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I would like to obtain the asymptotic expression for $\alpha \to \infty$ of the following integral $$I(\alpha)=\int_0^\infty\!dx\,x (1 - \cos[2\alpha K_0(x)]) = \int_0^\infty\!dx\, 2x \sin^2[\alpha K_0(x)]$$ where $K_0$ is the modified Besselfunktion of the second kind. I know that $$K_0(x) \sim \begin{cases} \log (2/x) -\gamma & x\ll1, \\ \sqrt{\frac{\pi}{2x}}\,e^{-x} & x \gg1. \end{cases}$$

The integral is finite for any $\alpha<\infty$: for small $x$ there is no problem because of the factor $x$. For large $x$ the Bessel-function approaches 0 exponentially fast and thus renders the integral finite.

The problem I have is that the integral is both oscillatory and dominated by intermediate $x$. From numerics, I would believe that $$ I(\alpha) \sim c_0 \log(\alpha) + c_1.$$

Do you have any ideas how to tackle integrals of this kind?

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Eep, it's nested... looks terribly difficult to do. –  J. M. Apr 25 '12 at 12:40
    
@J.M.: ok, so it is not just me struggling with this beast. Question: can one at least show that $ \lim_{\alpha\to\infty} I(\alpha)/\log(\alpha) = \text{const}$? –  Fabian Apr 25 '12 at 13:05
    
Nasty, but oddly fascinating. –  nbubis Apr 25 '12 at 14:12
    
Have you tried some stationary phase techniques? –  Alex R. Apr 25 '12 at 23:21
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2 Answers

up vote 4 down vote accepted

I think I found a way to apply the method of stationary phase (please give feedback if you think that the answer is not sound).

As Jon noted via partial integration the integral can be brought into the form $$I(\alpha) = \alpha \int_0^\infty\!dx\,x^2 K_1(x) \sin[2 \alpha K_0(x)].$$ Next I perform the substitution $y=2K_0(x)$ (note that $K_0$ is a monotonously falling function such that the inverse $K_0^{-1}$ is uniquely defined). I get $$ I(\alpha) = \frac{\alpha}2 \int_0^\infty\!dy\,\sin(\alpha y)[K_0^{-1}(y/2)]^2; $$ note that $K_1$ cancelled with the derivative of $K_0^{-1}$ via inverse function theorem and the fact that $K_0'(x)=-K_1(x)$.

This integral is almost ready for a stationary phase analysis, we replace $\sin(\alpha y) = \operatorname{Im} e^{i\alpha y}$ and note that the path of stationary phase starting from $y=0$ is along the imaginary axis. Thus, we substitute $y=i \zeta$ and get $$I(\alpha)=\frac{\alpha}2 \int_0^\infty \!d\zeta\, e^{-\alpha \zeta} \operatorname{Re}[K_0^{-1}(i\zeta/2)]^2 ;$$ here, we have analytically continued $K_0^{-1}$ into the complex plane.

The integral is dominated at $\zeta\approx 0$. Thus we can Taylor-expand $K_0^{-1}(i\zeta)$. We have $K_0(x) \sim e^{-x}$ thus $K_0^{-1}(y) \sim -\log y$. Using this asymptotic relation, we obtain $$I(\alpha)\sim \frac{\alpha}2 \int_0^\infty \!d\zeta\, e^{-\alpha \zeta} \overbrace{\operatorname{Re}[\log(i\zeta/2)]^2}^{\sim \log^2 (\zeta/2)} \sim \frac{\alpha}2 \int_0^\infty \!d\zeta\, e^{-\alpha \zeta}\log^2 (\zeta/2) \sim \tfrac12 \log^2\alpha. $$

I guess for the next term one has to work a bit harder and use $K_0^{-1}(y) \sim - \log y - \tfrac12 \log(-\log y)$. However, in principle it should be possible to extend this approach to next to leading terms.

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Hi Fabian, I played around with this a bit, trying to improve the bound using a better approximation for $K_0^{-1}$. I ran into the Lambert $W$ and things got ugly. I did verify your result. (+1) –  user26872 Jul 5 '12 at 1:53
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Firstly, you need to integrate by parts obtaining $$ I(\alpha)= \int_0^\infty\!dx\, 2x \sin^2[\alpha K_0(x)]= 2\alpha\int_0^\infty\!dx\,x^2K_1(x) \cos[\alpha K_0(x)]\sin[\alpha K_0(x)]. $$ Now, change variable to $y=\alpha x$ so that $$ I(\alpha)=\frac{1}{\alpha^2}\int_0^\infty\!dx\, y^2K_1\left(\frac{y}{\alpha}\right) \sin\left[2\alpha K_0\left(\frac{y}{\alpha}\right)\right]. $$ In the limit $y\rightarrow\infty$ one can take the asymptotic approximations $K_0(x)\sim\sqrt{\frac{\pi}{2x}}\,e^{-x} $ and $K_1(x)\sim\sqrt{\frac{\pi}{2x}}\,e^{-x} $ and you get $$ I(\alpha)\sim\frac{1}{\alpha^\frac{3}{2}}\int_0^\infty\!dy\, y^\frac{3}{2}\sqrt{\frac{\pi}{2}}e^{-\frac{y}{\alpha}}\sin\left[\sqrt{2}\alpha^\frac{3}{2}\sqrt{\frac{\pi}{y}}e^{-\frac{y}{\alpha}}\right]. $$ This integral can be rewritten using Lambert W function with the change of variable $z=\frac{e^{-\frac{y}{\alpha}}}{\sqrt{y}}$ so that $y=\frac{\alpha}{2}W\left(\frac{2}{\alpha z^2}\right)$ that for $\alpha\rightarrow\infty$ gives $y\sim\frac{1}{z^2}-\frac{2}{\alpha z^4}$. This will turn the integral into $$ I(\alpha)\sim\frac{\alpha^\frac{3}{2}}{4}\int_0^\infty\!dz\,\frac{\sin\left(\sqrt{2\pi}\alpha^\frac{3}{2}z\right)}{\frac{1}{W^2\left(\frac{2}{\alpha z^2}\right)}+\frac{1}{W^3\left(\frac{2}{\alpha z^2}\right)}}. $$ Then, we put $w=\sqrt{2\pi}\alpha^\frac{3}{2}z$ and we arrive at $$ I(\alpha)\sim\frac{1}{4\sqrt{2\pi}}\int_0^\infty\!dw\,\frac{\sin w}{\frac{1}{W^2\left(\frac{4\pi\alpha^2}{w^2}\right)}+\frac{1}{W^3\left(\frac{4\pi\alpha^2}{w^2}\right)}} $$ I checked the original integral proposed by OP and my approximation and I have got the following picture

enter image description here

so the approximation is really satisfactory and give a clear numerical proof of the finiteness of last approximate integral.

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I posted this because it appeared to me really nice. I am still working on it. Note that the last integral I get is finite. –  Jon Apr 26 '12 at 16:36
    
Hi @Jon, What about the wild behavior for small $y$? The $K$s are divergent there. These integrals don't appear to agree numerically. Also, have a look at your asymptotic behavior for $K_1$. –  user26872 Apr 26 '12 at 17:56
    
@oenamen: Thanks a lot for your comments. I have just checked the last integral numerically and Mathematica provides a finite value for increasing $\alpha$. Yes, I would check it further or finally I will remove it all. –  Jon Apr 26 '12 at 19:10
    
Glad to help. I mean the original integral and your integral do not seem to agree for large $\alpha$. I hope you can patch this up, I am very interested in the solution to this problem. –  user26872 Apr 26 '12 at 19:24
    
You are welcome. Also for me is really interesting. –  Jon Apr 26 '12 at 19:35
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