Does there exist a non-negative function $p(\lambda)\geq 0$ defined on $[0,1]$ such that $$\cosh(\sqrt{1+x^2})=\int^1_{0}p(\lambda) \cosh(\lambda x)d\lambda,~~~~ \forall x\in \mathbb{R} ~~~~~?$$
If we Taylor expand both sides and compare coefficients of $x^{2r}$, the above integral equation is equivalent to
$$\int_0^1p(\lambda) \lambda^{2r} d\lambda=(2r)!\sum_{n\geq r} \frac{\binom{n}{r}}{(2n)!}={}_0F_1(r+1/2,1/4), $$
where ${}_0F_1$ is the Generalized hypergeometric function. A necessary condition for the existence of a non-negative $p(\lambda)$ is obtained from Cauchy–Schwarz inequality, which requires that the LHS of the above equation is log-convex. Numerical results show that $\ln {}_0F_1(r+1/2,1/4)$ is indeed convex:
But I don't know how to go further from here. What is the general idea to tackle this kind of problems?