Question

For a real number \(a\), suppose \(\lim _{x \rightarrow a^{+}} f(x)=-\infty\) or \(\lim _{x \rightarrow a^{+}} f(x)=\infty .\) In these cases, the integral \(\int_{a}^{\infty} f(x) d x\) is improper for two reasons: \(\infty\) appears in the upper limit and \(f\) is unbounded at \(x=a .\) It can be shown that \(\int_{a}^{\infty} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{\infty} f(x) d x\) for any \(c>a .\) Use this result to evaluate the following improper integrals. $$\int_{1}^{\infty} \frac{d x}{x \sqrt{x-1}}$$

Short answer

Answer: The improper integral diverges.

Step-by-step solution

Split the integral into two parts

Given that \(\int_{a}^{\infty} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{\infty} f(x) d x\), we can choose any \(c > 1\) and rewrite the given improper integral as follows: $$\int_{1}^{\infty} \frac{d x}{x \sqrt{x-1}} = \int_{1}^{c} \frac{d x}{x \sqrt{x-1}} + \int_{c}^{\infty} \frac{d x}{x \sqrt{x-1}}$$

Evaluating the first integral

To evaluate \(\int_{1}^{c} \frac{d x}{x \sqrt{x-1}}\), we can use the substitution method. Let's set \(u = x-1\), so \(x = u+1\) and \(d x = d u\). After the substitution, the integral becomes: $$\int_{1}^{c} \frac{d x}{x \sqrt{x-1}} = \int_{0}^{c-1} \frac{d u}{(u+1) \sqrt{u}}$$ To evaluate this integral, use another substitution: let \(v = \sqrt{u}\), so \(u = v^2\) and \(d u = 2v d v\). We now have: $$\int_{0}^{c-1} \frac{d u}{(u+1) \sqrt{u}} = \int_{0}^{\sqrt{c-1}} \frac{2v d v}{(v^2+1) v} = 2\int_{0}^{\sqrt{c-1}} \frac{ d v}{v^2+1}$$ Now, the integral can be evaluated as: $$2\int_{0}^{\sqrt{c-1}} \frac{ d v}{v^2+1} = 2\left[\arctan(v)\right]_{0}^{\sqrt{c-1}} = 2\arctan(\sqrt{c-1})$$

Evaluating the second integral

Now we have to evaluate the second integral, \(\int_{c}^{\infty} \frac{d x}{x \sqrt{x-1}}\). Since the function is decreasing as \(x\) approaches infinity, we can use a comparison test. Given that \(\frac{1}{x \sqrt{x-1}} \leq \frac{1}{\sqrt{x-1}}\), comparing the function with another will result in an improper integral as follows: $$\int_{c}^{\infty} \frac{d x}{x \sqrt{x-1}} \leq \int_{c}^{\infty} \frac{d x}{\sqrt{x-1}}$$ Now, evaluate the right-hand side: $$\int_{c}^{\infty} \frac{d x}{\sqrt{x-1}} = 2\left[\sqrt{x-1}\right]_{c}^{\infty} = \lim_{x \to \infty} (2\sqrt{x-1} - 2\sqrt{c-1}) = \infty$$ Thus, the integral \(\int_{c}^{\infty} \frac{dx}{x\sqrt{x-1}}\) diverges.

Conclusion

As the second integral diverges, the original integral also diverges: $$\int_{1}^{\infty} \frac{d x}{x \sqrt{x-1}} = \int_{1}^{c} \frac{d x}{x \sqrt{x-1}} + \int_{c}^{\infty} \frac{d x}{x \sqrt{x-1}} = 2\arctan(\sqrt{c-1}) + \infty = \infty$$

Key concepts

Limit

The concept of a limit is fundamental in calculus, allowing us to analyze the behavior of a function as it approaches a particular point. When considering improper integrals, limits help us understand how a function behaves as it approaches infinity or as it nears a point where it becomes unbounded. For example, if a function has a limit of infinity as it approaches a specific value from the right, such as \((\lim _{x \rightarrow a^{+}} f(x)=\infty)\), it indicates the function grows without bound. This is critical when evaluating integrals over infinite intervals or where the function does not remain bounded.

Substitution Method

The substitution method is a powerful tool in calculus that simplifies the process of integration. By making a substitution, we aim to transform a complex integral into a more manageable one. In our exercise, to evaluate \(\int_{1}^{c} \frac{d x}{x \sqrt{x-1}}\), we began by letting \((u = x-1)\). This redefines the integral in terms of \(u\), modifying it to \((\int_{0}^{c-1} \frac{d u}{(u+1) \sqrt{u}})\). A further substitution, \((v = \sqrt{u})\), simplified it even more.

• The original integral's bounds change according to the substitution.
• The complexity of the integral decreases, often resulting in standard forms.
• Substitution can sometimes lead to integrals that involve basic functions, such as arctangent or logarithmic functions.

Comparison Test

The comparison test is utilized to ascertain whether improper integrals converge or diverge. This method involves comparing the function to another, simpler function whose behavior is known. In the context of our problem, we compared \(\frac{1}{x \sqrt{x-1}}\) to \(\frac{1}{\sqrt{x-1}}\). By establishing that \(\frac{1}{x \sqrt{x-1}} \leq \frac{1}{\sqrt{x-1}}\), we can use the behavior of the simpler comparison function to draw conclusions about the original integral.

• If the integral of the comparison function diverges, then the original integral also diverges.
• The test is a powerful technique, especially when direct evaluation of an integral is complex or impractical.
• This approach requires knowing or being able to determine the limit behavior of the comparison function.

Divergence

Divergence occurs when an integral does not converge to a finite value. For improper integrals, divergence indicates that the integral behavior tends towards infinity. In our example, \(\int_{c}^{\infty} \frac{d x}{x \sqrt{x-1}}\) diverges because the comparison test showed that this function, in comparison to a known divergent function, grows indefinitely.

• Understanding divergence is crucial for identifying when an integral goes to infinity.
• This is particularly important when dealing with infinite limits or unbounded functions.
• By determining divergence, we confirm that the entire integral does not produce a finite result, impacting practical calculations or predictions based on the integral.