Problem 20 - Biocalculus Calculus for the Life Sciences

Question

Determine whether each integral is convergent or divergent. Evaluate those that are convergent. \(\int_{1}^{\infty} \frac{\ln x}{x^{3}} d x\)

Step-by-Step Solution

Answer

The integral is convergent, and its value is \( \frac{1}{4} \).

Step 1: Identify the Type of Integral

The given integral is improper because it extends to infinity. Hence, we will evaluate its convergence from 1 to infinity, which we can express as \( \lim_{{b \to \infty}} \int_{{1}}^{{b}} \frac{{\ln x}}{{x^3}} \, dx \).

Step 2: Integrate the Function

To find \( \int \frac{{\ln x}}{{x^3}} \, dx \), use integration by parts: let \( u = \ln x \) and \( dv = \frac{{1}}{{x^3}} \, dx \). Then \( du = \frac{{1}}{{x}} \, dx \) and \( v = -\frac{{1}}{{2x^2}} \). The integration by parts formula, \( \int u \, dv = uv - \int v \, du \), gives us: \[ \int \frac{{\ln x}}{{x^3}} \, dx = -\frac{{\ln x}}{{2x^2}} - \int \left(-\frac{{1}}{{2x^2}} \cdot \frac{{1}}{{x}}\right) \, dx \].

Step 3: Simplify and Complete the Integral

Simplify the integral further: \( \int \frac{1}{2x^3} \, dx = \frac{-1}{4x^2} + C \). Thus, the indefinite integral becomes \[ -\frac{\ln x}{2x^2} + \frac{1}{4x^2} + C \].

Step 4: Evaluate the Definite Integral

Evaluate the definite integral using limits: \[ \lim_{b \to \infty} \left( -\frac{\ln x}{2x^2} + \frac{1}{4x^2} \right) \bigg|_1^b \]. Simplifying gives \( \lim_{b \to \infty} \left( -\frac{\ln b}{2b^2} + \frac{1}{4b^2} \right) - \left( -\frac{\ln 1}{2} + \frac{1}{4} \right) \).

Step 5: Evaluate the Limit at Infinity

Note that as \( b \to \infty \), both \( -\frac{\ln b}{2b^2} \) and \( \frac{1}{4b^2} \) approach zero because the logarithm grows slower than any polynomial. Therefore, the limit is \( 0 - \left( 0 + \frac{1}{4} \right) = -\frac{1}{4} \).

Step 6: Conclusion on Convergence

The integral converges and the value of the integral from 1 to infinity is \( \frac{1}{4} \).

Key Concepts

Convergence

When dealing with improper integrals, the first question to tackle is whether the integral is convergent or divergent. This means we need to find out if the integral reaches a finite value or not. Convergence is all about determining whether the area under the curve of the function can be computed up to a finite number.
To check this, we often set up the integral with limits approaching infinity, like with our integral, \( \int_{1}^{\infty} \frac{\ln x}{x^{3}} \, dx \) is evaluated by rewriting it with a limit: \[ \lim_{b \to \infty} \int_{1}^{b} \frac{\ln x}{x^{3}} \, dx \].
If this limit results in a finite number, we say the integral converges and can compute the actual value. But if it results in infinity, it means the integral is divergent, indicating the area under the curve is infinitely large.

• Convergent integrals have finite limits.
• Divergent integrals tend towards infinity.

Tackling these integrals often involves mathematical techniques like integration by parts or substitution to see if they settle to a finite number.

Integration by Parts

Integration by Parts is a nifty tool from calculus used to solve integrals, especially when they involve products of functions. It comes from the product rule for differentiation and is incredibly useful! The formula is:\[ \int u \, dv = uv - \int v \, du \].
This technique is all about picking parts of the integral to designate as \( u \) and \( dv \), hence the name "integration by parts". For our example, where we have \( \int \frac{\ln x}{x^3} \, dx \), the setup is crucial.
We choose \( u = \ln x \) and \( dv = \frac{1}{x^3} \, dx \). By differentiating \( u \) and integrating \( dv \), we find:

• \( du = \frac{1}{x} \, dx \)
• \( v = -\frac{1}{2x^2} \)

Plug these into the integration by parts formula to transform the integral into something more manageable! The choice of \( u \) and \( dv \) can make a world of difference, so it's something to practice a lot. Often, \( u \) is a function that simplifies when differentiated, like \( \ln x \) in this case.

Limits at Infinity

Limits at infinity explore what happens to functions as they stretch towards infinity. Notably, this concept reveals a function's behavior as its input grows endlessly large. It's crucial for determining convergence in improper integrals.
In our integral example, after using integration by parts and setting up the definite integral, we see parts of our function approaching infinity: \( \lim_{b \to \infty} \left( -\frac{\ln b}{2b^2} + \frac{1}{4b^2} \right) \).
Both expressions \(-\frac{\ln b}{2b^2}\) and \(\frac{1}{4b^2}\) head to zero as \( b \) becomes very large. Why? Because their denominators outgrow their numerators in the race to infinity. Remember:

• Polynomials grow quicker than logarithms.
• Fractions with larger denominators shrink towards zero.

This shrinking effect makes comparing functions to zero fundamental, forming the judgement on convergence. It's like seeing which passenger in the race reaches zero first! Understanding this behavior indicates if an integral converges, giving us insight into the overall size of the area under the curve.