Evaluating The Improper Integral ∫₀^∞ 1/(x³+x²+x+1) Dx A Step-by-Step Guide

Hey guys! Today, we're diving deep into the fascinating world of calculus, specifically tackling the evaluation of an improper integral. We're going to explore how to solve the integral ∫₀^∞ 1/(x³+x²+x+1) dx. This problem touches upon several key areas, including calculus, integration techniques, complex analysis, and the handling of improper integrals and fractions. So, buckle up and let's get started!

Understanding the Challenge

Before we jump into the solution, let's understand the nature of the problem. We're dealing with an improper integral because the upper limit of integration is infinity. This means we can't directly plug in infinity; we need to use a limit. Furthermore, the integrand, 1/(x³+x²+x+1), is a rational function, which hints at the need for techniques like partial fraction decomposition. The challenge here lies in finding a clever way to break down this integral into manageable parts. This particular problem requires a blend of algebraic manipulation and calculus techniques, making it a wonderfully instructive example. We'll need to strategically factor the denominator, apply partial fraction decomposition, and then evaluate the resulting simpler integrals. Don't worry if this sounds intimidating; we'll break it down step by step.

Factoring the Denominator: The First Crucial Step

The very first thing we need to do is to factor the denominator, x³ + x² + x + 1. Factoring this cubic polynomial will allow us to perform partial fraction decomposition, which is essential for simplifying the integral. We can factor by grouping: x³ + x² + x + 1 = x²(x + 1) + 1(x + 1) = (x² + 1)(x + 1). Now, our integral looks like this: ∫₀^∞ 1/((x + 1)(x² + 1)) dx. This factorization is a significant step forward. By expressing the denominator as a product of simpler polynomials, we pave the way for the next crucial technique: partial fraction decomposition. Recognizing and executing this factorization efficiently is often the key to unlocking the solution for integrals involving rational functions. It's like finding the secret key to unlock a treasure chest of mathematical insights!

Partial Fraction Decomposition: Breaking it Down

Now that we've factored the denominator, we can apply partial fraction decomposition. This technique allows us to break down a complex rational function into simpler fractions, which are much easier to integrate. We want to express 1/((x + 1)(x² + 1)) in the form A/(x + 1) + (Bx + C)/(x² + 1). To find A, B, and C, we multiply both sides by the common denominator (x + 1)(x² + 1), giving us: 1 = A(x² + 1) + (Bx + C)(x + 1). Expanding and collecting like terms, we have: 1 = (A + B)x² + (B + C)x + (A + C). By equating coefficients, we get the following system of equations:

• A + B = 0
• B + C = 0
• A + C = 1

Solving this system, we find A = 1/2, B = -1/2, and C = 1/2. Therefore, our integrand can be rewritten as:

1/((x + 1)(x² + 1)) = (1/2)/(x + 1) + (-1/2 x + 1/2)/(x² + 1). Partial fraction decomposition is a powerful tool in the calculus arsenal. It allows us to transform seemingly intractable rational functions into sums of simpler terms, each of which can be integrated using standard techniques. Mastering this technique is crucial for tackling a wide range of integration problems. It's like having a Swiss Army knife for calculus – versatile and indispensable!

Setting Up the Integral: Putting the Pieces Together

With the partial fraction decomposition in hand, we can now rewrite the original integral as a sum of simpler integrals:

∫₀^∞ 1/((x + 1)(x² + 1)) dx = (1/2) ∫₀^∞ 1/(x + 1) dx + (1/2) ∫₀^∞ (-x + 1)/(x² + 1) dx.

This step is critical because it transforms a single, complex integral into a sum of integrals that we can handle individually. We've effectively broken down a formidable challenge into smaller, more manageable tasks. This is a common strategy in problem-solving – divide and conquer! Now, let's focus on evaluating each of these integrals separately. We'll see how standard integration techniques can be applied to each term.

Evaluating the First Integral: A Simple Logarithm

The first integral is (1/2) ∫₀^∞ 1/(x + 1) dx. This is a straightforward integral. The antiderivative of 1/(x + 1) is ln|x + 1|. So, we have:

(1/2) ∫₀^∞ 1/(x + 1) dx = (1/2) [ln|x + 1|]₀^∞.

Remember, we're dealing with an improper integral, so we need to use a limit:

(1/2) [ln|x + 1|]₀^∞ = (1/2) lim(b→∞) [ln(b + 1) - ln(1)] = (1/2) lim(b→∞) ln(b + 1) = ∞.

Woah! Hold on! This integral diverges! It seems like we've hit a snag. But don't worry, this doesn't mean the entire original integral diverges. We need to carefully evaluate the other integral and see if things balance out. This result highlights the importance of treating improper integrals with caution. We can't just blindly apply the fundamental theorem of calculus; we must always consider the limits and potential divergences. This first integral, although simple in form, serves as a valuable reminder of the nuances involved in dealing with improper integrals.

Evaluating the Second Integral: Tangents and Logarithms

The second integral is (1/2) ∫₀^∞ (-x + 1)/(x² + 1) dx. We can split this into two integrals:

(1/2) ∫₀^∞ (-x + 1)/(x² + 1) dx = -(1/2) ∫₀^∞ x/(x² + 1) dx + (1/2) ∫₀^∞ 1/(x² + 1) dx.

The first part, -(1/2) ∫₀^∞ x/(x² + 1) dx, can be solved using a simple u-substitution. Let u = x² + 1, then du = 2x dx. So, the integral becomes:

-(1/2) ∫₀^∞ x/(x² + 1) dx = -(1/4) ∫₁^∞ 1/u du = -(1/4) [ln|u|]₁^∞ = -(1/4) [ln(x² + 1)]₀^∞ = -(1/4) lim(b→∞) [ln(b² + 1) - ln(1)] = -∞.

Again, we have a divergent integral! But let's not lose hope yet. The second part of the integral is:

(1/2) ∫₀^∞ 1/(x² + 1) dx = (1/2) [arctan(x)]₀^∞ = (1/2) [lim(b→∞) arctan(b) - arctan(0)] = (1/2) [π/2 - 0] = π/4.

This integral converges to π/4. Now, let's combine the results carefully. We had two divergent integrals, but we need to remember how they came about. The divergence we encountered in the integrals ∫₀^∞ 1/(x + 1) dx and ∫₀^∞ x/(x² + 1) dx indicates that we need to consider the original integral as a whole, rather than trying to assign separate values to each piece in isolation. This is a crucial lesson in dealing with improper integrals: sometimes, individual parts may diverge, but the whole integral may still converge.

Combining the Results: A Delicate Balance

We need to be very careful when combining the results of divergent integrals. Let's go back to the split integrals before we took the limits to infinity. We had:

(1/2) ∫₀^∞ 1/(x + 1) dx + (1/2) ∫₀^∞ (-x + 1)/(x² + 1) dx = (1/2) ∫₀^∞ 1/(x + 1) dx - (1/2) ∫₀^∞ x/(x² + 1) dx + (1/2) ∫₀^∞ 1/(x² + 1) dx.

Let's rewrite this using limits:

lim(b→∞) [(1/2) ∫₀^b 1/(x + 1) dx - (1/2) ∫₀^b x/(x² + 1) dx + (1/2) ∫₀^b 1/(x² + 1) dx].

Now, let's find the antiderivatives and evaluate them within the limit:

lim(b→∞) [(1/2) ln(x + 1) - (1/4) ln(x² + 1) + (1/2) arctan(x)]₀^b

= lim(b→∞) [(1/2) ln(b + 1) - (1/4) ln(b² + 1) + (1/2) arctan(b) - (1/2) ln(1) + (1/4) ln(1) - (1/2) arctan(0)]

= lim(b→∞) [(1/2) ln(b + 1) - (1/4) ln(b² + 1) + (1/2) arctan(b)].

We can rewrite the logarithmic terms as:

lim(b→∞) [(1/4) [2ln(b + 1) - ln(b² + 1)] + (1/2) arctan(b)]

= lim(b→∞) [(1/4) ln((b + 1)²/(b² + 1)) + (1/2) arctan(b)].

Now, let's analyze the limit of the logarithmic term:

lim(b→∞) ln((b + 1)²/(b² + 1)) = lim(b→∞) ln((b² + 2b + 1)/(b² + 1)) = ln(1) = 0.

And the limit of the arctangent term is:

lim(b→∞) arctan(b) = π/2.

So, putting it all together:

(1/4) * 0 + (1/2) * (π/2) = π/4.

The Final Answer: Convergence Achieved!

Therefore, the value of the integral ∫₀^∞ 1/(x³ + x² + x + 1) dx is π/4. We've successfully navigated through the complexities of this improper integral, employing techniques like factoring, partial fraction decomposition, and careful limit evaluation. This journey has not only given us a numerical answer but also a deeper understanding of the nuances of improper integrals and the importance of meticulous analysis. Remember, guys, even if parts of an integral seem to diverge, the whole can still converge beautifully!

Key Takeaways: Mastering the Art of Integration

• Factoring is Fundamental: Always start by factoring the denominator of a rational function. This simplifies the integrand and paves the way for partial fraction decomposition.
• Partial Fraction Decomposition is Your Friend: This technique is essential for breaking down complex rational functions into simpler, integrable terms.
• Limits are Crucial for Improper Integrals: Never forget to use limits when dealing with improper integrals. This ensures you're handling infinity correctly.
• Divergence Doesn't Always Mean Failure: Just because individual parts of an integral diverge doesn't mean the entire integral diverges. Careful combination and limit evaluation are key.
• Practice Makes Perfect: The more you practice these techniques, the more comfortable and confident you'll become in tackling complex integration problems.

So there you have it! We've successfully integrated ∫₀^∞ 1/(x³ + x² + x + 1) dx. I hope this comprehensive guide has been helpful. Keep practicing, keep exploring, and keep pushing your calculus skills to the limit! Happy integrating, guys!