Answer :
To determine the convergence of the improper integral [tex]\(\int_0^{\infty} \frac{\sin x + 2}{\log x} \, dx\)[/tex], we need to analyze the behavior of the integrand at the endpoints [tex]\(0\)[/tex] and [tex]\(\infty\)[/tex].
### Step 1: Examine the behavior near [tex]\(x = 0\)[/tex]
As [tex]\(x\)[/tex] approaches [tex]\(0\)[/tex], the term [tex]\(\log x\)[/tex] diverges to [tex]\(-\infty\)[/tex]. It is customary to perform a change of variable [tex]\(u = \log x\)[/tex], where [tex]\(x = e^u\)[/tex]. Consequently:
- When [tex]\(x \to 0^+\)[/tex], [tex]\(u \to -\infty\)[/tex].
- When [tex]\(x \to \infty\)[/tex], [tex]\(u \to \infty\)[/tex].
The differential [tex]\(dx = e^u \, du\)[/tex]. Substituting these into the integral, we get:
[tex]\[ \int_0^{\infty} \frac{\sin x + 2}{\log x} \, dx = \int_{-\infty}^{\infty} \frac{\sin(e^u) + 2}{u} \, e^u \, du \][/tex]
Since we are primarily concerned with potential convergence issues, we need to examine the integral's behavior at the ends of the integration range.
### Step 2: Examine the behavior near [tex]\(x = \infty\)[/tex]
As [tex]\(x \to \infty\)[/tex]:
- The behavior of [tex]\(\sin x\)[/tex] oscillates between [tex]\(-1\)[/tex] and [tex]\(1\)[/tex].
- The term [tex]\(2\)[/tex] is constant.
- The term [tex]\(\log x\)[/tex] diverges to [tex]\(\infty\)[/tex].
Given the presence of a constant and an oscillating term divided by a logarithmic term, let us separately consider the integral's dominant converging part:
[tex]\[ \int_1^{\infty} \frac{2}{\log x} \, dx \][/tex]
Changing variables [tex]\( u = \log x \)[/tex]:
[tex]\[ du = \frac{dx}{x} \implies dx = e^u du \][/tex]
Rewriting the integral:
[tex]\[ \int_1^{\infty} \frac{2}{\log x} \, dx = \int_{0}^{\infty} \frac{2}{u} \, e^u \, du \][/tex]
The integral [tex]\(\int_{0}^{\infty} \frac{2}{u} \, e^u \, du\)[/tex] diverges because [tex]\( \frac{2}{u} \)[/tex] and [tex]\( e^u \)[/tex] grow without bound.
### Step 3: Considering [tex]\(\sin x + 2\)[/tex] near [tex]\(x = 0\)[/tex]
Near [tex]\(x = 0\)[/tex]:
- [tex]\(\log x \to -\infty\)[/tex].
- Since [tex]\(\sin x\)[/tex] behaves linearly, being negligible before divergence, the term [tex]\(2\)[/tex] dominates.
Combined, the key term becomes:
[tex]\[ \int_{0}^{1} \frac{2}{\log x} \, dx \][/tex]
Rewriting with the substitution [tex]\( u = \log x \)[/tex]:
\[
dx = \frac{e^u}{u}du, so \, integral becomes \int_{-\infty}^0 \frac {2}{u}e^u du= diverges.
### Conclusion
Given detailed observations demonstrate nonlinear asymptotic results as x->0 and x->∞ including dominant divergent term for [tex]\(sin x\)[/tex].
Thus, the integral [tex]\(\int_0^{\infty} \frac{\sin x + 2}{\log x} \, dx\)[/tex] \textbf{diverges}.
### Step 1: Examine the behavior near [tex]\(x = 0\)[/tex]
As [tex]\(x\)[/tex] approaches [tex]\(0\)[/tex], the term [tex]\(\log x\)[/tex] diverges to [tex]\(-\infty\)[/tex]. It is customary to perform a change of variable [tex]\(u = \log x\)[/tex], where [tex]\(x = e^u\)[/tex]. Consequently:
- When [tex]\(x \to 0^+\)[/tex], [tex]\(u \to -\infty\)[/tex].
- When [tex]\(x \to \infty\)[/tex], [tex]\(u \to \infty\)[/tex].
The differential [tex]\(dx = e^u \, du\)[/tex]. Substituting these into the integral, we get:
[tex]\[ \int_0^{\infty} \frac{\sin x + 2}{\log x} \, dx = \int_{-\infty}^{\infty} \frac{\sin(e^u) + 2}{u} \, e^u \, du \][/tex]
Since we are primarily concerned with potential convergence issues, we need to examine the integral's behavior at the ends of the integration range.
### Step 2: Examine the behavior near [tex]\(x = \infty\)[/tex]
As [tex]\(x \to \infty\)[/tex]:
- The behavior of [tex]\(\sin x\)[/tex] oscillates between [tex]\(-1\)[/tex] and [tex]\(1\)[/tex].
- The term [tex]\(2\)[/tex] is constant.
- The term [tex]\(\log x\)[/tex] diverges to [tex]\(\infty\)[/tex].
Given the presence of a constant and an oscillating term divided by a logarithmic term, let us separately consider the integral's dominant converging part:
[tex]\[ \int_1^{\infty} \frac{2}{\log x} \, dx \][/tex]
Changing variables [tex]\( u = \log x \)[/tex]:
[tex]\[ du = \frac{dx}{x} \implies dx = e^u du \][/tex]
Rewriting the integral:
[tex]\[ \int_1^{\infty} \frac{2}{\log x} \, dx = \int_{0}^{\infty} \frac{2}{u} \, e^u \, du \][/tex]
The integral [tex]\(\int_{0}^{\infty} \frac{2}{u} \, e^u \, du\)[/tex] diverges because [tex]\( \frac{2}{u} \)[/tex] and [tex]\( e^u \)[/tex] grow without bound.
### Step 3: Considering [tex]\(\sin x + 2\)[/tex] near [tex]\(x = 0\)[/tex]
Near [tex]\(x = 0\)[/tex]:
- [tex]\(\log x \to -\infty\)[/tex].
- Since [tex]\(\sin x\)[/tex] behaves linearly, being negligible before divergence, the term [tex]\(2\)[/tex] dominates.
Combined, the key term becomes:
[tex]\[ \int_{0}^{1} \frac{2}{\log x} \, dx \][/tex]
Rewriting with the substitution [tex]\( u = \log x \)[/tex]:
\[
dx = \frac{e^u}{u}du, so \, integral becomes \int_{-\infty}^0 \frac {2}{u}e^u du= diverges.
### Conclusion
Given detailed observations demonstrate nonlinear asymptotic results as x->0 and x->∞ including dominant divergent term for [tex]\(sin x\)[/tex].
Thus, the integral [tex]\(\int_0^{\infty} \frac{\sin x + 2}{\log x} \, dx\)[/tex] \textbf{diverges}.
