To determine whether the given improper integral

$\int_{1}^{\infty} \frac{2 + e^{-x}}{x} \, dx$

is convergent or divergent using the comparison theorem, follow these steps:

Step 1: Identify the Dominant Term in the Integrand

The integrand is:

$f(x) = \frac{2 + e^{-x}}{x}.$

As $x \to \infty$, the term $e^{-x}$ decays exponentially to zero. Hence, for large $x$, the expression $2 + e^{-x}$ behaves like $2$. So we can approximate:

$\frac{2 + e^{-x}}{x} \sim \frac{2}{x}.$

This suggests that we compare the given integral to:

$\int_{1}^{\infty} \frac{2}{x} \, dx.$

Step 2: Compare the Integrals Using the Comparison Theorem

• Comparison Theorem states: If $0 \leq f(x) \leq g(x)$ for all $x \geq 1$ and $\int_{1}^{\infty} g(x) \, dx$ converges, then $\int_{1}^{\infty} f(x) \, dx$ also converges.

Here:

$f(x) = \frac{2 + e^{-x}}{x} \quad \text{and} \quad g(x) = \frac{2}{x}.$

For $x \geq 1$, we know that $e^{-x} \geq 0$, so:

$2 \leq 2 + e^{-x}.$

Thus:

$\frac{2}{x} \leq \frac{2 + e^{-x}}{x}.$

This shows $g(x) \leq f(x)$ for $x \geq 1$.

Step 3: Behavior of the Comparison Integral

Now, consider the integral of $g(x)$:

$\int_{1}^{\infty} \frac{2}{x} \, dx = 2 \int_{1}^{\infty} \frac{1}{x} \, dx.$

The integral of $1/x$ is:

$\int \frac{1}{x} \, dx = \ln |x| + C.$

Thus:

$\int_{1}^{\infty} \frac{2}{x} \, dx = 2 \left[ \ln x \right]_{1}^{\infty}.$

As $x \to \infty$, $\ln x \to \infty$, so:

$\int_{1}^{\infty} \frac{2}{x} \, dx = \infty.$

Step 4: Conclusion Using the Comparison Theorem

Since $\int_{1}^{\infty} \frac{2}{x} \, dx$ diverges and $\frac{2 + e^{-x}}{x} \geq \frac{2}{x}$, the given integral:

$\int_{1}^{\infty} \frac{2 + e^{-x}}{x} \, dx$

also diverges by the Comparison Theorem.

Final Answer:

The integral diverges.

Would you like further clarification or a breakdown of any specific step? Let me know!

Related Questions:

1. What is the Comparison Theorem for improper integrals?
2. How does $e^{-x}$ behave as $x \to \infty$?
3. How do you determine the dominant term in a function for large $x$?
4. Why does $\int_{1}^{\infty} \frac{1}{x} \, dx$ diverge?
5. How can the Limit Comparison Test be applied here?

Tip:

For improper integrals, always identify the dominant term as $x \to \infty$ to determine which comparison function to use.