# Help needed for decay property of the following integral

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!

• 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. Commented Jun 20 at 20:54
• @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. Commented Jun 21 at 9:12
• 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. Commented Jun 21 at 10:17
• @Svyatoslav Mathematica (paid) unforunately also cannot evaluate this integral numerically. Indeed, I already tried so and could not get the numerics to converge... Commented Jun 21 at 11:15
• If my asymptotic is wrong, I'm afraid I cannot bring any valuable contribution to your problem. Anyway, I wish you good luck. Commented Jun 24 at 11:51

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 $$a/x^c$$). $t=1$" />
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...