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 $\pi(x)$.

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