Question:
What is
$$
\int_0^{\infty} e^{\kappa (1+u^2)^{\alpha/2} \cos(\alpha\arctan u)} du
$$
where $\kappa >0$ and $\alpha$ is a real between 1 and 2?
Alternatively, what is a good approximation of this integral in terms of $\alpha$ and $\kappa$?
Some ideas
(skipping the details of the derivations to keep it short.)
One can express the integrand as (writing $arctan$ via the $log$ function, etc): $$ e^{\kappa (1+u^2)^{\alpha/2} \cos(\alpha\arctan u)} = e^{\frac{\kappa}{2} ((1+iu)^{\alpha} + (1-iu)^{\alpha})} $$ Now for $u \in (0,1)$, Taylor expansions of both $(1+iu)^{\alpha}$ and $(1-iu)^{\alpha}$ yield: \begin{equation}\label{leftInterval} e^{\kappa (1+u^2)^{\alpha/2} \cos(\alpha\arctan u)} = e^{\kappa \sum_{n=1}^{+\infty} (-1)^n \tbinom{\alpha}{2n} u^{2n}} \end{equation} where $\tbinom{\alpha}{2n}$ is the binomial coefficient: $$ \tbinom{\alpha}{2n} = \frac{\alpha (\alpha-1)\dots(\alpha-2n+1)}{(2n)!} $$ For $u \in (1, +\infty)$, with $v=\frac{1}{u}$ and using Taylor expansions on $v < 1$, one can derive: $$ e^{\kappa (1+u^2)^{\alpha/2} \cos(\alpha\arctan u)} = e^{\kappa u^{\alpha} \cos\!\big(\frac{\pi \alpha}{2}\big) \sum_{n=0}^{+\infty} \tbinom{\alpha}{2n} \frac{(-1)^n}{u^{2n}}} \times e^{\kappa u^{\alpha} \sin\!\big(\frac{\pi \alpha}{2}\big) \sum_{n=0}^{+\infty} \tbinom{\alpha}{2n+1} \frac{(-1)^n}{u^{2n+1}}} $$
Then, one is left with integrating/approximating these expressions on $(0,1)$ and $(1,+\infty)$ but how to do that?