To solve the integral $\int_{1}^{\infty} \frac{\ln x}{x} \, dx$, let's proceed with the integration step by step.

Step 1: Recognize the integral form

The given integral is: $\int_{1}^{\infty} \frac{\ln x}{x} \, dx$

Step 2: Perform substitution

Let $u = \ln x$. Then, the differential $du = \frac{1}{x} dx$.

This transforms the integral into: $\int_{0}^{\infty} u \, du$ (Notice that when $x = 1$, $u = \ln(1) = 0$, and as $x$ approaches infinity, $u$ also approaches infinity.)

Step 3: Integrate

Now, integrate $u$ with respect to $u$: $\int u \, du = \frac{u^2}{2}$

Step 4: Apply the limits

Substitute the limits $u = 0$ and $u \rightarrow \infty$ into the integrated function: $\left[ \frac{u^2}{2} \right]_{0}^{\infty} = \lim_{u \to \infty} \frac{u^2}{2} - \frac{0^2}{2}$

As $u$ approaches infinity, $\frac{u^2}{2}$ also approaches infinity.

Conclusion

The integral diverges, meaning: $\int_{1}^{\infty} \frac{\ln x}{x} \, dx = \infty$ So, the integral does not converge to a finite value.

Would you like further details on this process or have any questions?

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Related Questions:

1. How can we determine if an integral converges or diverges?
2. What are common techniques for solving improper integrals?
3. How does substitution simplify integrals with logarithmic functions?
4. Can you provide an example of a convergent improper integral?
5. What are the applications of improper integrals in physics?

Tip: When dealing with improper integrals, always check the behavior of the integrand at the limits of integration to assess convergence.