Investigate the convergence of the integral: integral

Investigate the convergence of the integral: $$ \int_0^\infty \frac{x^a \arctan x}{c+x^b} \, dx \quad (b>0, c>0) $$

This integral is improper at both $x = 0$ and $x = \infty$. We must analyze convergence at both endpoints separately.

Step 1: Split the integral

$$\int_0^\infty \frac{x^a \arctan x}{c+x^b} \, dx = \int_0^1 \frac{x^a \arctan x}{c+x^b} \, dx + \int_1^\infty \frac{x^a \arctan x}{c+x^b} \, dx$$

For the integral to converge, BOTH parts must converge.

Step 2: Analyze convergence at $x = 0$

For small $x$ (as $x \to 0^+$):

• $\arctan x \approx x$ (Taylor series)
• $c + x^b \approx c$ (since $c > 0$ and $x^b \to 0$)

Therefore:

$$\frac{x^a \arctan x}{c+x^b} \approx \frac{x^a \cdot x}{c} = \frac{x^{a+1}}{c}$$

The integral $\int_0^1 x^{a+1} \, dx$ converges if and only if $a+1 > -1$, i.e., **$a > -2$**.

Step 3: Analyze convergence at $x = \infty$

For large $x$ (as $x \to \infty$):

• $\arctan x \to \frac{\pi}{2}$ (bounded)
• $c + x^b \approx x^b$ (dominant term)

Therefore:

$$\frac{x^a \arctan x}{c+x^b} \approx \frac{x^a \cdot \frac{\pi}{2}}{x^b} = \frac{\pi}{2} \cdot x^{a-b}$$

The integral $\int_1^\infty x^{a-b} \, dx$ converges if and only if $a-b < -1$, i.e., **$a < b-1$**.

Step 4: Combine conditions

For the integral to converge, we need BOTH:

1. Convergence at $x = 0$: $a > -2$
2. Convergence at $x = \infty$: $a < b - 1$

These conditions are compatible if and only if:

$$-2 < a < b - 1$$

This requires $b - 1 > -2$, i.e., $b > -1$.

Since we're given $b > 0$, we automatically have $b > -1$. ✓

Final Answer

$$\boxed{\text{The integral converges if and only if } -2 < a < b-1}$$

Given the constraints $b > 0$ and $c > 0$, this is equivalent to:

$$\boxed{a \in (-2, b-1)}$$

Special cases and boundary analysis

Case 1: $a = -2$

At $x = 0$: $\frac{x^{-2} \cdot x}{c} = \frac{x^{-1}}{c} = \frac{1}{cx}$

This gives $\int_0^1 \frac{dx}{x}$, which **diverges**. ✗

Case 2: $a = b-1$

At $x = \infty$: $\frac{x^{b-1}}{x^b} = \frac{1}{x}$

This gives $\int_1^\infty \frac{dx}{x}$, which **diverges**. ✗

Case 3: $a = 0, b = 2$

Condition: $-2 < 0 < 1$? Yes! ✓

The integral $\int_0^\infty \frac{\arctan x}{c+x^2} \, dx$ converges.

Case 4: $a = 1, b = 3$

Condition: $-2 < 1 < 2$? Yes! ✓

The integral $\int_0^\infty \frac{x \arctan x}{c+x^3} \, dx$ converges.

Case 5: $a = 1, b = 2$

Condition: $-2 < 1 < 1$? No! (Need $1 < 1$) ✗

The integral $\int_0^\infty \frac{x \arctan x}{c+x^2} \, dx$ diverges at infinity.

Detailed verification at $x = 0$

The integrand near $x = 0$ is:

$$\frac{x^a \arctan x}{c+x^b}$$

For $x$ near 0:

$$\arctan x = x - \frac{x^3}{3} + O(x^5) \approx x$$

$$c + x^b \approx c$$

So:

$$\frac{x^a \arctan x}{c+x^b} \approx \frac{x^{a+1}}{c}$$

The integral $\int_0^1 x^{a+1} \, dx$ converges if $a+1 > -1$.

Detailed verification at $x = \infty$

The integrand for large $x$ is:

$$\frac{x^a \arctan x}{c+x^b}$$

For large $x$:

$$\arctan x \to \frac{\pi}{2}$$

$$c + x^b = x^b\left(\frac{c}{x^b} + 1\right) \approx x^b$$

So:

$$\frac{x^a \arctan x}{c+x^b} \approx \frac{x^a \cdot \frac{\pi}{2}}{x^b} = \frac{\pi}{2x^{b-a}}$$

The integral $\int_1^\infty \frac{1}{x^{b-a}} \, dx$ converges if $b - a > 1$.

Graphical interpretation

The convergence region is:

$$\{(a,b) : b > 0 \text{ and } -2 < a < b-1\}$$

In the $(a,b)$ plane:

• Lower boundary: $a = -2$ (horizontal line)
• Upper boundary: $a = b - 1$ (line with slope 1)
• Left boundary: $b = 0$ (not included)

The convergence region is the area between $a = -2$ and $a = b - 1$ for $b > 0$.

Summary

1. Split the integral at $x = 1$ to analyze both endpoints
2. At $x = 0$: Integrand $\sim x^{a+1}$, converges if $a > -2$
3. At $x = \infty$: Integrand $\sim x^{a-b}$, converges if $a < b-1$
4. Combined condition: $-2 < a < b - 1$
5. The interval is non-empty since $b > 0 \implies b - 1 > -1 > -2$

Key lessons:

• Integrals on $(0, \infty)$ require checking both endpoints
• Use Taylor series near 0: $\arctan x \approx x$
• Use limits at $\infty$: $\arctan x \to \frac{\pi}{2}$
• For rational functions with powers, identify dominant terms
• The convergence region forms an interval in the parameter $a$
• Both boundaries are excluded (strict inequalities)
• Parameters $b$ and $c$ affect the convergence condition differently