Logarithmic integral

Remarks

The integrand tends to $0$ as $t \to 0^+$, so there is no problem at the lower end.

Example: the substitution $t = e^u$ gives $\operatorname{li}(x) = \operatorname{Ei}(\ln x)$, where $\operatorname{Ei}$ is the exponential integral. The function $\operatorname{li}$ has exactly one positive zero, at $x = \mu \approx 1.4513$, the Ramanujan–Soldner constant, and $\operatorname{li}(2) \approx 1.045$.

Non-example: without the principal value the integral diverges for $x > 1$, because $\frac{1}{\ln t} \sim \frac{1}{t - 1}$ near $t = 1$ and $\int \frac{dt}{\abs{t - 1}}$ diverges.

The prime number theorem is equivalent to $\pi(x) \sim \operatorname{li}(x)$, where $\pi$ is the prime-counting function. Integration by parts gives $\operatorname{li}(x) \sim \frac{x}{\ln x}$ as $x \to \infty$.