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So the integral is like this: $$\int_1^\infty \frac{\cos xt}{(x^2-1)\left[\left(\ln\left|\frac{1-x}{1+x}\right|\right)^2+\pi^2\right]}\mathrm{d}x$$

The question is how to get the asymptotic form of this integral when $t$ is very large.

The integrand is a decaying oscilating function:

enter image description here

The denominator of the integrand is zero when $x\to 1$.

Honestly, I don't know whether this integral will converge, for it reminds me of the integral:

$$\int_0^\infty \frac{1}{x^2}\mathrm{d}x$$ which is $\infty$.

I hope it is a meaningful integral and the asymptotic can be obtained.

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This is more of a comment, but by mimicking the procedure in this answer near the endpoint $x=1$ we can show that

$$ \int_1^\infty \frac{\cos xt}{(x^2-1)\left[\left(\ln\frac{x-1}{x+1}\right)^2+\pi^2\right]}\,dx \approx \operatorname{Re} \left\{ \frac{e^{it}}{2} \int_0^\infty \frac{e^{-yt}}{y \left[\left(\log \frac{iy}{2}\right)^2 + \pi^2\right]}\,dy\right\} $$

to first order as $t \to \infty$. Numerically these agree very well; below is a plot of the left integral in red and the approximation in blue.

enter image description here

This image was created with the following code in Mathematica. The program complains about the singularities in the integrands but ultimately it probably handles them alright.

pointSpacing = 1/5; (* decrease this for a higher resolution plot *)
Show[
 ListPlot[
  Table[{t, 
    Re[E^(I t)/2 NIntegrate[
        E^(-y t)/(y (Log[I y/2]^2 + Pi^2)), {y, 0, \[Infinity]}] // 
      Quiet]},
   {t, 30, 50, pointSpacing}],
  Joined -> True, PlotRange -> All],
 ListPlot[
  Table[{t, 
    NIntegrate[
      Cos[x t]/((x^2 - 1) (Log[(x - 1)/(x + 1)]^2 + Pi^2)), {x, 
       1, \[Infinity]}, MaxRecursion -> 20] // Quiet},
   {t, 30, 50, pointSpacing}],
  PlotStyle -> Red, Joined -> True]
 ]

Because of the slow $y \to 0$ logarithmic singularity in the denominator we lose a lot of numerical accuracy if we replace the denominator by a simpler approximation. To first order the denominator is

$$ \sim \frac{1}{y(\log y)^2} $$

near $y=0$, so if we replace the denominator with this in the integrand (and replace the integration interval $(0,\infty)$ with $(0,c)$ for some $0 < c < 1$) then we should still obtain the correct leading order behavior (but, again, it will be a worse approximation numerically). Then, since the integral is real, we can use $\operatorname{Re} e^{it} = \cos t$ to obtain the approximation

$$ \int_1^\infty \frac{\cos xt}{(x^2-1)\left[\left(\ln\frac{x-1}{x+1}\right)^2+\pi^2\right]}\,dx \approx \frac{\cos t}{2} \int_0^c \frac{e^{-yt}}{y (\log y)^2}\,dy $$

as $t \to \infty$. Now we can use a result of Erdélyi (see this answer) to conclude that the leading order behavior should be

$$ \int_1^\infty \frac{\cos xt}{(x^2-1)\left[\left(\ln\frac{x-1}{x+1}\right)^2+\pi^2\right]}\,dx \approx \frac{\cos t}{2\log t} $$

as $t \to \infty$. It is important to note here that by ignoring complex contributions from the approximating integral we no longer accurately approximate the phase of the integral. If higher-order corrections to this approximation were computed, terms of the form $o(1/\log t)\sin t$ will appear compensate for this.

Below is a plot of this $\cos t/2\log t$ approximation in blue versus the original integral in red. As mentioned earlier, this is a worse approximation than before, but at least it captures the correct logarithmic decrease of the amplitude of the oscillation.

enter image description here

This image was created with the following code.

pointSpacing = 1/2; (*decrease this for a higher res plot*)
Show[
 Plot[Cos[t]/(2 Log[t]),
  {t, 570, 600},
  PlotRange -> All],
 ListPlot[
  Table[{t, 
    NIntegrate[
      Cos[x t]/((x^2 - 1) (Log[(x - 1)/(x + 1)]^2 + Pi^2)), {x, 
       1, \[Infinity]}, MaxRecursion -> 20] // Quiet},
   {t, 570, 600, pointSpacing}],
  PlotStyle -> Red, Joined -> True]
 ]
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  • $\begingroup$ Thank you very much for your "comment", I will look into it carefully soon. At this moment I would like to know how you obtain the numerical results? Using mathematica? Can you put the code snipit for future reader's convenience? What is the $t$ value you choose, I want to reproduce it myself. Thank you! $\endgroup$ – 喵喵是我的猫猫 Jun 8 '15 at 4:46
  • $\begingroup$ Haha, I forgot I started the post with that. I ended up writing more than I expected :) Yes to Mathematica. I accidentally closed the window without saving the code but I'll write it up again this afternoon and include it in the post. $\endgroup$ – Antonio Vargas Jun 8 '15 at 11:18
  • $\begingroup$ @buzhidao Included the code in the answer. $\endgroup$ – Antonio Vargas Jun 8 '15 at 23:19
  • $\begingroup$ Thank you very much, I'll reproduce it and follow your answer carefully! $\endgroup$ – 喵喵是我的猫猫 Jun 9 '15 at 4:51
  • $\begingroup$ I've reproduce all your calculations. Here is some questions:(1).If I expand the integrand by $1/\ln y$, the second leading term will not give a term of $\sin t /\ln t$, instead it give a $\sin t /(\ln t)^2$ term. (2). I think the $\pi ^2$ is essential, how to keep track of this? For if you delete this term in the first code snippt, you can see huge difference(3). How can I get the phase information, can you give me more hint? $\endgroup$ – 喵喵是我的猫猫 Jun 9 '15 at 14:32
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This is not a full answer to the problem, but only concerns the question of whether the integral converges.

With change of variable $x=1+y$ : $$\int_1^\infty \frac{\cos xt}{(x^2-1)\left[\left(\ln\left|\frac{1-x}{1+x}\right|\right)^2+\pi^2\right]}\mathrm{d}x = \int_0^\infty \frac{\cos (1+y)t}{y(1+2y)\left[\left(\ln\left|\frac{y}{2+y}\right|\right)^2+\pi^2\right]}\mathrm{d}y $$ In $y>0$ close to $0$ : $$ \frac{\cos (1+y)t}{y(1+2y)\left[\left(\ln\left|\frac{y}{2+y}\right|\right)^2+\pi^2\right]} \sim \frac{\cos(t)}{y\left(\ln(y) \right)^2}$$ $\int \frac{1}{y\left(\ln(y) \right)^2}dy = -\frac{1}{\ln(y)}$ which tends to $0$ when $y$ tends to $0$. So, the integral is convergent for the lower bound.

$\int_1^\infty \frac{\cos xt}{(x^2-1)\left[\left(\ln\left|\frac{1-x}{1+x}\right|\right)^2+\pi^2\right]}\mathrm{d}x$ obviously is convergent in $x$ tending to infinity.

So, the answer is : The integral is convergent.

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  • $\begingroup$ Very nice argument, I am glad that it coverages. $\endgroup$ – 喵喵是我的猫猫 Jun 7 '15 at 18:09

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