Let us analyze the given integrals and determine whether they converge or diverge:

Integrals:

1. $\int_0^1 \frac{e^{x^2} - 1}{\sqrt{\tan{\sqrt{x}}}} \, dx$
2. $\int_0^1 \frac{\sin^2{x}}{x^3} \, dx$
3. $\int_0^1 \frac{e^{\sqrt{x}} - 1}{x} \, dx$

We will evaluate the behavior of each integral at the given bounds.

Step 1: Analyze $\int_0^1 \frac{e^{x^2} - 1}{\sqrt{\tan{\sqrt{x}}}} \, dx$

• As $x \to 0^+$:

  • The numerator $e^{x^2} - 1 \approx x^2$ (using Taylor expansion of $e^{x^2}$).
  • The denominator $\sqrt{\tan{\sqrt{x}}}$ approximates $\sqrt{\sqrt{x}} = x^{1/4}$.
  • The integrand behaves like $\frac{x^2}{x^{1/4}} = x^{7/4}$, which is integrable at 0.

• As $x \to 1^-$:

  • Both $e^{x^2} - 1$ and $\tan{\sqrt{x}}$ are finite. The integrand remains finite.

Conclusion: This integral converges.

Step 2: Analyze $\int_0^1 \frac{\sin^2{x}}{x^3} \, dx$

• As $x \to 0^+$:
  • The numerator $\sin^2{x} \sim x^2$ (using $\sin{x} \approx x$).
  • The denominator $x^3$ dominates $x^2$, and the integrand behaves like $\frac{x^2}{x^3} = \frac{1}{x}$, which is not integrable at 0 (diverges as $\ln{x}$).

Conclusion: This integral diverges.

Step 3: Analyze $\int_0^1 \frac{e^{\sqrt{x}} - 1}{x} \, dx$

• As $x \to 0^+$:
  • The numerator $e^{\sqrt{x}} - 1 \approx \sqrt{x}$ (using Taylor expansion of $e^{\sqrt{x}}$).
  • The denominator $x$ dominates $\sqrt{x}$, and the integrand behaves like $\frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}$, which is not integrable at 0.

Conclusion: This integral diverges.

Final Answers:

1. $\int_0^1 \frac{e^{x^2} - 1}{\sqrt{\tan{\sqrt{x}}}} \, dx$ — Converges.
2. $\int_0^1 \frac{\sin^2{x}}{x^3} \, dx$ — Diverges.
3. $\int_0^1 \frac{e^{\sqrt{x}} - 1}{x} \, dx$ — Diverges.

Would you like further explanations or steps for any specific case?

Related Questions:

1. What is the mathematical concept of convergence for improper integrals?
2. How can Taylor expansions help analyze limits of functions?
3. How does the behavior near singular points affect integrals?
4. What are the key differences between convergent and divergent integrals?
5. How can substitution simplify the evaluation of improper integrals?

Tip:

Always check the behavior of the integrand at critical points (like $x \to 0$ or $x \to \infty$) to determine convergence.