Question

Determine whether the improper integrals converge or diverge. If possible, determine the value of the integrals that converge.\(\int_{0}^{1} \frac{\ln x}{\sqrt{x}} d x\)

Short answer

The integral converges, and its value is -4.

Step-by-step solution

Recognize the type of integral

This is an improper integral since the integrand \( \frac{\ln x}{\sqrt{x}} \) is not defined at the lower limit \( x = 0 \) due to the natural logarithm. We need to examine the behavior as \( x \to 0^{+} \).

Set up the improper integral

Express the given integral \( \int_{0}^{1} \frac{\ln x}{\sqrt{x}} dx \) as a limit: \[ \lim_{a \to 0^{+}} \int_{a}^{1} \frac{\ln x}{\sqrt{x}} dx. \]

Evaluate the integral

Determine \( \int \frac{\ln x}{\sqrt{x}} dx \) using integration by parts. Let \( u = \ln x \) and \( dv = \frac{1}{\sqrt{x}} dx \), then \( du = \frac{1}{x} dx \) and \( v = 2\sqrt{x} \). Using integration by parts formula \( \int u \, dv = uv - \int v \, du \), we get:\[ 2\sqrt{x} \ln x - \int 2 \cdot \frac{1}{\sqrt{x}} dx = 2\sqrt{x} \ln x - 4\sqrt{x} \].

Estimate the limit

Plug the evaluated integral back into the limit, \[ \lim_{a \to 0^{+}} \left[ 2\sqrt{x} \ln x - 4\sqrt{x} \right]_{a}^{1} = \lim_{a \to 0^{+}} \left( (2\ln 1 - 4) - (2\sqrt{a}\ln a - 4\sqrt{a}) \right) \].

Analyze convergence and calculate the limit

As \( a \to 0^{+} \), the term \( 2\sqrt{a} \ln a \) dominates the behavior. Analyze this term: \[ \lim_{a \to 0^{+}} 2\sqrt{a} \ln a \]it can be rewritten using substitution \( x = \sqrt{a} \) which leads to \( a \ln a \to 0 \) as \( a \to 0^{+} \). Therefore, \[ \lim_{a \to 0^{+}} 2\sqrt{a}\ln a = 0 \]. This means the integral \( \int_{0}^{1} \frac{\ln x}{\sqrt{x}} dx \) converges to \( -4 \).

Conclusion

The improper integral \( \int_{0}^{1} \frac{\ln x}{\sqrt{x}} dx \) converges and the evaluated value is \( -4 \).

Key concepts

Convergence and Divergence

Improper integrals present us with a fascinating challenge: determining whether they converge or diverge. For an improper integral to converge, its limit must exist and be finite. In this context, think of convergence as the integral magically adding up to a finite number, even when the function becomes infinite at some point within the integration region or the limits themselves are infinite. Divergence, on the other hand, occurs when the function refuses to settle at any finite value, no matter how hard we try to sum it up.

In our specific example, the integral \( \int_{0}^{1} \frac{\ln x}{\sqrt{x}} \, dx \)is improper because the logarithm becomes undefined at zero. We approached this by considering the limit as \( x \to 0^+ \). By setting up the limit of the integral from \( a \) to 1 as \( a \to 0^+ \), we could carefully analyze the behavior of the function. Through this process, we determined that the integral converges to \(-4\), which means, quite counterintuitively, this seemingly divergent expression resolves into a neat numerical value.

Integration by Parts

Integration by parts is a crucial method in calculus used to integrate products of functions. The technique is akin to the reverse of the product rule for differentiation, and it's particularly powerful when dealing with integrals involving logarithmic or polynomial terms.

In our problem, integration by parts helped us tackle the troublesome integrand \( \frac{\ln x}{\sqrt{x}} \). We chose \( u = \ln x \) and \( dv = \frac{1}{\sqrt{x}} \, dx \). Then, we computed \( du = \frac{1}{x} \, dx \) and \( v = 2\sqrt{x} \). Applying the formula \[ \int u \, dv = uv - \int v \, du \] transformed the original integral into a simpler form: \[ 2\sqrt{x} \ln x - 4\sqrt{x}. \] By effectively reorganizing the components, integration by parts enabled us to simplify the integral and further evaluate its behavior as \( x \to 0^+ \).

Limit of a Function

The limit of a function is at the heart of calculus, serving as a tool to understand the behavior of functions as they approach a particular point. This concept is especially useful for analyzing improper integrals, where traditional integration methods falter due to undefined terms or infinite boundaries.

In considering \( \lim_{a \to 0^+} \left( 2\sqrt{a} \ln a \right) \) in our problem, we faced a classic case of an indeterminate form. To resolve this, a substitution (\( x = \sqrt{a} \)) broke it down into an expression whose value approaches zero as \( a \to 0^+ \). Through these clever manipulations, we sidestepped the originally undefined behavior and paved a path to comprehending the integral's definitive behavior by arriving at a limit of zero. Therefore, this ensured the convergence of our integral to \(-4\). Knowing how to compute limits effectively is like holding a key that unlocks the intricacies of integrals where traditional rules shy away.