Investigate the convergence of the integral: integral

Investigate the convergence of the integral: $$ \int_{-1}^\infty \frac{dx}{x^2+\sqrt[y]{x^4+1}} $$

This integral is improper at infinity and depends on the parameter $y$. We need to analyze the behavior of the integrand for large $x$.

Step 1: Check for singularities in $[-1, \infty)$

The integrand is:

$$f(x) = \frac{1}{x^2 + (x^4+1)^{1/y}}$$

Denominator:

$$x^2 + (x^4+1)^{1/y}$$

For $x \geq -1$:

• $x^2 \geq 0$
• $x^4 + 1 \geq 1 > 0$, so $(x^4+1)^{1/y}$ is well-defined and positive (for $y \neq 0$)

The denominator is always positive, so there are no singularities on $[-1, \infty)$. The integral is only improper at $x = \infty$.

Step 2: Analyze behavior at infinity

For large $x$ (specifically $x \gg 1$):

Term 1: $x^2$

Term 2: $(x^4+1)^{1/y}$

For large $x$: $x^4 + 1 \approx x^4$, so:

$$(x^4+1)^{1/y} \approx (x^4)^{1/y} = x^{4/y}$$

Therefore, the denominator behaves like:

$$x^2 + x^{4/y}$$

Step 3: Determine dominant term

The dominant term depends on the relative magnitudes of $x^2$ and $x^{4/y}$:

Case A: $\frac{4}{y} > 2$, i.e., $y < 2$

When $y < 2$: $x^{4/y} > x^2$ for large $x$, so the dominant term is $x^{4/y}$.

The integrand behaves like:

$$f(x) \sim \frac{1}{x^{4/y}} \quad \text{as } x \to \infty$$

Case B: $\frac{4}{y} < 2$, i.e., $y > 2$

When $y > 2$: $x^2 > x^{4/y}$ for large $x$, so the dominant term is $x^2$.

The integrand behaves like:

$$f(x) \sim \frac{1}{x^2} \quad \text{as } x \to \infty$$

Case C: $\frac{4}{y} = 2$, i.e., $y = 2$

When $y = 2$: Both terms are $x^2$, so:

$$f(x) \sim \frac{1}{2x^2} \quad \text{as } x \to \infty$$

Step 4: Apply convergence criterion

We know that $\int_a^\infty \frac{dx}{x^p}$ converges if and only if $p > 1$.

Case A: $y < 2$

The integrand behaves like $\frac{1}{x^{4/y}}$.

Converges if: $\frac{4}{y} > 1 \iff y < 4$

So for $y < 2$: always $y < 4$, so the integral **converges**. ✓

Case B: $y > 2$

The integrand behaves like $\frac{1}{x^2}$.

Since $2 > 1$, the integral **converges**. ✓

Case C: $y = 2$

The integrand behaves like $\frac{1}{2x^2}$.

Since $2 > 1$, the integral **converges**. ✓

Step 5: Special cases requiring closer inspection

Wait, let's reconsider Case A more carefully.

For $y < 2$, we have $\frac{4}{y} > 2$.

The convergence criterion is $\frac{4}{y} > 1$, which means $y > 4$ is needed for convergence when the dominant term is $x^{4/y}$.

But we're in the case $y < 2$, so we need to check if $y > 4$ and $y < 2$ can both be satisfied. They cannot!

Let me reconsider...

Step 4 (Corrected): Convergence analysis

If $y < 2$:

Dominant term is $x^{4/y}$ (since $4/y > 2$).

For convergence, need: $\frac{4}{y} > 1 \implies y > 4$

But $y < 2$ and $y > 4$ cannot both be true!

Therefore: For $0 < y < 2$, we have $\frac{4}{y} > 2$ but also $\frac{4}{y} < \infty$.

Specifically:

- When $y = 1$: $\frac{4}{y} = 4 > 1$ → **converges** ✓

- When $y = 0.5$: $\frac{4}{y} = 8 > 1$ → **converges** ✓

So for $0 < y < 2$: integral converges if $\frac{4}{y} > 1$, i.e., $y > 4$ is NOT satisfied, but wait...

Let me recalculate: For convergence of $\int \frac{dx}{x^p}$, we need $p > 1$.

If $p = \frac{4}{y}$, then we need $\frac{4}{y} > 1$, which gives $4 > y$, i.e., $y < 4$.

For $y < 2$:

We have $\frac{4}{y} > 2 > 1$, so the integral **converges**. ✓

For $y = 2$:

We have $\frac{4}{y} = 2 > 1$, so the integral **converges**. ✓

For $y > 2$:

Dominant term is $x^2$, and $2 > 1$, so the integral **converges**. ✓

Conclusion

$$\boxed{\text{The integral converges for all } y > 0}$$

Actually, wait. Let me verify the case when $y$ is very large or very small.

Step 6: Verification of extreme cases

Case: $y \to 0^+$

$(x^4+1)^{1/y} = (x^4+1)^{\infty} \to \infty$ very rapidly.

The integrand $\frac{1}{x^2 + (x^4+1)^{1/y}} \to 0$ very rapidly, so convergence is even stronger.

Case: $y \to \infty$

$(x^4+1)^{1/y} = (x^4+1)^0 \to 1$.

The integrand $\frac{1}{x^2 + 1}$.

We know $\int_{-1}^\infty \frac{dx}{x^2+1}$ converges (it's $[\arctan x]_{-1}^\infty$, which is finite).

Case: $y = 4$

$(x^4+1)^{1/4} \approx x$ for large $x$.

Integrand $\sim \frac{1}{x^2 + x} \sim \frac{1}{x^2}$ for large $x$ (since $x^2 > x$).

This converges. ✓

Final Answer

For $y > 0$, the denominator is:

$$x^2 + (x^4+1)^{1/y}$$

For large $x$, this behaves like:

$$\max(x^2, x^{4/y})$$

• If $\frac{4}{y} > 2$ (i.e., $y < 2$): dominated by $x^{4/y}$, and $\frac{4}{y} > 2 > 1$, so converges
• If $\frac{4}{y} = 2$ (i.e., $y = 2$): both terms equal, behaves like $x^2$, so converges
• If $\frac{4}{y} < 2$ (i.e., $y > 2$): dominated by $x^2$, and $2 > 1$, so converges

$$\boxed{\text{The integral CONVERGES for all } y > 0}$$

Summary

1. The integral is only improper at infinity (no interior singularities)
2. For large $x$: $(x^4+1)^{1/y} \sim x^{4/y}$
3. Denominator: $x^2 + x^{4/y}$ dominated by the larger power
4. For all $y > 0$: either $x^{4/y}$ or $x^2$ dominates, both with exponent $> 1$
5. By the comparison test: integral converges for all $y > 0$

Key lessons:

• For sums in denominators, the larger power dominates at infinity
• Parameters in exponents affect which term dominates
• For all reasonable values of $y > 0$, we get sufficient decay
• The minimum decay rate is $x^{-2}$ (when $y \geq 2$ or $y \to \infty$)
• The maximum decay rate occurs as $y \to 0^+$ (super-exponential)