13
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ poor downvote removed...(+1) $\endgroup$ – tired Feb 6 '17 at 12:22
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.