Evaluation of the following definite integral $$
\int_{-\infty}^\infty \frac{\sin\left(\sqrt{2-\sqrt{2}+\frac{k^2}{\sqrt{2}}}\,\,\right)}{\sqrt{2-\sqrt{2}+\frac{k^2}{\sqrt{2}}}} \cos(k) \,dk
$$
I arrived at this form of the integral while evaluating the causal propagator corresponding to a second order linear partial differential equation in position and time that resembles the Klein-Gordon equation of relativistic quantum mechanics in form. I'm having difficulty in solving this integral. I tried to do this using integration by parts and I may have to use the concept of branch points (which I'm not quite familiar with). A guidance on how to go about evaluating this further would be appreciated.
 A: I would like to submit a partial answer which provides an explicit result. As noted in the comments, this integrand is even. After some rearrangements, we obtain 
$$\int_{-\infty}^\infty\frac{\sin\left(\sqrt{2-\sqrt{2}+\frac{k^2}{\sqrt{2}}}\,\,\right)}{\sqrt{2-\sqrt{2}+\frac{k^2}{\sqrt{2}}}}\cos(k)\,dk=2\sqrt[4]{2}\int_0^\infty\frac{\sin\left(\frac{1}{\sqrt[4]{2}}\sqrt{2\sqrt{2}-2+k^2}\,\right)}{\sqrt{2\sqrt{2}-2+k^2}}\cos(k)\,dk.$$ 
This latter integral is of a form found in Gradshteyn and Ryzhik, specifically formula $3.876.1$ (Note: I only have access to the Fifth Edition, so I'm unsure of its location in newer editions).

$$3.876.1\;\;\int_0^\infty\frac{\sin{\left(p\sqrt{x^2+a^2}\right)}}{\sqrt{x^2+a^2}}\cos(bx)\,dx=\cases{\frac{\pi}2 J_0\left(a\sqrt{p^2-b^2}\right), & \text{if $0<b<p$}\\[2ex] 0, & \text{if $b>p>0$}}$$

In the case of our particular integral, $a=\sqrt{2\sqrt{2}-2}$, $b=1$, and $p=\frac1{\sqrt[4]{2}}$. Since $1>\frac1{\sqrt[4]{2}}>0$, our integral falls under the second category; thus, the result that we seek to prove is the following:

$$\int_0^\infty\frac{\sin\left(\frac{1}{\sqrt[4]{2}}\sqrt{2\sqrt{2}-2+k^2}\,\right)}{\sqrt{2\sqrt{2}-2+k^2}}\cos(k)\,dk\stackrel{?}{=}0.$$

