Investigate the convergence of the integral: integral

Investigate the convergence of the integral: $$ \int_1^\infty \frac{2x^2-3x+2}{\sqrt{x^8+3x^6+5}} \, dx $$

This integral is improper at infinity. We need to analyze the behavior of the integrand for large $x$.

Step 1: Analyze the behavior at infinity

For large $x$, we need to determine the dominant terms in the numerator and denominator.

Numerator:

$$2x^2 - 3x + 2$$

For large $x$, the dominant term is $2x^2$, so:

$$2x^2 - 3x + 2 \sim 2x^2$$

Denominator:

$$\sqrt{x^8 + 3x^6 + 5}$$

For large $x$, the dominant term under the square root is $x^8$, so:

$$\sqrt{x^8 + 3x^6 + 5} \sim \sqrt{x^8} = x^4$$

Step 2: Asymptotic behavior

Therefore, for large $x$:

$$\frac{2x^2-3x+2}{\sqrt{x^8+3x^6+5}} \sim \frac{2x^2}{x^4} = \frac{2}{x^2}$$

Step 3: More precise analysis

Let's be more rigorous. Factor out the highest powers:

Numerator:

$$2x^2 - 3x + 2 = x^2\left(2 - \frac{3}{x} + \frac{2}{x^2}\right)$$

Denominator:

$$\sqrt{x^8 + 3x^6 + 5} = \sqrt{x^8\left(1 + \frac{3}{x^2} + \frac{5}{x^8}\right)} = x^4\sqrt{1 + \frac{3}{x^2} + \frac{5}{x^8}}$$

Therefore:

$$\frac{2x^2-3x+2}{\sqrt{x^8+3x^6+5}} = \frac{x^2\left(2 - \frac{3}{x} + \frac{2}{x^2}\right)}{x^4\sqrt{1 + \frac{3}{x^2} + \frac{5}{x^8}}}$$

$$= \frac{1}{x^2} \cdot \frac{2 - \frac{3}{x} + \frac{2}{x^2}}{\sqrt{1 + \frac{3}{x^2} + \frac{5}{x^8}}}$$

Step 4: Limit comparison test

As $x \to \infty$:

$$\frac{2 - \frac{3}{x} + \frac{2}{x^2}}{\sqrt{1 + \frac{3}{x^2} + \frac{5}{x^8}}} \to \frac{2 - 0 + 0}{\sqrt{1 + 0 + 0}} = \frac{2}{1} = 2$$

Therefore:

$$\frac{2x^2-3x+2}{\sqrt{x^8+3x^6+5}} \sim \frac{2}{x^2} \quad \text{as } x \to \infty$$

Step 5: Apply the limit comparison test

We compare with $g(x) = \frac{1}{x^2}$.

Calculate:

$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{\frac{2x^2-3x+2}{\sqrt{x^8+3x^6+5}}}{\frac{1}{x^2}}$$

$$= \lim_{x \to \infty} \frac{x^2(2x^2-3x+2)}{\sqrt{x^8+3x^6+5}}$$

$$= \lim_{x \to \infty} \frac{2x^4-3x^3+2x^2}{\sqrt{x^8+3x^6+5}}$$

Factor:

$$= \lim_{x \to \infty} \frac{x^4(2-\frac{3}{x}+\frac{2}{x^2})}{x^4\sqrt{1+\frac{3}{x^2}+\frac{5}{x^8}}}$$

$$= \lim_{x \to \infty} \frac{2-\frac{3}{x}+\frac{2}{x^2}}{\sqrt{1+\frac{3}{x^2}+\frac{5}{x^8}}} = \frac{2}{1} = 2$$

Since the limit is a positive finite number ($L = 2$), and we know that:

$$\int_1^\infty \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_1^\infty = 0 - (-1) = 1 \quad \text{(converges)}$$

By the limit comparison test, our integral also converges.

Conclusion

$$\boxed{\text{The integral } \int_1^\infty \frac{2x^2-3x+2}{\sqrt{x^8+3x^6+5}} \, dx \text{ CONVERGES}}$$

Why it converges

The key insight is that for large $x$:

$$\frac{\text{polynomial of degree 2}}{\sqrt{\text{polynomial of degree 8}}} = \frac{x^2}{x^4} = \frac{1}{x^2}$$

Since $\int_1^\infty \frac{1}{x^p} \, dx$ converges for $p > 1$, and here $p = 2 > 1$, the integral converges.

General principle

For rational functions at infinity:

$$\int_a^\infty \frac{P(x)}{\sqrt{Q(x)}} \, dx$$

where $P(x)$ is a polynomial of degree $m$ and $Q(x)$ is a polynomial of degree $n$:

The integral behaves like:

$$\int_a^\infty \frac{x^m}{x^{n/2}} \, dx = \int_a^\infty x^{m - n/2} \, dx$$

This converges if and only if:

$$m - \frac{n}{2} < -1 \quad \iff \quad m < \frac{n}{2} - 1 \quad \iff \quad 2m < n - 2 \quad \iff \quad n > 2m + 2$$

In our case:

• $m = 2$ (degree of numerator)
• $n = 8$ (degree of polynomial under square root)

Check: $8 > 2(2) + 2 = 6$ ✓

Therefore, the integral converges.

Alternative: Direct comparison test

For $x \geq 1$, we have:

• $2x^2 - 3x + 2 \leq 2x^2 - 3x + 2x^2 = 4x^2 - 3x \leq 4x^2$ (for large $x$)
• $\sqrt{x^8 + 3x^6 + 5} \geq \sqrt{x^8} = x^4$

Therefore:

$$\frac{2x^2-3x+2}{\sqrt{x^8+3x^6+5}} \leq \frac{4x^2}{x^4} = \frac{4}{x^2}$$

Since $\int_1^\infty \frac{4}{x^2} \, dx = 4$ converges, by the comparison test, our integral also converges.

Numerical estimate

Since the integral behaves like $\int_1^\infty \frac{2}{x^2} \, dx = 2$ for large $x$, we expect the value to be on the order of 2 (though the exact value will differ due to lower-order terms).

Summary

1. Identify dominant terms: numerator $\sim 2x^2$, denominator $\sim x^4$
2. Asymptotic behavior: $f(x) \sim \frac{2}{x^2}$
3. Apply limit comparison test with $g(x) = \frac{1}{x^2}$
4. Since $\int_1^\infty \frac{1}{x^2} \, dx$ converges and $\lim_{x \to \infty} \frac{f(x)}{g(x)} = 2$ (finite and positive)
5. Conclude: the integral converges
6. General criterion: For polynomials, need degree(denominator) > 2·degree(numerator) + 2

Key lessons:

• For rational functions, identify the dominant terms at infinity
• The asymptotic behavior determines convergence
• Limit comparison test is ideal when you can identify similar behavior
• For $\frac{x^m}{x^n}$ form, convergence requires $n - m > 1$
• Square roots in denominators halve the effective degree
• The criterion $n > 2m + 2$ when comparing $\frac{x^m}{\sqrt{x^n}}$