To analyze the convergence of the integral

[ \int_{0}^{1} \frac{\cos(\pi x)}{\sqrt[4]{x} + \sqrt[3]{x^2} + \sqrt{x}} , dx, ]

we need to examine the behavior of the integrand near the boundaries of the interval, specifically as $x \to 0^+$ and $x \to 1^-$.

Step 1: Boundary Analysis at $x = 0$

Near $x = 0$, we look at the behavior of each term in the denominator:

• $\sqrt[4]{x} = x^{1/4}$,
• $\sqrt[3]{x^2} = x^{2/3}$,
• $\sqrt{x} = x^{1/2}$.

The smallest exponent here is $x^{1/4}$, meaning $x^{1/4}$ dominates the denominator as $x \to 0^+$. Thus, the integrand behaves like

$\frac{\cos(\pi x)}{x^{1/4}} \approx \frac{1}{x^{1/4}}.$

To check convergence near $x = 0$, we need to see if the integral

$\int_{0}^{1} \frac{1}{x^{1/4}} \, dx$

converges. This integral is of the form $\int \frac{1}{x^{p}} \, dx$ with $p = 1/4$. Since $p < 1$, this integral converges near $x = 0$.

Step 2: Boundary Analysis at $x = 1$

As $x \to 1$, each term $\sqrt[4]{x}$, $\sqrt[3]{x^2}$, and $\sqrt{x}$ approaches $1$. Therefore, the denominator approaches a constant, and the integrand $\frac{\cos(\pi x)}{\text{constant}}$ remains bounded as $x \to 1$.

Conclusion

Since the integrand is integrable near both $x = 0$ and $x = 1$, the integral converges on $[0, 1]$.

Would you like more details on any specific step, or do you have questions?

Here are five questions related to this analysis:

1. How would the convergence change if the exponent of $x$ in the denominator terms were altered?
2. What role does $\cos(\pi x)$ play in the convergence of the integral?
3. Can you explain why $p < 1$ ensures convergence for $\int \frac{1}{x^p} \, dx$ as $x \to 0$?
4. What would happen if the limits of integration were extended beyond $[0, 1]$?
5. How would the analysis change if we replaced $\cos(\pi x)$ with $\sin(\pi x)$?

Tip: When analyzing integrals for convergence, always examine the behavior of the integrand near points where it might become unbounded, such as near zero or at infinity.