We are tasked with checking the convergence of the integral

$I = \int_0^{\infty} \frac{1}{e^x + 1} \, dx.$

Step 1: Investigate behavior near $x = 0$

At $x = 0$, the function $\frac{1}{e^x + 1}$ is well-behaved. In fact:

$\frac{1}{e^x + 1} \bigg|_{x = 0} = \frac{1}{1 + 1} = \frac{1}{2}.$

Since the function is continuous near $x = 0$, the integral from 0 to some positive finite number is bounded.

Step 2: Investigate behavior as $x \to \infty$

As $x \to \infty$, we have:

$\lim_{x \to \infty} \frac{1}{e^x + 1} = 0.$

For large $x$, $e^x$ grows exponentially, so the function $\frac{1}{e^x + 1}$ decays rapidly like $e^{-x}$. Thus, for large $x$, the integrand behaves similarly to $e^{-x}$, and we know the integral

$\int_1^{\infty} e^{-x} \, dx$

converges (its value is 1).

Step 3: Conclusion

Since the integrand $\frac{1}{e^x + 1}$ behaves nicely near $x = 0$ and decays exponentially as $x \to \infty$, the improper integral converges.

The integral is therefore convergent.

Would you like the actual value of this integral or further details on this process? Here are some related questions:

1. What are the exact steps to calculate the value of this integral?
2. How does exponential decay affect the convergence of integrals?
3. Can you compare this integral to others with similar behavior at infinity?
4. What are some common improper integrals that do not converge?
5. How can we generalize this method to check the convergence of other integrals?

Tip: Always check the behavior of the integrand at the endpoints (0 and ∞) to determine convergence.