The problem asks to study the convergence of the improper integral as the parameter $\alpha \in \mathbb{R}$ varies:

$\int_0^{+\infty} \frac{(\sinh x)^\alpha}{x^{2\alpha}(5 + e^{x/2})} \, dx$

Steps for Analyzing Convergence:

To study the convergence of this improper integral, we should consider the behavior of the integrand at two critical points:

1. As $x \to 0$ (behavior near 0).
2. As $x \to +\infty$ (behavior at infinity).

1. Behavior as $x \to 0$:

• Near $x = 0$, the hyperbolic sine function $\sinh x$ behaves as $\sinh x \approx x$.
• Therefore, $(\sinh x)^\alpha \approx x^\alpha$ for small $x$.
• The denominator for small $x$ behaves as $x^{2\alpha} \cdot 5$ (since $e^{x/2} \approx 1$ for small $x$).

Thus, near $x = 0$, the integrand behaves as: $\frac{x^\alpha}{x^{2\alpha} \cdot 5} = \frac{1}{5 x^\alpha}$

For the integral to converge at $x = 0$, we require that: $\int_0^\epsilon \frac{1}{x^\alpha} \, dx$ converges. This integral converges if $\alpha < 1$.

2. Behavior as $x \to +\infty$:

• As $x \to +\infty$, $\sinh x \sim \frac{1}{2} e^x$.
• Thus, $(\sinh x)^\alpha \sim \left(\frac{1}{2} e^x\right)^\alpha = \frac{1}{2^\alpha} e^{\alpha x}$.
• The denominator for large $x$ behaves as $x^{2\alpha} \cdot e^{x/2}$.

Thus, for large $x$, the integrand behaves as: $\frac{e^{\alpha x}}{x^{2\alpha} e^{x/2}} = \frac{e^{(\alpha - 1/2)x}}{x^{2\alpha}}$

For the integral to converge as $x \to +\infty$, we require that the exponential term decays to 0. This happens if: $\alpha - \frac{1}{2} < 0 \quad \text{or} \quad \alpha < \frac{1}{2}$

Conclusion:

• Near $x = 0$, we require $\alpha < 1$ for convergence.
• Near $x = +\infty$, we require $\alpha < \frac{1}{2}$ for convergence.

Therefore, the integral converges for $\alpha < \frac{1}{2}$.

Would you like more details or have any questions on specific steps?

Related Questions:

1. How can we derive the asymptotic behavior of $\sinh(x)$ for large $x$?
2. What happens if $\alpha = \frac{1}{2}$? Does the integral converge or diverge?
3. Can you explain why $\sinh(x)$ behaves like $e^x/2$ for large $x$?
4. How does the behavior change if we replace the term $e^{x/2}$ with a different exponential?
5. What would happen if the power of $x$ in the denominator were changed?

Tip:

Always check the behavior of improper integrals at both ends of the integration range, especially when dealing with infinity.