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This integral looks almost like a Fourier-integral, if it wasn't for the limits: $$\int^\infty_{\sqrt{1+1/p^2}}dx\left(\sqrt{\frac{x^2-1/p^2}{x^2-1}}-1\right)e^{-itpx},$$ where $t$ is some positive parameter. I want to determine how the integral decays when $p\rightarrow \infty$. (I don't actually think that for any $t>0$ this changes the behaviour, but just in case, I included it. Indeed, for $t=0$, the integral goes to 1 for $p\rightarrow \infty$.)

I can't seem to approach this problem (neither using pen and paper nor Mathematica) since the standard tricks for fourier-transform fail as far as I can see. (One can extend this to the negative side and see that due to the symmetry in $x\rightarrow -x$, this is just the complex conjugate, so has the same modulus. But there is still a gap in between...)

Any help would be appreciated!

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    $\begingroup$ I got (not rigorously) $$I(p)=\int^\infty_{\sqrt{1+1/p^2}}\left(\sqrt{\frac{x^2-1/p^2}{x^2-1}}-1\right)e^{-itpx}dx=e^{-it\sqrt{p^2+1}-\frac{\pi i}4}\left(\sqrt\frac\pi{2tp}-\frac1p\right)+O\big(p^{-\frac32}\big)$$ Random check with WA (free option) mostly confirms the answer, though the accuracy is not fully satisfactory. The explanation could be that, handling a strongly oscillating function, higher terms can bring substantial contribution to its real/imaginary part. $\endgroup$
    – Svyatoslav
    Commented Jun 20 at 20:54
  • $\begingroup$ @Svyatoslav How did you obtain this result - even without rigor? Did you do some integration by parts? If so, then could you elaborate on the "rest" ($\mathcal{O}(...)$) - which seems to have a similar problem when trying to determine the asymptotics. $\endgroup$
    – Confuse-ray30
    Commented Jun 21 at 9:12
  • $\begingroup$ I did not mention that the obtained asymptotics was valid for $tp\gg1$. In the case $p\gg1;\, tp\ll1$ we have to use another approach, and the answer will be different. Yes, I can post the heuristic solution, but could you check the answer numerically first? WA (free option) does not allow to do it properly: I can see that the approximation generally works, but cannot understand the approximation accuracy. $\endgroup$
    – Svyatoslav
    Commented Jun 21 at 10:17
  • $\begingroup$ @Svyatoslav Mathematica (paid) unforunately also cannot evaluate this integral numerically. Indeed, I already tried so and could not get the numerics to converge... $\endgroup$
    – Confuse-ray30
    Commented Jun 21 at 11:15
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    $\begingroup$ If my asymptotic is wrong, I'm afraid I cannot bring any valuable contribution to your problem. Anyway, I wish you good luck. $\endgroup$
    – Svyatoslav
    Commented Jun 24 at 11:51

1 Answer 1

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This is a temporary answer done using numerics. I calculated the modulus of the integral in question for varying $p$ at fixed $t$. Results for the leading order asymptotic (i.e. the smallest exponent in an $\frac{1}{p}$ expansion) below:

It seems the exponent is indeed $t$ independent.

For $t=0$, the modulus of the integral is convergent to a constant.

For $t=1$, we have numerically around $c=0.436$ (using a fit of the form $a/x^c$). <span class=$t=1$" />

For $t=2$ we have $c=0.432$ ...

And lastly, for $t=1/2$, we have $c=0.433$. So it seems not quite a square root... ...

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