I want to show that the following equality does not hold:

\begin{equation}\label{at3} \frac{\lambda^2-1}{2}x^2-\int_{-\infty}^{\infty}\!\!\int_{-\infty}^{\infty}K(y_1,y_2,x)\ln g(y_1,y_2)dy_2dy_1=\mbox{constant}\ \ ,\ \ \forall x\in\mathbb{R}, \end{equation} where $\lambda>1$, and \begin{equation}\label{at2} g(y_1,y_2)=\int_{-1}^{1}e^{-\frac{(\lambda^2-1)u^2}{2}+\lambda y_1u}\cosh\left(y_2\sqrt{1-u^2}\right)dF(u), \end{equation} in which $F(u)$ is an unknown cumulative distribution function (CDF) for $U$ with the support $[-1,1]$. It could be discrete or continuous or a mixture. Finally, \begin{align} K(y_1,y_2,x)=e^{-\frac{(\lambda^2-1)}{2}x^2}e^{\frac{-y_1^2-y_2^2+2\lambda y_1x}{2}}\cos \left(y_2\sqrt{x^2-1}\right). \end{align} In order to show that the first equality is not valid, I tried to see the scaling behaviour of the double integral term in $x$. I think that the double integral term is at most $O(x)$, so that it cannot cancel the effect of the first term $\frac{\lambda^2-1}{2}x^2$. However, I was not successful.

I highly appreciate your help about this problem.

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