ln x,\) we get $$\begin{aligned}\int \ln x \, d x &=\int y e^{y} d y \\\&=y e^{y}-e^{y}+C \\\&=x \ln x-x+C \end{aligned}$$ For the integral of \(\cos ^{-1} x\) we get $$\begin{aligned}\int \cos ^{-1} x d x &=x \cos ^{-1} x-\int \cos y \, d y \\\&=x \cos ^{-1} x-\sin y+C \\\&=x \cos ^{-1} x-\sin \left(\cos ^{-1} x\right)+C\end{aligned}$$ Use the formula $$\int f^{-1}(x) d x=x f^{-1}(x)-\int f(y) d y$$ Express your answers in terms of \(x.\) $$\int \log _{2} x d x$$

In Exercises \(33-38,\) perform long division on the integrand, write the proper fraction as a sum of partial fractions, and then evaluate the integral. $$\int \frac{2 x^{3}-2 x^{2}+1}{x^{2}-x} d x$$

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