In a physics paper (here) I found this variant of the Bessel function of the first kind.

$$ \tag{1} Z(g) ~=~ \frac{1}{\sqrt{g}} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^{-\frac{1}{2g} \sin^2 x} \, dx = \frac{\pi}{\sqrt{g}}e^{-\frac{1}{4g}}I_0(\tfrac{1}{4g}). $$

Later in the paper they provide an "expansion around 0" that I have trouble to understand:

\begin{eqnarray*} Z(g)|_{y=0} &=& \int_{-\frac{\pi}{2\sqrt{g}}}^{\frac{\pi}{2\sqrt{g}}} e^{-\tfrac{1}{2g} \sin^2 (y g^{1/2})} \, dy \\ &=& \int_{-\frac{\pi}{2\sqrt{g}}}^{\frac{\pi}{2\sqrt{g}}} \bigg[ e^{-\tfrac{y^2}{2}} + \frac{1}{6}y^2 e^{-\tfrac{y^2}{2}} + \dots \bigg] dy \\ &=&\tag{5} \sqrt{2\pi} \sum_{k=0}^\infty \frac{\Gamma(k+\tfrac{1}{2})^2 }{\Gamma(k+1) \Gamma(\tfrac{1}{2})^2} (2g)^k. \end{eqnarray*}

The moments of the Gaussian are known by Wick formula. I am not sure how he is getting all the coefficients in this expansion.

One possible starting point is that $\sin x \approx x$ for $x\ll 1$ so that

$$ \frac{1}{2g} \sin^2 (y g^{1/2}) \approx \frac{y^2}{2} \left(\frac{\sin (y g^{1/2})}{y g^{1/2}} \right)^2 \approx \frac{y^2}{2} \left(1 - \frac{g y^2 }{3!} + \dots \right)^2 $$

Then we have take the exponent of this and integrate, so I am not sure how they computed all the terms.


Comment to the question (v2):

Yes, the authors of Ref. 1 are cheating. They are not using Wick's theorem (although one in principle could do so). They know that the modified Bessel function $I_0$ of first kind has an asymptotic series expansion in terms of a generalized hypergeometric function$^1$

$$e^{\frac{1}{4g}} \frac{Z(g)}{\sqrt{2\pi}}~=~\frac{1}{\sqrt{2\pi g}}\int_0^{\pi}d\theta~ e^{\frac{1}{4g}\cos\theta}~=~\sqrt{\frac{\pi}{2g}} I_0\left( \frac{1}{4g}\right)$$

$$\tag{A} ~\sim~ e^{\frac{1}{4g}} {}_2 F_{0}(\frac{1}{2},\frac{1}{2};2g)+e^{-\frac{1}{4g}} {}_2 F_{0}(\frac{1}{2},\frac{1}{2};-2g) \quad\text{for} \quad g\to 0.$$


  1. A. Cherman, P. Koroteev and M. Unsal, arXiv:1410.0388.


$^1$ In the asymptotic series expansion (A) we have been cavalier about branch-cuts and Stokes sectors. But that's sort of the main theme of Ref. 1, so we will not elaborate further in this answer.


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