While solving an exercise in analytic number theory, I ran into difficulty of estimating an integral of the form $\displaystyle\int_{1}^{x} \frac{\pi(t)}{t} dt$ where $\pi(x)$ is the prime counting function.
I am interested in understanding how to estimate this integral (as a function of $x$ of course), whether it is a big Oh estimate, or something more precise. Any references are appreciated.
Similarly, how would one estimate $\displaystyle\int_{1}^{x} \frac{\pi(t)}{t^2} dt$ as a function of $x$?
Thanks!
I am posting to add to the answers already appearing in the comments, and so that the question does not remain unanswered.
By the prime number theorem $$\int_1^x \frac{\pi(t)}{t}dt = \int_1^x \frac{1}{\log t}dt+O\left(\int_1^x \frac{1}{\log^2 t}dt\right)$$ $$=\frac{x}{\log x}+O\left(\frac{x}{\log^2 x}\right).$$ Using Chebyshev's bounds instead of the PNT allows us to show that the integral is $O\left(\frac{x}{\log x}\right).$
To be more precise, we may note that $$\int_1^x \frac{\pi(t)}{t} dt=\sum_{p\leq x} \log(x/p).$$ This is a basic example of Riesz weights. Then we have that $$\sum_{p\leq x} \log x =\log x \text{li}(x)+O\left(xe^{-c\sqrt{\log x}}\right)$$ where $\text{li}(x)=\int_2^x \frac{1}{\log t}dt$, and $$\sum_{p\leq x} \log p = \theta(x)=x+O\left(xe^{-c\sqrt{\log x}}\right).$$ Hence $$\int_1^x \frac{\pi(t)}{t} dt=\log x \text{li}(x)-x +O\left(xe^{-c\sqrt{\log x}}\right)$$ which implies that $$\int_1^x \frac{\pi(t)}{t} dt=\frac{x}{\log x}+\frac{2x}{\log^2 x}+\cdots .$$