The integral is like this:

$$ \int_0^\infty \mathrm{d} x \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} $$

The short time asymptotics is like this (some constant maybe missing):

$$ \sim \frac{1}{\ln t} $$

I don't know how to get this...

Edit: what's the meaning of short time?

The source I am reading says:

At short times the integral rises or falls sharply as $\sim \frac{1}{\ln t}$

My guess that the meaning of short time is the time smaller than the first maximum or minimum of the integral.

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    $\begingroup$ By "short time" do you mean $t \to 0^+$? $\endgroup$ – Antonio Vargas Jan 30 '16 at 16:47
  • $\begingroup$ @AntonioVargas actually t is of order of 1.However, according to the context, t=0+ may be also applys..... $\endgroup$ – an offer can't refuse Jan 30 '16 at 19:12
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    $\begingroup$ As $t \to 0^+$ the integral tends to $\pi/2$. If $t = O(1)$, in what sense do you mean "asymptotics" for the integral? In most cases an asymptotic involves a limit. If you're not interested in taking a limit, then it would be best if you could clarify the question. $\endgroup$ – Antonio Vargas Jan 30 '16 at 19:18
  • $\begingroup$ By the way, here is a plot generated by Mathematica of your integral for $1 < t < 10$: i.stack.imgur.com/cqT7Y.png $\endgroup$ – Antonio Vargas Jan 30 '16 at 19:21
  • $\begingroup$ Funnily enough, if I plot the integral in Mathematica as Plot[NIntegrate[ Cos[2 t Cosh[Pi x/2]]/(1 + x^2), {x, 0, Infinity}], {t, 0, 0.1}] I obtain that for $t\to 0^+$ the integral approaches $\pi/4$, not $\pi/2$... $\endgroup$ – Pierpaolo Vivo Jan 31 '16 at 0:18

For $x$ real we have

$$ \left| \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} \right| \leq \frac{1}{1+x^2}, $$

we know that

$$ \frac{1}{1+x^2} \in L^1([0,\infty)), $$

and for fixed $x \geq 0$ we calculate

$$ \lim_{t \to 0^+} \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} = \frac{1}{1+x^2}. $$

Therefore we may conclude that

$$ \lim_{t \to 0^+} \int_0^\infty dx \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} = \int_0^\infty \frac{dx}{1+x^2} = \frac{\pi}{2}. $$

by the Dominated Convergence Theorem.

  • $\begingroup$ But at least you need give a dependence on $t$... You mean there is no dependence on $t$? $\endgroup$ – an offer can't refuse Jan 31 '16 at 2:47
  • $\begingroup$ @buzhidao, Yes and no. Once you've taken a limit there is no more $t$ dependence. You seem to be interested in the higher-order terms in the asymptotic as well though, i.e. an approximation for $$\int_0^\infty dx \frac{\cos[2t\cosh(\frac{\pi x}{2})]}{1+x^2} - \frac{\pi}{2}$$ when $t$ is small. This would indeed have some $t$ dependence. You may want to edit your question to reflect this. $\endgroup$ – Antonio Vargas Jan 31 '16 at 2:52
  • $\begingroup$ Yes, I am really interested in the $t$ dependence... $\endgroup$ – an offer can't refuse Jan 31 '16 at 3:01

here is an idea: i will look at the asymptotics of the simpler $$I= \int_0^\infty \frac{\cos(te^x)}{1+x^2}\, dx \text{ for small } t. $$

observe that the graph of $y = \frac{\cos(te^x)}{1+x^2}$ oscillates between the envelopes $y = \pm \frac 1{1+x^2}$ starts at $(0,1)$ and the first $x$ -intercept is $\ln(\pi/2) - \ln(t).$ there after the intercepts are spaced by $\ln 2, \ln 3. \cdots.$

therefore $$I= \int_0^\infty \frac{\cos(te^x)}{1+x^2}\, dx=\int_0^{\ln(\pi/(2t)}\frac{\cos(te^x)}{1+x^2}\, dx +\cdots = \frac12 \times 1 \times \ln\left(\frac{\pi}{2t}\right) + \cdots$$ the estimate being the area of the triangle with base $ \ln\left(\frac{\pi}{2t}\right)$ and height $1.$


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