This question is a more specific question related to https://mathoverflow.net/questions/69900/asymptotics-for-the-number-of-ways-to-sum-primes-such-that-the-sum-is-n

I am looking for a lower bound (that is tight as possible) for the integral $\int_{2}^n e^{\sqrt{x/\log{x}}} dx.$ Unfortunatelly, $e^{\sqrt{x/\log{x}}}$ is not integrable, so one has to bound it first and then evalute the integral of the bounded function. For example $$\int_{2}^n e^{\sqrt{x/\log{x}}} \geq \int_{2}^n e^\sqrt[3]{x} dx = O(n^{\frac{2}{3}}e^{\sqrt[3]{n}}) $$

The bound obtained above is inferior to $e^{\sqrt{n/\log{n}}}$, so I should either find a better bound for the integrated function, or perhaps use another (unknown to me) trick to bound the stated integral.

Any suggestions are appreciated!

  • 3
    $\begingroup$ The statement that $e^{\sqrt{x/\log{x}}}$ is not integrable at least misleading (if not wrong)... $\endgroup$ – Fabian Jul 20 '11 at 20:52

I can give you better than upper and lower bounds. We can find a precise asymptotic.

Proposition: I claim that $$ \int_2^n \exp \left(\sqrt{\frac{x}{\log x}}\right)\sim 2\sqrt{n\log n}\exp\left(\sqrt{\frac{n}{\log n}}\right)$$ as $n\rightarrow \infty$.

Proof: Let $$f(x)=2\sqrt{x\log x}\exp\left(\sqrt{\frac{x}{\log x}}\right)$$ be the claimed asymptotic function. Then we need to prove that the limit $$\lim_{n\rightarrow\infty} \frac{\int_2^n \exp \left(\sqrt{\frac{x}{\log x}}\right)}{f(n)}$$ is equal to $1$.

To do this, we apply l'Hopitals rule. Computing we find $$f^{'}(x)=\frac{1}{\sqrt{x\log x}}\left(\log x+1\right)\exp\left(\sqrt{\frac{x}{\log x}}\right)$$ $$+2\sqrt{x\log x}\exp\left(\sqrt{\frac{x}{\log x}}\right)\left(\frac{1}{2}\sqrt{\frac{\log x}{x}}\left(\frac{1}{\log x}-\frac{1}{\left(\log x\right)^{2}}\right)\right).$$ Notice that we can rewrite this as $$f^{'}(n)=\left(1+o(1)\right)\exp\left(\sqrt{\frac{n}{\log n}}\right).$$

The fundamental theorem of calculus tells us that $$\frac{d}{dn}\int_{2}^{n}\exp\left(\sqrt{\frac{x}{\log x}}\right)dx=\exp\left(\sqrt{\frac{n}{\log n}}\right).$$ Hence l'Hopitals rule implies the original limit is $1$, and the result is proven.

  • $\begingroup$ That's very nice! $\endgroup$ – Jernej Jul 20 '11 at 21:12

If one writes $I_n$ for the integral from $2$ to $n$, obvious nonasymptotic bounds are $$ \exp\left(\sqrt{(n-1)/\log(n)}\right)\le I_n\le (n-2)\exp\left(\sqrt{n/\log(n)}\right), $$ for every $n\ge3$. In particular, $\displaystyle\log\left(I_n\right)=(1+o(1))\sqrt{n/\log(n)}$.


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