# Asymptotic lower bound for the prime counting function, $\pi(x)?$

Consider a function that attempts to count primes up to a given $$x$$:$$\varphi(x)=\int_2^x \frac{1}{\log(t)}e^{-\frac{1}{\sqrt{t}}}~dt$$

Is $$\varphi(x)$$ an asymptotic lower bound to the prime counting function, $$\pi(x)?$$

If you take away the square root on the $$t$$ the function does only slightly better than $$Li(x)$$ and I'm interested in a lower bound that is also asymptotic.

Just to provide some numerical results, $$\varphi(10^9)$$ is less than $$\pi(10^9)$$ by approximately $$1,750$$ which is approximately the difference between $$Li(x)$$ and $$\pi(x).$$

\begin{align*} \phi(x) = \int_2^x \frac1{\log t}e^{-1/\sqrt t}\,dt &= \int_2^x \frac1{\log t} \bigg( 1 - \frac1{\sqrt t} + O\bigg( \frac1t \bigg) \bigg)\,dt \\ &= \mathop{\rm li}(x) - \int_2^x \frac1{\sqrt t\log t}\,dt + O(\log\log x). \end{align*} The integral has order of magnitude $$\sqrt x/\log x$$. Since $$\pi(x)-\mathop{\rm li}(x)$$ can be as small as $$-\sqrt x\log\log\log x/\log x$$ by Littlewood's result, this function $$\phi(x)$$ is not eventually a lower bound for $$\pi(x)$$.