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Let $\alpha>1$. I would like to find a closed form or an upper bound of

$$f\left(\alpha\right)=\int_{e}^{\infty}\frac{t^{1/2}}{\log^{1/2}\left(t\right)}\alpha^{-t/\log\left(t\right)}dt.$$

For the closed form I'm very skeptical but I have trouble also for an upper bound. I tried, manipulating a bit, to integrate w.r.t. $\alpha$ since $$\frac{\partial}{\partial\alpha}\alpha^{-t/\log\left(t\right)}=-\frac{t}{\alpha\log\left(t\right)}\alpha^{-t/\log\left(t\right)}$$ but it seems quite useless and at this moment I didn't see a good way to proceed. Maybe it is interesting to see, using some trivial substitutions, that $$f\left(\alpha\right)=\int_{e}^{\infty}\frac{\left(e^{3/2}\right)^{-W_{-1}\left(-1/v\right)}}{v\left(-W_{-1}\left(-\frac{1}{v}\right)\right){}^{1/2}}\frac{W_{-1}\left(-\frac{1}{v}\right)}{W_{-1}\left(-\frac{1}{v}\right)+1}\alpha^{-v}dv$$ $$=\int_{e}^{\infty}g\left(w\right)\alpha^{-v}dv$$ where $W_{-1}\left(x\right)$ is the Lambert $W$ function. So it seems that $f(\alpha)$ is somehow connected to the Mellin transform of $g(w).$

Thank you.

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  • $\begingroup$ poor downvote removed...(+1) $\endgroup$ – tired Feb 6 '17 at 12:22
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    $\begingroup$ concerning Lambert W, you can also write $$ \int_e^{\infty}\frac{-W_{-1}(-1/q)^2}{1+W_{-1}(-1/q)}\sqrt{q}e^{-\log(\alpha )q}dq $$ $\endgroup$ – tired Feb 6 '17 at 14:30
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    $\begingroup$ Togehter with theorem (1) from here arxiv.org/pdf/1601.04895.pdf i think we might find a upper bound $\endgroup$ – tired Feb 6 '17 at 14:47
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    $\begingroup$ @tired Yes I solved it using that bounds, thank you again! I hope there are good enough for my problem. I'm working on it. $\endgroup$ – Marco Cantarini Feb 8 '17 at 11:30
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    $\begingroup$ i obtained a super stupid bound by observing that $\sqrt{x}\geq\frac{\sqrt{x}}{\sqrt{\log(x)}}$ and $\frac{x}{\log(x)}>\sqrt{x}$ for $x \in (e,\infty)$. this gives $$ f(\alpha)<\int_{e}^{\infty}\sqrt{x}e^{-\log(\alpha)\sqrt{x}} $$ $\endgroup$ – tired Feb 8 '17 at 13:16
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A naif but probably efficient approach is to exploit the fact that the logarithm function is approximately constant on short intervals and $$ \frac{1}{\sqrt{N}}\int_{e^N}^{e^{N+1}}\sqrt{t}\,\alpha^{-t/N}\,dt =\frac{N\sqrt{\pi}}{2\log(\alpha)^{3/2}}\,\text{Erf}\left(\sqrt{\frac{e^N\log\alpha}{N}}\right)$$ can be efficiently approximated through the continued fraction for the error function.
We may also consider this fact: through the Laplace transform $$ \int_{0}^{+\infty}\sqrt{t}\exp\left(-\frac{t\log\alpha}{N}\right)\,dt = \int_{0}^{+\infty}\mathcal{L}^{-1}\left(\frac{1}{\sqrt{t}}\right)\,\mathcal{L}\left(t \exp\left(-\frac{t\log\alpha}{N}\right)\right)\,ds $$ we get the following integral: $$ \int_{0}^{+\infty}\frac{N^2}{\sqrt{\pi s}(Ns+\log\alpha)^2}\,ds =\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{1}{(s^2+\frac{\log\alpha}{N})^2}\,ds$$ that is simple to estimate in terms of $N$ and $\alpha$. The original integral is a weigthed sum of these integrals, that according to my computations should behave like $$\exp\left(-\log(\alpha)^{3/2}\right).$$ But I am probably over-complicating things, and we may recover the same bound by just applying a modified version of Laplace's method to the original integral.

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  • $\begingroup$ what $\alpha$ are you considering? for large $\alpha$ we would expect something $\sim \frac{Const}{\alpha^e \sqrt{\log(\alpha)}}$ for the original integral $\endgroup$ – tired Feb 6 '17 at 19:11
  • $\begingroup$ @tired: I plotted $\frac{\log\log(1/f(\alpha))}{\log\log\alpha}$ and its seems to converge to $\frac{3}{2}$ for $\alpha\gg e$. $\endgroup$ – Jack D'Aurizio Feb 6 '17 at 19:14
  • $\begingroup$ that is strange...i check tomorrow..but i don't see what exactly should dismiss a laplace style approach in this parameter domain $\endgroup$ – tired Feb 6 '17 at 19:34
  • $\begingroup$ @MarcoCantarini: the factor $\sqrt{\frac{t}{\log t}}$ makes me think about quadratic forms. Are you going to break an unconditional barrier I was not able to break? If so, please let me know. I have to book strippers for a decent party. $\endgroup$ – Jack D'Aurizio Feb 6 '17 at 20:52
  • $\begingroup$ Well, I'm working on an ambitious project. If all goes well, I will propose to you a party with math problems, beer and strippers. If I fail, I will propose to you a party with hard math problems, a lot of beer and a lot of strippers. $\endgroup$ – Marco Cantarini Feb 6 '17 at 21:00

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