Sure! Here's a sample problem involving the logarithmic integral:
Problem: Evaluate the following integral using the logarithmic integral function:
\[ \int_{1}^{e} \frac{1}{x \log(x)} dx \]
Solution:
To solve this integral, we first notice that the integrand is of the form \(\frac{1}{x \log(x)}\). We can rewrite this as \(\frac{1}{\log(x)}\cdot \frac{1}{x}\), and then the substitution \(u = \log(x)\) is natural. Thus, \(du = \frac{1}{x} dx\), which transforms the integral into
\[ \int_{0}^{1} \frac{1}{u} du. \]
This integral is simply \(\log(u)\), so we find that the original integral evaluates to \(\log(\log(x))\), evaluated from \(1\) to \(e\). Thus,
\[ \int_{1}^{e} \frac{1}{x \log(x)} dx = \log(\log(e)) - \log(\log(1)) = \log(1) - \log(0) = \infty. \]
Therefore, the integral diverges.