I am doing work on analytic number theory, and I am currently looking at the Prime Number Theorem, that is
$$\pi(x) \sim Li(x)$$
Some of my sources say that I can do integration by parts on the integral that defines Li$(x)$, that is $\int_2^x \frac{dt}{\ln(t)}$ and then take the limit as $x \rightarrow \infty$, to see that Li$(x)$ is asymptotic to $\frac{x}{\ln (x)}$ and so $\pi(x) \sim \frac{x}{\ln (x)}$.
I'm sure this is pretty easy, and I've done the integration by parts, but I am quite rusty so I'm not entirely sure how the calculation would go.
By integration by parts I get soemthing like:
$$\int \frac{1}{\ln(x)} dx = \frac{x}{\ln (x)} + \int \frac{1}{\ln(x)^2} dx$$
It's been a while, so I don't rememeber if the following is valid:
Let $I :=\int \frac{1}{\ln(x)} dx$, so we have:
$$I = \frac{x}{\ln (x)} + \frac{I}{\ln(x)}$$
so $I(1+\ln(x)) = x$, so $I=\frac{x}{1+\ln(x)}$, and as $x \rightarrow \infty$ this is $\frac{x}{\ln(x)}$,
so Li$(x) \sim \frac{x}{\ln (x)}$.
I'm not at all confident this is correct, so I was wondering if you can show me what I am doing wrong.
Many thanks.
Best Answer
This is wrong, because if you call
$$I = \int \frac{1}{\ln(x)}\ \text{d}x$$
Then
$$\int \frac{1}{\ln^2(x)}\ \text{d}x \neq \frac{I}{\ln(x)}$$
That is FALSE.