To determine whether the given integral

$\int_0^\infty \frac{e^x}{e^{\alpha x} + x^2} \, dx$

converges when $\alpha > 1$, we will use the comparison test.

Step 1: Analyze the behavior of the integrand for large $x$

• For large $x$, the term $e^{\alpha x}$ in the denominator dominates $x^2$ because exponential growth is much faster than polynomial growth.
• Therefore, for large $x$, the integrand behaves approximately like: $\frac{e^x}{e^{\alpha x}} = e^{(1-\alpha)x}.$

Since $\alpha > 1$, the exponent $1-\alpha$ is negative, and $e^{(1-\alpha)x}$ decays exponentially as $x \to \infty$. The integral of this type of function converges.

Step 2: Analyze the behavior of the integrand for small $x$

• For small $x$, the denominator $e^{\alpha x} + x^2 \approx 1 + x^2$ (since $e^{\alpha x} \approx 1$ for small $x$).
• The integrand behaves like: $\frac{e^x}{1 + x^2}.$ This function is finite and integrable over the range $0 \leq x \leq 1$.

Step 3: Use a formal comparison test

We can split the integral into two regions:

1. $x \in [0, 1]$: The integrand is finite and bounded because $\frac{e^x}{e^{\alpha x} + x^2} \leq \frac{e^x}{1 + x^2}$, which is integrable over this range.
2. $x \in [1, \infty)$: For large $x$, we compare $\frac{e^x}{e^{\alpha x} + x^2}$ with $e^{(1-\alpha)x}$. Since $e^{(1-\alpha)x}$ is integrable over $[1, \infty)$ when $\alpha > 1$, the original integrand is smaller than or comparable to an integrable function.

Thus, by the comparison test, the given integral converges when $\alpha > 1$.

Would you like me to elaborate on any part of this solution? Here are five related questions to explore:

1. How does the exponential growth rate compare to polynomial growth?
2. What is the comparison test for improper integrals?
3. Why does $e^{(1-\alpha)x}$ lead to convergence when $\alpha > 1$?
4. How does the behavior of the integrand near $x = 0$ affect convergence?
5. Can this integral be evaluated explicitly, or is only convergence important?

Tip: Always analyze the asymptotic behavior (both for large and small $x$) when testing for the convergence of improper integr