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To determine if the given integrals are convergent or divergent using the Comparison Theorem, let's analyze each one step-by-step.

### (a) \(\int_{4}^{\infty} \frac{dx}{\ln x - 1}\)

We need to find a function \(g(x)\) such that:

1. \(0 \leq \frac{1}{\ln x - 1} \leq g(x)\) for all \(x \geq 4\)
2. The integral \(\int_{4}^{\infty} g(x) \, dx\) is convergent or divergent.

Let's choose \(g(x) = \frac{1}{\ln x}\). For \(x \geq 4\), \(\ln x - 1 < \ln x\), hence \(\frac{1}{\ln x - 1} > \frac{1}{\ln x}\). 

Now, let's examine the integral \(\int_{4}^{\infty} \frac{1}{\ln x} \, dx\):

Using the substitution \(u = \ln x\), \(du = \frac{1}{x} dx\):

\[
\int_{4}^{\infty} \frac{1}{\ln x} \, dx = \int_{\ln 4}^{\infty} \frac{1}{u} \, du = \left. \ln |u| \right|_{\ln 4}^{\infty}
\]

The integral \(\int_{\ln 4}^{\infty} \frac{1}{u} \, du\) diverges because \(\ln |u|\) approaches infinity as \(u \to \infty\).

Since \(\int_{4}^{\infty} \frac{1}{\ln x} \, dx\) diverges and \(\frac{1}{\ln x - 1} > \frac{1}{\ln x}\) for \(x \geq 4\), by the Comparison Theorem, the integral \(\int_{4}^{\infty} \frac{dx}{\ln x - 1}\) also diverges.

### (b) \(\int_{1}^{\infty} e^{-x^2 / 2} \, dx\)

We need to find a function \(g(x)\) such that:

1. \(0 \leq e^{-x^2 / 2} \leq g(x)\) for all \(x \geq 1\)
2. The integral \(\int_{1}^{\infty} g(x) \, dx\) is convergent or divergent.

Let's choose \(g(x) = e^{-x}\). Notice that for \(x \geq 1\):

\[
e^{-x^2 / 2} \leq e^{-x}
\]

because \(x^2 / 2 \geq x\) for \(x \geq 1\).

Now, let's examine the integral \(\int_{1}^{\infty} e^{-x} \, dx\):

\[
\int_{1}^{\infty} e^{-x} \, dx = \left. -e^{-x} \right|_{1}^{\infty} = 0 - (-e^{-1}) = \frac{1}{e}
\]

The integral \(\int_{1}^{\infty} e^{-x} \, dx\) converges.

Since \(\int_{1}^{\infty} e^{-x} \, dx\) converges and \(e^{-x^2 / 2} \leq e^{-x}\) for \(x \geq 1\), by the Comparison Theorem, the integral \(\int_{1}^{\infty} e^{-x^2 / 2} \, dx\) also converges.

### Summary

- (a) \(\int_{4}^{\infty} \frac{dx}{\ln x - 1}\) diverges.
- (b) \(\int_{1}^{\infty} e^{-x^2 / 2} \, dx\) converges.

Do you want details on any part of the solution, or do you have any other questions?

Here are some more questions you might ask:
1. Can you explain the Comparison Theorem in more detail?
2. How do you determine which function to compare an integral to?
3. What are some common divergent and convergent integrals to know?
4. Can you solve an example of an improper integral using a different method?
5. What are other techniques for determining the convergence of an integral?

**Tip:** When using the Comparison Theorem, always ensure that the function you compare to is simpler and that you know whether its integral converges or diverges.

Explore the convergence of improper integrals using the Comparison Theorem. Detailed analysis of two integrals: \(\int_{4}^{\infty} \frac{dx}{\ln x - 1}\) diverges, and \(\int_{1}^{\infty} e^{-x^2 / 2} \, dx\) converges. Suitable for advanced undergraduate students and anyone interested in calculus and integration.

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